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Amazon Top 100 Coding Problems

Amazon L6/L7 Engineering Manager Interview Preparation

This curated collection features the most frequently asked coding problems in Amazon engineering manager interviews, specifically designed for L6/L7 positions. The problems are organized by pattern type and difficulty, focusing on practical algorithms that relate to real-world systems and leadership scenarios.

Introduction

This document contains 100 carefully selected coding problems that represent the most common interview questions for Amazon L6 and L7 Engineering Manager positions. Each problem includes:

  • Problem statement with clear constraints
  • Example input/output for verification
  • Optimal solution approach with detailed explanation
  • Time and space complexity analysis
  • Production-ready Python implementation
  • Amazon business context connecting to real systems
  • Management insights for technical leadership
  • Similar problem patterns for deeper understanding

Difficulty Distribution

  • Easy: 20 problems (20%) - Foundation concepts
  • Medium: 70 problems (70%) - Core competency
  • Hard: 10 problems (10%) - Advanced problem solving

Problem Categories Overview

Category Problems Focus Area
Arrays & Strings 20 Data manipulation, optimization
Two Pointers 10 Efficient searching, space optimization
Sliding Window 10 Stream processing, real-time analysis
Hash Tables 10 Fast lookups, caching strategies
Trees & Graphs 15 Hierarchical data, network analysis
Dynamic Programming 10 Optimization, resource planning
Stack & Queue 10 Processing pipelines, state management
Binary Search 5 Efficient searching, system scaling
Design Problems 5 System architecture, scalability
Mathematical 5 Algorithmic thinking, edge cases

Category 1: Arrays & Strings (20 Problems)

Easy Difficulty (6 problems)

1. Two Sum (LeetCode #1)

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def two_sum(nums, target):
    """Find two numbers that add up to target."""
    num_map = {}
    for i, num in enumerate(nums):
        complement = target - num
        if complement in num_map:
            return [num_map[complement], i]
        num_map[num] = i
    return []

# Example: nums = [2,7,11,15], target = 9 → [0,1]
- Time: O(n), Space: O(n) - Amazon Context: Inventory matching, price optimization - Manager Insight: Explain hash table trade-offs to team - Similar: Two Sum II, 3Sum, 4Sum

2. Valid Palindrome (LeetCode #125)

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def is_palindrome(s):
    """Check if string is a valid palindrome."""
    left, right = 0, len(s) - 1

    while left < right:
        while left < right and not s[left].isalnum():
            left += 1
        while left < right and not s[right].isalnum():
            right -= 1

        if s[left].lower() != s[right].lower():
            return False

        left += 1
        right -= 1

    return True

# Example: "A man, a plan, a canal: Panama" → True
- Time: O(n), Space: O(1) - Amazon Context: Data validation, input sanitization - Manager Insight: Discuss edge case handling importance - Similar: Valid Palindrome II, Longest Palindromic Substring

3. Best Time to Buy and Sell Stock (LeetCode #121)

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def max_profit(prices):
    """Maximum profit from single buy/sell."""
    min_price = float('inf')
    max_profit = 0

    for price in prices:
        min_price = min(min_price, price)
        max_profit = max(max_profit, price - min_price)

    return max_profit

# Example: [7,1,5,3,6,4] → 5 (buy at 1, sell at 6)
- Time: O(n), Space: O(1) - Amazon Context: Cost optimization, resource allocation - Manager Insight: Connect to business metrics and KPIs - Similar: Best Time to Buy and Sell Stock II, III, IV

4. Valid Parentheses (LeetCode #20)

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def is_valid(s):
    """Check if parentheses are properly matched."""
    stack = []
    mapping = {')': '(', '}': '{', ']': '['}

    for char in s:
        if char in mapping:
            if not stack or stack.pop() != mapping[char]:
                return False
        else:
            stack.append(char)

    return not stack

# Example: "()[]{}" → True, "([)]" → False
- Time: O(n), Space: O(n) - Amazon Context: Configuration validation, JSON parsing - Manager Insight: Discuss stack-based parsing in systems - Similar: Generate Parentheses, Longest Valid Parentheses

5. Longest Common Prefix (LeetCode #14)

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def longest_common_prefix(strs):
    """Find longest common prefix among strings."""
    if not strs:
        return ""

    prefix = strs[0]
    for string in strs[1:]:
        while not string.startswith(prefix):
            prefix = prefix[:-1]
            if not prefix:
                return ""

    return prefix

# Example: ["flower","flow","flight"] → "fl"
- Time: O(S) where S is sum of all characters, Space: O(1) - Amazon Context: URL routing, path matching - Manager Insight: Optimization for large datasets - Similar: Longest Common Subsequence, Find Common Characters

6. Merge Sorted Array (LeetCode #88)

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def merge(nums1, m, nums2, n):
    """Merge nums2 into nums1 in-place."""
    p1, p2 = m - 1, n - 1
    write_index = m + n - 1

    while p1 >= 0 and p2 >= 0:
        if nums1[p1] > nums2[p2]:
            nums1[write_index] = nums1[p1]
            p1 -= 1
        else:
            nums1[write_index] = nums2[p2]
            p2 -= 1
        write_index -= 1

    # Copy remaining elements from nums2
    while p2 >= 0:
        nums1[write_index] = nums2[p2]
        p2 -= 1
        write_index -= 1

# Example: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3 → [1,2,2,3,5,6]
- Time: O(m + n), Space: O(1) - Amazon Context: Data merging, batch processing - Manager Insight: In-place algorithms for memory efficiency - Similar: Merge Two Sorted Lists, Merge k Sorted Arrays

Medium Difficulty (12 problems)

7. Maximum Subarray (LeetCode #53)

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def max_subarray(nums):
    """Find maximum sum of contiguous subarray (Kadane's Algorithm)."""
    max_sum = current_sum = nums[0]

    for num in nums[1:]:
        current_sum = max(num, current_sum + num)
        max_sum = max(max_sum, current_sum)

    return max_sum

# Example: [-2,1,-3,4,-1,2,1,-5,4] → 6 ([4,-1,2,1])
- Time: O(n), Space: O(1) - Amazon Context: Performance metrics analysis, trend detection - Manager Insight: Dynamic programming fundamentals - Similar: Maximum Product Subarray, Maximum Subarray Sum with One Deletion

8. Product of Array Except Self (LeetCode #238)

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def product_except_self(nums):
    """Product of all elements except current."""
    n = len(nums)
    result = [1] * n

    # Left products
    for i in range(1, n):
        result[i] = result[i-1] * nums[i-1]

    # Right products
    right = 1
    for i in range(n-1, -1, -1):
        result[i] *= right
        right *= nums[i]

    return result

# Example: [1,2,3,4] → [24,12,8,6]
- Time: O(n), Space: O(1) excluding output - Amazon Context: Metrics aggregation, business intelligence - Manager Insight: Space-time trade-offs in large systems - Similar: Product of the Last K Numbers, Maximum Product Subarray

9. Container With Most Water (LeetCode #11)

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def max_area(height):
    """Find container that holds most water."""
    left, right = 0, len(height) - 1
    max_water = 0

    while left < right:
        width = right - left
        current_water = min(height[left], height[right]) * width
        max_water = max(max_water, current_water)

        if height[left] < height[right]:
            left += 1
        else:
            right -= 1

    return max_water

# Example: [1,8,6,2,5,4,8,3,7] → 49
- Time: O(n), Space: O(1) - Amazon Context: Resource optimization, capacity planning - Manager Insight: Greedy algorithm principles - Similar: Trapping Rain Water, Largest Rectangle in Histogram

10. 3Sum (LeetCode #15)

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def three_sum(nums):
    """Find all unique triplets that sum to zero."""
    nums.sort()
    result = []

    for i in range(len(nums) - 2):
        if i > 0 and nums[i] == nums[i-1]:
            continue

        left, right = i + 1, len(nums) - 1
        while left < right:
            total = nums[i] + nums[left] + nums[right]
            if total == 0:
                result.append([nums[i], nums[left], nums[right]])
                while left < right and nums[left] == nums[left+1]:
                    left += 1
                while left < right and nums[right] == nums[right-1]:
                    right -= 1
                left += 1
                right -= 1
            elif total < 0:
                left += 1
            else:
                right -= 1

    return result

# Example: [-1,0,1,2,-1,-4] → [[-1,-1,2],[-1,0,1]]
- Time: O(n²), Space: O(1) excluding output - Amazon Context: Team composition, resource allocation - Manager Insight: Complexity management in algorithms - Similar: 3Sum Closest, 4Sum, 3Sum Smaller

11. Group Anagrams (LeetCode #49)

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def group_anagrams(strs):
    """Group strings that are anagrams."""
    from collections import defaultdict

    groups = defaultdict(list)
    for s in strs:
        key = ''.join(sorted(s))
        groups[key].append(s)

    return list(groups.values())

# Example: ["eat","tea","tan","ate","nat","bat"] → [["eat","tea","ate"],["tan","nat"],["bat"]]
- Time: O(n * k log k), Space: O(n * k) - Amazon Context: Data deduplication, categorization - Manager Insight: Hash function design choices - Similar: Group Shifted Strings, Find All Anagrams

12. Longest Substring Without Repeating Characters (LeetCode #3)

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def length_of_longest_substring(s):
    """Length of longest substring without repeating chars."""
    char_map = {}
    left = max_length = 0

    for right, char in enumerate(s):
        if char in char_map and char_map[char] >= left:
            left = char_map[char] + 1
        char_map[char] = right
        max_length = max(max_length, right - left + 1)

    return max_length

# Example: "abcabcbb" → 3 ("abc")
- Time: O(n), Space: O(min(m,n)) - Amazon Context: Session management, caching strategies - Manager Insight: Sliding window pattern applications - Similar: Longest Repeating Character Replacement, Minimum Window Substring

13. String to Integer (atoi) (LeetCode #8)

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def my_atoi(s):
    """Convert string to integer with constraints."""
    i = 0
    sign = 1
    result = 0

    # Skip whitespace
    while i < len(s) and s[i] == ' ':
        i += 1

    # Handle sign
    if i < len(s) and s[i] in '+-':
        sign = -1 if s[i] == '-' else 1
        i += 1

    # Convert digits
    while i < len(s) and s[i].isdigit():
        result = result * 10 + int(s[i])
        i += 1

    result *= sign

    # Handle overflow
    return max(-2**31, min(2**31 - 1, result))

# Example: "   -42" → -42, "4193 with words" → 4193
- Time: O(n), Space: O(1) - Amazon Context: Input validation, data parsing - Manager Insight: Robust error handling patterns - Similar: String to Integer (atoi), Reverse Integer

14. Longest Palindromic Substring (LeetCode #5)

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def longest_palindrome(s):
    """Find longest palindromic substring."""
    def expand_around_center(left, right):
        while left >= 0 and right < len(s) and s[left] == s[right]:
            left -= 1
            right += 1
        return right - left - 1

    start = end = 0

    for i in range(len(s)):
        len1 = expand_around_center(i, i)      # odd length
        len2 = expand_around_center(i, i + 1)  # even length
        max_len = max(len1, len2)

        if max_len > end - start:
            start = i - (max_len - 1) // 2
            end = i + max_len // 2

    return s[start:end + 1]

# Example: "babad" → "bab" or "aba"
- Time: O(n²), Space: O(1) - Amazon Context: Pattern matching, data analysis - Manager Insight: Center expansion technique - Similar: Palindromic Substrings, Shortest Palindrome

15. ZigZag Conversion (LeetCode #6)

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def convert(s, num_rows):
    """Convert string to zigzag pattern."""
    if num_rows == 1:
        return s

    rows = [''] * min(num_rows, len(s))
    current_row = 0
    going_down = False

    for char in s:
        rows[current_row] += char

        if current_row == 0 or current_row == num_rows - 1:
            going_down = not going_down

        current_row += 1 if going_down else -1

    return ''.join(rows)

# Example: "PAYPALISHIRING", numRows = 3 → "PAHNAPLSIIGYIR"
- Time: O(n), Space: O(n) - Amazon Context: Data formatting, report generation - Manager Insight: Pattern simulation techniques - Similar: Diagonal Traverse, Spiral Matrix

16. Integer to Roman (LeetCode #12)

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def int_to_roman(num):
    """Convert integer to Roman numeral."""
    values = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1]
    symbols = ["M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"]

    result = ""
    for i in range(len(values)):
        count = num // values[i]
        result += symbols[i] * count
        num %= values[i]

    return result

# Example: 1994 → "MCMXCIV"
- Time: O(1), Space: O(1) - Amazon Context: Data representation, format conversion - Manager Insight: Greedy algorithm applications - Similar: Roman to Integer, Integer to English Words

17. Generate Parentheses (LeetCode #22)

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def generate_parentheses(n):
    """Generate all valid parentheses combinations."""
    result = []

    def backtrack(current, open_count, close_count):
        if len(current) == 2 * n:
            result.append(current)
            return

        if open_count < n:
            backtrack(current + "(", open_count + 1, close_count)

        if close_count < open_count:
            backtrack(current + ")", open_count, close_count + 1)

    backtrack("", 0, 0)
    return result

# Example: n = 3 → ["((()))","(()())","(())()","()(())","()()()"]
- Time: O(4^n / √n), Space: O(4^n / √n) - Amazon Context: Configuration generation, valid combinations - Manager Insight: Backtracking algorithm patterns - Similar: Valid Parentheses, Remove Invalid Parentheses

18. Letter Combinations of Phone Number (LeetCode #17)

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def letter_combinations(digits):
    """Generate letter combinations from phone number."""
    if not digits:
        return []

    mapping = {
        '2': 'abc', '3': 'def', '4': 'ghi', '5': 'jkl',
        '6': 'mno', '7': 'pqrs', '8': 'tuv', '9': 'wxyz'
    }

    result = []

    def backtrack(index, path):
        if index == len(digits):
            result.append(path)
            return

        letters = mapping[digits[index]]
        for letter in letters:
            backtrack(index + 1, path + letter)

    backtrack(0, "")
    return result

# Example: "23" → ["ad","ae","af","bd","be","bf","cd","ce","cf"]
- Time: O(3^m × 4^n), Space: O(3^m × 4^n) - Amazon Context: Search suggestions, autocompletion - Manager Insight: Combinatorial problem solving - Similar: Generate Parentheses, Combination Sum

Hard Difficulty (2 problems)

19. Minimum Window Substring (LeetCode #76)

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def min_window(s, t):
    """Find minimum window containing all chars of t."""
    from collections import Counter, defaultdict

    if not s or not t:
        return ""

    dict_t = Counter(t)
    required = len(dict_t)
    formed = 0
    window_counts = defaultdict(int)

    l, r = 0, 0
    ans = float('inf'), None, None

    while r < len(s):
        character = s[r]
        window_counts[character] += 1

        if character in dict_t and window_counts[character] == dict_t[character]:
            formed += 1

        while l <= r and formed == required:
            character = s[l]

            if r - l + 1 < ans[0]:
                ans = (r - l + 1, l, r)

            window_counts[character] -= 1
            if character in dict_t and window_counts[character] < dict_t[character]:
                formed -= 1

            l += 1

        r += 1

    return "" if ans[0] == float('inf') else s[ans[1]: ans[2] + 1]

# Example: s = "ADOBECODEBANC", t = "ABC" → "BANC"
- Time: O(|s| + |t|), Space: O(|s| + |t|) - Amazon Context: Log analysis, pattern extraction - Manager Insight: Advanced sliding window techniques - Similar: Substring with Concatenation of All Words, Find All Anagrams

20. Text Justification (LeetCode #68)

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def full_justify(words, max_width):
    """Justify text to fit max width."""
    result = []
    i = 0

    while i < len(words):
        # Find words that fit in current line
        j = i
        line_length = len(words[i])

        while j + 1 < len(words) and line_length + 1 + len(words[j + 1]) <= max_width:
            j += 1
            line_length += 1 + len(words[j])

        # Build justified line
        if j == len(words) - 1 or j == i:  # Last line or single word
            line = ' '.join(words[i:j+1])
            line += ' ' * (max_width - len(line))
        else:
            spaces_needed = max_width - sum(len(word) for word in words[i:j+1])
            gaps = j - i

            line = ""
            for k in range(i, j):
                line += words[k]
                if k < j:
                    spaces = spaces_needed // gaps
                    if k - i < spaces_needed % gaps:
                        spaces += 1
                    line += ' ' * spaces
            line += words[j]

        result.append(line)
        i = j + 1

    return result

# Example: words = ["This", "is", "an", "example"], maxWidth = 16
# → ["This    is    an","example         "]
- Time: O(n), Space: O(1) excluding output - Amazon Context: Report formatting, document generation - Manager Insight: Complex string manipulation requirements - Similar: Reorder Data in Log Files, Valid Word Abbreviation

Category 2: Two Pointers (10 Problems)

Easy Difficulty (4 problems)

21. Two Sum II - Input Array Is Sorted (LeetCode #167)

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def two_sum(numbers, target):
    """Find two numbers in sorted array that add to target."""
    left, right = 0, len(numbers) - 1

    while left < right:
        current_sum = numbers[left] + numbers[right]
        if current_sum == target:
            return [left + 1, right + 1]  # 1-indexed
        elif current_sum < target:
            left += 1
        else:
            right -= 1

    return []

# Example: numbers = [2,7,11,15], target = 9 → [1,2]
- Time: O(n), Space: O(1) - Amazon Context: Efficient searching in sorted data - Manager Insight: Leverage data properties for optimization - Similar: Two Sum, 3Sum, Pair with Target Sum

22. Remove Duplicates from Sorted Array (LeetCode #26)

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def remove_duplicates(nums):
    """Remove duplicates in-place from sorted array."""
    if not nums:
        return 0

    write_index = 1

    for read_index in range(1, len(nums)):
        if nums[read_index] != nums[read_index - 1]:
            nums[write_index] = nums[read_index]
            write_index += 1

    return write_index

# Example: [1,1,2] → length = 2, nums = [1,2,_]
- Time: O(n), Space: O(1) - Amazon Context: Data deduplication, cleanup processes - Manager Insight: In-place algorithms for memory efficiency - Similar: Remove Element, Remove Duplicates from Sorted Array II

23. Move Zeroes (LeetCode #283)

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def move_zeroes(nums):
    """Move all zeros to end while maintaining relative order."""
    write_index = 0

    for read_index in range(len(nums)):
        if nums[read_index] != 0:
            nums[write_index] = nums[read_index]
            write_index += 1

    while write_index < len(nums):
        nums[write_index] = 0
        write_index += 1

# Example: [0,1,0,3,12] → [1,3,12,0,0]
- Time: O(n), Space: O(1) - Amazon Context: Data filtering, null value handling - Manager Insight: Stable partitioning algorithms - Similar: Remove Element, Partition Array

24. Squares of a Sorted Array (LeetCode #977)

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def sorted_squares(nums):
    """Return sorted squares of array elements."""
    result = [0] * len(nums)
    left, right = 0, len(nums) - 1
    write_index = len(nums) - 1

    while left <= right:
        left_square = nums[left] ** 2
        right_square = nums[right] ** 2

        if left_square > right_square:
            result[write_index] = left_square
            left += 1
        else:
            result[write_index] = right_square
            right -= 1

        write_index -= 1

    return result

# Example: [-4,-1,0,3,10] → [0,1,9,16,100]
- Time: O(n), Space: O(n) - Amazon Context: Data transformation, metrics processing - Manager Insight: Handling negative values efficiently - Similar: Sort Colors, Merge Sorted Array

Medium Difficulty (5 problems)

25. 3Sum Closest (LeetCode #16)

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def three_sum_closest(nums, target):
    """Find three numbers whose sum is closest to target."""
    nums.sort()
    closest_sum = float('inf')

    for i in range(len(nums) - 2):
        left, right = i + 1, len(nums) - 1

        while left < right:
            current_sum = nums[i] + nums[left] + nums[right]

            if abs(current_sum - target) < abs(closest_sum - target):
                closest_sum = current_sum

            if current_sum < target:
                left += 1
            elif current_sum > target:
                right -= 1
            else:
                return current_sum

    return closest_sum

# Example: nums = [-1,2,1,-4], target = 1 → 2
- Time: O(n²), Space: O(1) - Amazon Context: Optimization problems, resource allocation - Manager Insight: Approximation algorithms in constraints - Similar: 3Sum, 4Sum Closest, Closest Binary Search Tree Value

26. Sort Colors (LeetCode #75)

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def sort_colors(nums):
    """Sort array with only 0s, 1s, and 2s."""
    left, current, right = 0, 0, len(nums) - 1

    while current <= right:
        if nums[current] == 0:
            nums[left], nums[current] = nums[current], nums[left]
            left += 1
            current += 1
        elif nums[current] == 1:
            current += 1
        else:  # nums[current] == 2
            nums[current], nums[right] = nums[right], nums[current]
            right -= 1
            # Don't increment current as we need to check swapped element

# Example: [2,0,2,1,1,0] → [0,0,1,1,2,2]
- Time: O(n), Space: O(1) - Amazon Context: Data categorization, priority sorting - Manager Insight: Dutch national flag algorithm - Similar: Partition Array, Move Zeroes

27. Container With Most Water (LeetCode #11)

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def max_area(height):
    """Find container that holds most water."""
    left, right = 0, len(height) - 1
    max_water = 0

    while left < right:
        width = right - left
        current_water = min(height[left], height[right]) * width
        max_water = max(max_water, current_water)

        if height[left] < height[right]:
            left += 1
        else:
            right -= 1

    return max_water

# Example: [1,8,6,2,5,4,8,3,7] → 49
- Time: O(n), Space: O(1) - Amazon Context: Resource optimization, capacity planning - Manager Insight: Greedy approach vs brute force trade-offs - Similar: Trapping Rain Water, Largest Rectangle

28. Remove Nth Node From End of List (LeetCode #19)

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def remove_nth_from_end(head, n):
    """Remove nth node from end of linked list."""
    dummy = ListNode(0)
    dummy.next = head
    first = second = dummy

    # Move first pointer n+1 steps ahead
    for _ in range(n + 1):
        first = first.next

    # Move both pointers until first reaches end
    while first:
        first = first.next
        second = second.next

    # Remove the nth node
    second.next = second.next.next

    return dummy.next

# Example: [1,2,3,4,5], n = 2 → [1,2,3,5]
- Time: O(n), Space: O(1) - Amazon Context: Data structure manipulation, list operations - Manager Insight: One-pass algorithms using two pointers - Similar: Middle of Linked List, Delete Node in Linked List

29. Partition Labels (LeetCode #763)

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def partition_labels(s):
    """Partition string into as many parts as possible."""
    last_occurrence = {char: i for i, char in enumerate(s)}
    partitions = []
    start = end = 0

    for i, char in enumerate(s):
        end = max(end, last_occurrence[char])

        if i == end:
            partitions.append(end - start + 1)
            start = i + 1

    return partitions

# Example: "ababcbacadefegdehijhklij" → [9,7,8]
- Time: O(n), Space: O(1) - limited alphabet - Amazon Context: Data segmentation, batch processing - Manager Insight: Greedy partitioning strategies - Similar: Merge Intervals, Non-overlapping Intervals

Hard Difficulty (1 problem)

30. Trapping Rain Water (LeetCode #42)

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def trap(height):
    """Calculate trapped rainwater."""
    left, right = 0, len(height) - 1
    left_max = right_max = water = 0

    while left < right:
        if height[left] < height[right]:
            if height[left] >= left_max:
                left_max = height[left]
            else:
                water += left_max - height[left]
            left += 1
        else:
            if height[right] >= right_max:
                right_max = height[right]
            else:
                water += right_max - height[right]
            right -= 1

    return water

# Example: [0,1,0,2,1,0,1,3,2,1,2,1] → 6
- Time: O(n), Space: O(1) - Amazon Context: Capacity planning, resource utilization - Manager Insight: Complex optimization with constraints - Similar: Container With Most Water, Largest Rectangle

Category 3: Sliding Window (10 Problems)

Easy Difficulty (3 problems)

31. Maximum Average Subarray I (LeetCode #643)

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def find_max_average(nums, k):
    """Find maximum average of contiguous subarray of length k."""
    current_sum = sum(nums[:k])
    max_sum = current_sum

    for i in range(k, len(nums)):
        current_sum = current_sum - nums[i-k] + nums[i]
        max_sum = max(max_sum, current_sum)

    return max_sum / k

# Example: nums = [1,12,-5,-6,50,3], k = 4 → 12.75
- Time: O(n), Space: O(1) - Amazon Context: Performance metrics, moving averages - Manager Insight: Fixed-size window optimization - Similar: Moving Average from Data Stream, Max Sum of Rectangle

32. Contains Duplicate II (LeetCode #219)

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def contains_nearby_duplicate(nums, k):
    """Check if duplicate exists within k distance."""
    seen = {}

    for i, num in enumerate(nums):
        if num in seen and i - seen[num] <= k:
            return True
        seen[num] = i

    return False

# Example: nums = [1,2,3,1], k = 3 → True
- Time: O(n), Space: O(min(n,k)) - Amazon Context: Fraud detection, duplicate checking - Manager Insight: Time-window based validation - Similar: Contains Duplicate, Contains Duplicate III

33. Minimum Size Subarray Sum (LeetCode #209)

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def min_subarray_len(target, nums):
    """Minimum length subarray with sum >= target."""
    left = current_sum = 0
    min_length = float('inf')

    for right in range(len(nums)):
        current_sum += nums[right]

        while current_sum >= target:
            min_length = min(min_length, right - left + 1)
            current_sum -= nums[left]
            left += 1

    return min_length if min_length != float('inf') else 0

# Example: target = 7, nums = [2,3,1,2,4,3] → 2
- Time: O(n), Space: O(1) - Amazon Context: Resource allocation, threshold optimization - Manager Insight: Variable-size window techniques - Similar: Maximum Size Subarray Sum Equals k, Subarray Sum Equals K

Medium Difficulty (6 problems)

34. Longest Substring Without Repeating Characters (LeetCode #3)

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def length_of_longest_substring(s):
    """Length of longest substring without repeating chars."""
    char_map = {}
    left = max_length = 0

    for right, char in enumerate(s):
        if char in char_map and char_map[char] >= left:
            left = char_map[char] + 1
        char_map[char] = right
        max_length = max(max_length, right - left + 1)

    return max_length

# Example: "abcabcbb" → 3
- Time: O(n), Space: O(min(m,n)) - Amazon Context: Session management, unique access patterns - Manager Insight: Character frequency tracking - Similar: Longest Repeating Character Replacement, Fruit Into Baskets

35. Longest Repeating Character Replacement (LeetCode #424)

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def character_replacement(s, k):
    """Longest substring with same character after k replacements."""
    from collections import defaultdict

    char_count = defaultdict(int)
    left = max_length = max_freq = 0

    for right in range(len(s)):
        char_count[s[right]] += 1
        max_freq = max(max_freq, char_count[s[right]])

        # If window size - most frequent char > k, shrink window
        if right - left + 1 - max_freq > k:
            char_count[s[left]] -= 1
            left += 1

        max_length = max(max_length, right - left + 1)

    return max_length

# Example: s = "ABAB", k = 2 → 4
- Time: O(n), Space: O(1) - Amazon Context: Data normalization, error correction - Manager Insight: Frequency-based optimization - Similar: Max Consecutive Ones III, Longest Substring with At Most K Distinct

36. Permutation in String (LeetCode #567)

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def check_inclusion(s1, s2):
    """Check if s2 contains permutation of s1."""
    from collections import Counter

    if len(s1) > len(s2):
        return False

    s1_count = Counter(s1)
    window_count = Counter(s2[:len(s1)])

    if window_count == s1_count:
        return True

    for i in range(len(s1), len(s2)):
        # Add new character
        window_count[s2[i]] += 1

        # Remove old character
        old_char = s2[i - len(s1)]
        window_count[old_char] -= 1
        if window_count[old_char] == 0:
            del window_count[old_char]

        if window_count == s1_count:
            return True

    return False

# Example: s1 = "ab", s2 = "eidbaooo" → True
- Time: O(n), Space: O(k) - Amazon Context: Pattern matching, anagram detection - Manager Insight: Fixed-window string matching - Similar: Find All Anagrams in a String, Substring with Concatenation

37. Max Consecutive Ones III (LeetCode #1004)

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def longest_ones(nums, k):
    """Longest subarray with at most k zeros flipped."""
    left = zeros_count = max_length = 0

    for right in range(len(nums)):
        if nums[right] == 0:
            zeros_count += 1

        while zeros_count > k:
            if nums[left] == 0:
                zeros_count -= 1
            left += 1

        max_length = max(max_length, right - left + 1)

    return max_length

# Example: nums = [1,1,1,0,0,0,1,1,1,1,0], k = 2 → 6
- Time: O(n), Space: O(1) - Amazon Context: System reliability, uptime optimization - Manager Insight: Constraint-based window expansion - Similar: Max Consecutive Ones, Longest Repeating Character Replacement

38. Fruit Into Baskets (LeetCode #904)

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def total_fruit(fruits):
    """Maximum fruits you can collect with 2 baskets."""
    from collections import defaultdict

    basket = defaultdict(int)
    left = max_fruits = 0

    for right in range(len(fruits)):
        basket[fruits[right]] += 1

        while len(basket) > 2:
            basket[fruits[left]] -= 1
            if basket[fruits[left]] == 0:
                del basket[fruits[left]]
            left += 1

        max_fruits = max(max_fruits, right - left + 1)

    return max_fruits

# Example: fruits = [1,2,1] → 3
- Time: O(n), Space: O(1) - Amazon Context: Resource optimization, capacity constraints - Manager Insight: At-most-K distinct elements pattern - Similar: Longest Substring with At Most Two Distinct Characters

39. Subarrays with K Different Integers (LeetCode #992)

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def subarrays_with_k_distinct(nums, k):
    """Count subarrays with exactly k distinct integers."""
    def at_most_k(k):
        from collections import defaultdict
        count = defaultdict(int)
        left = result = 0

        for right in range(len(nums)):
            if count[nums[right]] == 0:
                k -= 1
            count[nums[right]] += 1

            while k < 0:
                count[nums[left]] -= 1
                if count[nums[left]] == 0:
                    k += 1
                left += 1

            result += right - left + 1

        return result

    return at_most_k(k) - at_most_k(k - 1)

# Example: nums = [1,2,1,2,3], k = 2 → 7
- Time: O(n), Space: O(k) - Amazon Context: Analytics, distinct value counting - Manager Insight: Exactly-K problems using at-most-K - Similar: Binary Subarrays With Sum, Count Number of Nice Subarrays

Hard Difficulty (1 problem)

40. Sliding Window Maximum (LeetCode #239)

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def max_sliding_window(nums, k):
    """Maximum in each sliding window of size k."""
    from collections import deque

    dq = deque()
    result = []

    for i, num in enumerate(nums):
        # Remove elements outside window
        while dq and dq[0] < i - k + 1:
            dq.popleft()

        # Remove smaller elements
        while dq and nums[dq[-1]] < num:
            dq.pop()

        dq.append(i)

        # Add to result if window is complete
        if i >= k - 1:
            result.append(nums[dq[0]])

    return result

# Example: nums = [1,3,-1,-3,5,3,6,7], k = 3 → [3,3,5,5,6,7]
- Time: O(n), Space: O(k) - Amazon Context: Real-time metrics, monitoring dashboards - Manager Insight: Monotonic deque optimization - Similar: Sliding Window Minimum, Constrained Subsequence Sum

Category 4: Hash Tables (10 Problems)

Easy Difficulty (4 problems)

41. Contains Duplicate (LeetCode #217)

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def contains_duplicate(nums):
    """Check if array contains duplicates."""
    return len(nums) != len(set(nums))

# Alternative O(n) space approach:
def contains_duplicate_optimized(nums):
    seen = set()
    for num in nums:
        if num in seen:
            return True
        seen.add(num)
    return False

# Example: [1,2,3,1] → True
- Time: O(n), Space: O(n) - Amazon Context: Data validation, duplicate detection - Manager Insight: Set operations for uniqueness checks - Similar: Contains Duplicate II, Contains Duplicate III

42. Valid Anagram (LeetCode #242)

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def is_anagram(s, t):
    """Check if two strings are anagrams."""
    from collections import Counter
    return Counter(s) == Counter(t)

# Alternative sorting approach:
def is_anagram_sort(s, t):
    return sorted(s) == sorted(t)

# Example: s = "anagram", t = "nagaram" → True
- Time: O(n), Space: O(1) - limited alphabet - Amazon Context: Text analysis, content comparison - Manager Insight: Character frequency comparison - Similar: Group Anagrams, Find All Anagrams

43. Two Sum (LeetCode #1)

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def two_sum(nums, target):
    """Find two numbers that add up to target."""
    num_map = {}
    for i, num in enumerate(nums):
        complement = target - num
        if complement in num_map:
            return [num_map[complement], i]
        num_map[num] = i
    return []

# Example: nums = [2,7,11,15], target = 9 → [0,1]
- Time: O(n), Space: O(n) - Amazon Context: Pair finding, complement lookup - Manager Insight: Hash map for O(1) lookups - Similar: Two Sum II, 3Sum, 4Sum

44. Happy Number (LeetCode #202)

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def is_happy(n):
    """Check if number is happy."""
    def get_sum_of_squares(num):
        total = 0
        while num:
            digit = num % 10
            total += digit * digit
            num //= 10
        return total

    seen = set()
    while n != 1 and n not in seen:
        seen.add(n)
        n = get_sum_of_squares(n)

    return n == 1

# Example: 19 → True (1→1, 9→81, 8→64, 1→1, 4→16, 6→36, 3→9, 6→36...)
- Time: O(log n), Space: O(log n) - Amazon Context: Validation algorithms, cycle detection - Manager Insight: Cycle detection using hash sets - Similar: Linked List Cycle, Find the Duplicate Number

Medium Difficulty (5 problems)

45. Group Anagrams (LeetCode #49)

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def group_anagrams(strs):
    """Group strings that are anagrams."""
    from collections import defaultdict

    groups = defaultdict(list)
    for s in strs:
        key = ''.join(sorted(s))
        groups[key].append(s)

    return list(groups.values())

# Example: ["eat","tea","tan","ate","nat","bat"] → [["eat","tea","ate"],["tan","nat"],["bat"]]
- Time: O(n * k log k), Space: O(n * k) - Amazon Context: Data categorization, content grouping - Manager Insight: Canonical form for grouping - Similar: Valid Anagram, Find All Anagrams

46. Top K Frequent Elements (LeetCode #347)

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def top_k_frequent(nums, k):
    """Find k most frequent elements."""
    from collections import Counter
    import heapq

    count = Counter(nums)
    return heapq.nlargest(k, count.keys(), key=count.get)

# Alternative bucket sort approach:
def top_k_frequent_bucket(nums, k):
    from collections import Counter
    count = Counter(nums)
    buckets = [[] for _ in range(len(nums) + 1)]

    for num, freq in count.items():
        buckets[freq].append(num)

    result = []
    for i in range(len(buckets) - 1, -1, -1):
        for num in buckets[i]:
            result.append(num)
            if len(result) == k:
                return result

# Example: nums = [1,1,1,2,2,3], k = 2 → [1,2]
- Time: O(n log k), Space: O(n) - Amazon Context: Analytics, trending analysis - Manager Insight: Frequency-based ranking systems - Similar: Kth Largest Element, Top K Frequent Words

47. Longest Consecutive Sequence (LeetCode #128)

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def longest_consecutive(nums):
    """Find length of longest consecutive sequence."""
    num_set = set(nums)
    longest = 0

    for num in num_set:
        # Only start counting from sequence start
        if num - 1 not in num_set:
            current_num = num
            current_streak = 1

            while current_num + 1 in num_set:
                current_num += 1
                current_streak += 1

            longest = max(longest, current_streak)

    return longest

# Example: [100,4,200,1,3,2] → 4 ([1,2,3,4])
- Time: O(n), Space: O(n) - Amazon Context: Sequence analysis, data continuity - Manager Insight: Set-based sequence detection - Similar: Binary Tree Longest Consecutive Sequence

48. Subarray Sum Equals K (LeetCode #560)

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def subarray_sum(nums, k):
    """Count subarrays with sum equal to k."""
    from collections import defaultdict

    count = 0
    prefix_sum = 0
    sum_count = defaultdict(int)
    sum_count[0] = 1  # Empty prefix

    for num in nums:
        prefix_sum += num
        count += sum_count[prefix_sum - k]
        sum_count[prefix_sum] += 1

    return count

# Example: nums = [1,1,1], k = 2 → 2
- Time: O(n), Space: O(n) - Amazon Context: Financial analysis, target sum finding - Manager Insight: Prefix sum with hash map optimization - Similar: Two Sum, Continuous Subarray Sum

49. 4Sum II (LeetCode #454)

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def four_sum_count(nums1, nums2, nums3, nums4):
    """Count tuples with sum equal to 0."""
    from collections import defaultdict

    # Count sums of pairs from first two arrays
    sum_count = defaultdict(int)
    for a in nums1:
        for b in nums2:
            sum_count[a + b] += 1

    count = 0
    # Check if complement exists in remaining arrays
    for c in nums3:
        for d in nums4:
            count += sum_count[-(c + d)]

    return count

# Example: nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2] → 2
- Time: O(n²), Space: O(n²) - Amazon Context: Multi-dimensional analysis, correlation finding - Manager Insight: Hash map to reduce complexity from O(n⁴) to O(n²) - Similar: 4Sum, Two Sum, 3Sum

Hard Difficulty (1 problem)

50. First Missing Positive (LeetCode #41)

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def first_missing_positive(nums):
    """Find smallest missing positive integer."""
    n = len(nums)

    # Mark presence of numbers 1 to n
    for i in range(n):
        while 1 <= nums[i] <= n and nums[nums[i] - 1] != nums[i]:
            nums[nums[i] - 1], nums[i] = nums[i], nums[nums[i] - 1]

    # Find first missing positive
    for i in range(n):
        if nums[i] != i + 1:
            return i + 1

    return n + 1

# Example: [3,4,-1,1] → 2
- Time: O(n), Space: O(1) - Amazon Context: Resource tracking, ID generation - Manager Insight: Array as hash map using indices - Similar: Missing Number, Find All Numbers Disappeared

Category 5: Trees & Graphs (15 Problems)

Easy Difficulty (6 problems)

51. Maximum Depth of Binary Tree (LeetCode #104)

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def max_depth(root):
    """Find maximum depth of binary tree."""
    if not root:
        return 0
    return 1 + max(max_depth(root.left), max_depth(root.right))

# Iterative approach:
def max_depth_iterative(root):
    if not root:
        return 0

    from collections import deque
    queue = deque([(root, 1)])
    max_depth = 0

    while queue:
        node, depth = queue.popleft()
        max_depth = max(max_depth, depth)

        if node.left:
            queue.append((node.left, depth + 1))
        if node.right:
            queue.append((node.right, depth + 1))

    return max_depth

# Example: [3,9,20,null,null,15,7] → 3
- Time: O(n), Space: O(h) where h is height - Amazon Context: System hierarchy analysis, org chart depth - Manager Insight: Tree traversal fundamentals - Similar: Minimum Depth of Binary Tree, Balanced Binary Tree

52. Same Tree (LeetCode #100)

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def is_same_tree(p, q):
    """Check if two trees are the same."""
    if not p and not q:
        return True
    if not p or not q:
        return False
    return (p.val == q.val and 
            is_same_tree(p.left, q.left) and 
            is_same_tree(p.right, q.right))

# Example: p = [1,2,3], q = [1,2,3] → True
- Time: O(n), Space: O(h) - Amazon Context: Configuration comparison, system validation - Manager Insight: Structural equality checking - Similar: Symmetric Tree, Subtree of Another Tree

53. Invert Binary Tree (LeetCode #226)

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def invert_tree(root):
    """Invert a binary tree."""
    if not root:
        return None

    root.left, root.right = root.right, root.left
    invert_tree(root.left)
    invert_tree(root.right)

    return root

# Example: [4,2,7,1,3,6,9] → [4,7,2,9,6,3,1]
- Time: O(n), Space: O(h) - Amazon Context: Data transformation, mirror operations - Manager Insight: In-place tree manipulation - Similar: Symmetric Tree, Binary Tree Upside Down

54. Path Sum (LeetCode #112)

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def has_path_sum(root, target_sum):
    """Check if path from root to leaf sums to target."""
    if not root:
        return False

    if not root.left and not root.right:
        return root.val == target_sum

    return (has_path_sum(root.left, target_sum - root.val) or
            has_path_sum(root.right, target_sum - root.val))

# Example: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22 → True
- Time: O(n), Space: O(h) - Amazon Context: Budget validation, resource tracking - Manager Insight: Path-based decision making - Similar: Path Sum II, Path Sum III

55. Symmetric Tree (LeetCode #101)

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def is_symmetric(root):
    """Check if tree is symmetric."""
    def is_mirror(t1, t2):
        if not t1 and not t2:
            return True
        if not t1 or not t2:
            return False
        return (t1.val == t2.val and 
                is_mirror(t1.left, t2.right) and 
                is_mirror(t1.right, t2.left))

    return is_mirror(root, root)

# Example: [1,2,2,3,4,4,3] → True
- Time: O(n), Space: O(h) - Amazon Context: Data validation, symmetry checking - Manager Insight: Mirror image comparison - Similar: Same Tree, Invert Binary Tree

56. Binary Tree Level Order Traversal (LeetCode #102)

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def level_order(root):
    """Level order traversal of binary tree."""
    if not root:
        return []

    from collections import deque
    result = []
    queue = deque([root])

    while queue:
        level_size = len(queue)
        level = []

        for _ in range(level_size):
            node = queue.popleft()
            level.append(node.val)

            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)

        result.append(level)

    return result

# Example: [3,9,20,null,null,15,7] → [[3],[9,20],[15,7]]
- Time: O(n), Space: O(w) where w is max width - Amazon Context: Hierarchical reporting, level-wise processing - Manager Insight: BFS traversal pattern - Similar: Binary Tree Level Order Traversal II, Binary Tree Zigzag

Medium Difficulty (7 problems)

57. Validate Binary Search Tree (LeetCode #98)

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def is_valid_bst(root):
    """Validate binary search tree."""
    def validate(node, min_val, max_val):
        if not node:
            return True

        if node.val <= min_val or node.val >= max_val:
            return False

        return (validate(node.left, min_val, node.val) and
                validate(node.right, node.val, max_val))

    return validate(root, float('-inf'), float('inf'))

# Example: [2,1,3] → True, [5,1,4,null,null,3,6] → False
- Time: O(n), Space: O(h) - Amazon Context: Data integrity validation, ordered system verification - Manager Insight: Property validation with constraints - Similar: Recover Binary Search Tree, Convert Sorted Array to BST

58. Lowest Common Ancestor of a Binary Tree (LeetCode #236)

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def lowest_common_ancestor(root, p, q):
    """Find LCA in binary tree."""
    if not root or root == p or root == q:
        return root

    left = lowest_common_ancestor(root.left, p, q)
    right = lowest_common_ancestor(root.right, p, q)

    if left and right:
        return root

    return left or right

# Example: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1 → 3
- Time: O(n), Space: O(h) - Amazon Context: Hierarchy management, common parent finding - Manager Insight: Bottom-up tree search - Similar: Lowest Common Ancestor of BST, Lowest Common Ancestor III

59. Binary Tree Maximum Path Sum (LeetCode #124)

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def max_path_sum(root):
    """Find maximum path sum in binary tree."""
    max_sum = float('-inf')

    def max_gain(node):
        nonlocal max_sum
        if not node:
            return 0

        left_gain = max(max_gain(node.left), 0)
        right_gain = max(max_gain(node.right), 0)

        current_max = node.val + left_gain + right_gain
        max_sum = max(max_sum, current_max)

        return node.val + max(left_gain, right_gain)

    max_gain(root)
    return max_sum

# Example: [1,2,3] → 6, [-10,9,20,null,null,15,7] → 42
- Time: O(n), Space: O(h) - Amazon Context: Optimization analysis, maximum value paths - Manager Insight: Complex optimization with choices - Similar: Diameter of Binary Tree, Binary Tree Maximum Width

60. Construct Binary Tree from Preorder and Inorder (LeetCode #105)

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def build_tree(preorder, inorder):
    """Build tree from preorder and inorder traversals."""
    if not preorder or not inorder:
        return None

    root = TreeNode(preorder[0])
    mid = inorder.index(preorder[0])

    root.left = build_tree(preorder[1:mid+1], inorder[:mid])
    root.right = build_tree(preorder[mid+1:], inorder[mid+1:])

    return root

# Optimized version with index map:
def build_tree_optimized(preorder, inorder):
    inorder_map = {val: i for i, val in enumerate(inorder)}
    preorder_idx = [0]

    def helper(left, right):
        if left > right:
            return None

        root_val = preorder[preorder_idx[0]]
        preorder_idx[0] += 1
        root = TreeNode(root_val)

        mid = inorder_map[root_val]
        root.left = helper(left, mid - 1)
        root.right = helper(mid + 1, right)

        return root

    return helper(0, len(inorder) - 1)

# Example: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7] → tree
- Time: O(n), Space: O(n) - Amazon Context: System reconstruction, data restoration - Manager Insight: Tree construction from traversals - Similar: Construct Binary Tree from Inorder and Postorder

61. Serialize and Deserialize Binary Tree (LeetCode #297)

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def serialize(root):
    """Serialize binary tree to string."""
    def preorder(node):
        if not node:
            vals.append("#")
        else:
            vals.append(str(node.val))
            preorder(node.left)
            preorder(node.right)

    vals = []
    preorder(root)
    return ','.join(vals)

def deserialize(data):
    """Deserialize string to binary tree."""
    def build_tree():
        val = next(vals)
        if val == "#":
            return None
        node = TreeNode(int(val))
        node.left = build_tree()
        node.right = build_tree()
        return node

    vals = iter(data.split(','))
    return build_tree()

# Example: [1,2,3,null,null,4,5] → "1,2,#,#,3,4,#,#,5,#,#"
- Time: O(n), Space: O(n) - Amazon Context: Data persistence, caching strategies - Manager Insight: Tree serialization protocols - Similar: Serialize and Deserialize BST, Encode and Decode Strings

62. Number of Islands (LeetCode #200)

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def num_islands(grid):
    """Count number of islands in grid."""
    if not grid or not grid[0]:
        return 0

    rows, cols = len(grid), len(grid[0])
    islands = 0

    def dfs(r, c):
        if (r < 0 or r >= rows or c < 0 or c >= cols or 
            grid[r][c] != '1'):
            return

        grid[r][c] = '0'  # Mark as visited

        # Visit all adjacent cells
        dfs(r-1, c)
        dfs(r+1, c)
        dfs(r, c-1)
        dfs(r, c+1)

    for r in range(rows):
        for c in range(cols):
            if grid[r][c] == '1':
                islands += 1
                dfs(r, c)

    return islands

# Example: grid = [["1","1","0"],["1","1","0"],["0","0","1"]] → 2
- Time: O(m*n), Space: O(m*n) - Amazon Context: Network analysis, cluster detection - Manager Insight: Connected components in graphs - Similar: Number of Connected Components, Max Area of Island

63. Clone Graph (LeetCode #133)

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def clone_graph(node):
    """Clone an undirected graph."""
    if not node:
        return None

    visited = {}

    def dfs(node):
        if node in visited:
            return visited[node]

        clone = Node(node.val)
        visited[node] = clone

        for neighbor in node.neighbors:
            clone.neighbors.append(dfs(neighbor))

        return clone

    return dfs(node)

# BFS approach:
def clone_graph_bfs(node):
    if not node:
        return None

    from collections import deque
    visited = {node: Node(node.val)}
    queue = deque([node])

    while queue:
        current = queue.popleft()

        for neighbor in current.neighbors:
            if neighbor not in visited:
                visited[neighbor] = Node(neighbor.val)
                queue.append(neighbor)

            visited[current].neighbors.append(visited[neighbor])

    return visited[node]

# Example: adjList = [[2,4],[1,3],[2,4],[1,3]] → cloned graph
- Time: O(V + E), Space: O(V) - Amazon Context: System replication, backup creation - Manager Insight: Graph traversal and cloning - Similar: Copy List with Random Pointer, Clone N-ary Tree

Hard Difficulty (2 problems)

64. Binary Tree Maximum Path Sum (Already covered above)

65. Word Ladder (LeetCode #127)

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def ladder_length(begin_word, end_word, word_list):
    """Find length of shortest transformation sequence."""
    if end_word not in word_list:
        return 0

    from collections import deque, defaultdict

    # Build adjacency list
    neighbors = defaultdict(list)
    word_list.append(begin_word)

    for word in word_list:
        for i in range(len(word)):
            pattern = word[:i] + "*" + word[i+1:]
            neighbors[pattern].append(word)

    # BFS
    visited = {begin_word}
    queue = deque([begin_word])
    level = 1

    while queue:
        for _ in range(len(queue)):
            word = queue.popleft()

            if word == end_word:
                return level

            for i in range(len(word)):
                pattern = word[:i] + "*" + word[i+1:]

                for neighbor in neighbors[pattern]:
                    if neighbor not in visited:
                        visited.add(neighbor)
                        queue.append(neighbor)

        level += 1

    return 0

# Example: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"] → 5
- Time: O(M² × N), Space: O(M² × N) - Amazon Context: System migration paths, dependency resolution - Manager Insight: Shortest path in unweighted graphs - Similar: Word Ladder II, Minimum Genetic Mutation

Category 6: Dynamic Programming (10 Problems)

Easy Difficulty (3 problems)

66. Climbing Stairs (LeetCode #70)

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def climb_stairs(n):
    """Number of ways to climb n stairs."""
    if n <= 2:
        return n

    prev2, prev1 = 1, 2
    for i in range(3, n + 1):
        current = prev1 + prev2
        prev2, prev1 = prev1, current

    return prev1

# Example: n = 3 → 3 (1+1+1, 1+2, 2+1)
- Time: O(n), Space: O(1) - Amazon Context: Resource scaling, capacity planning - Manager Insight: Basic DP optimization from recursion - Similar: Fibonacci Number, Min Cost Climbing Stairs

67. House Robber (LeetCode #198)

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def rob(nums):
    """Maximum money that can be robbed."""
    if not nums:
        return 0
    if len(nums) == 1:
        return nums[0]

    prev2, prev1 = nums[0], max(nums[0], nums[1])

    for i in range(2, len(nums)):
        current = max(prev1, prev2 + nums[i])
        prev2, prev1 = prev1, current

    return prev1

# Example: [2,7,9,3,1] → 12 (2+9+1)
- Time: O(n), Space: O(1) - Amazon Context: Resource optimization with constraints - Manager Insight: State-based decision making - Similar: House Robber II, House Robber III

68. Maximum Subarray (LeetCode #53)

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def max_subarray(nums):
    """Find maximum sum of contiguous subarray (Kadane's Algorithm)."""
    max_sum = current_sum = nums[0]

    for num in nums[1:]:
        current_sum = max(num, current_sum + num)
        max_sum = max(max_sum, current_sum)

    return max_sum

# Example: [-2,1,-3,4,-1,2,1,-5,4] → 6
- Time: O(n), Space: O(1) - Amazon Context: Performance metrics analysis - Manager Insight: Optimal substructure principle - Similar: Maximum Product Subarray, Maximum Sum Circular Subarray

Medium Difficulty (6 problems)

69. Coin Change (LeetCode #322)

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def coin_change(coins, amount):
    """Minimum coins to make amount."""
    dp = [float('inf')] * (amount + 1)
    dp[0] = 0

    for i in range(1, amount + 1):
        for coin in coins:
            if coin <= i:
                dp[i] = min(dp[i], dp[i - coin] + 1)

    return dp[amount] if dp[amount] != float('inf') else -1

# Example: coins = [1,3,4], amount = 6 → 2 (3+3)
- Time: O(amount × coins), Space: O(amount) - Amazon Context: Cost optimization, resource allocation - Manager Insight: Unbounded knapsack variant - Similar: Coin Change 2, Combination Sum IV

70. Longest Increasing Subsequence (LeetCode #300)

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def length_of_lis(nums):
    """Length of longest increasing subsequence."""
    if not nums:
        return 0

    dp = [1] * len(nums)

    for i in range(1, len(nums)):
        for j in range(i):
            if nums[j] < nums[i]:
                dp[i] = max(dp[i], dp[j] + 1)

    return max(dp)

# O(n log n) solution using binary search:
def length_of_lis_optimized(nums):
    import bisect
    tails = []

    for num in nums:
        pos = bisect.bisect_left(tails, num)
        if pos == len(tails):
            tails.append(num)
        else:
            tails[pos] = num

    return len(tails)

# Example: [10,9,2,5,3,7,101,18] → 4
- Time: O(n²) or O(n log n), Space: O(n) - Amazon Context: Growth analysis, trend detection - Manager Insight: Patience sorting algorithm - Similar: Russian Doll Envelopes, Number of LIS

71. Word Break (LeetCode #139)

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def word_break(s, word_dict):
    """Check if string can be segmented using dictionary."""
    dp = [False] * (len(s) + 1)
    dp[0] = True
    word_set = set(word_dict)

    for i in range(1, len(s) + 1):
        for j in range(i):
            if dp[j] and s[j:i] in word_set:
                dp[i] = True
                break

    return dp[len(s)]

# Example: s = "leetcode", wordDict = ["leet","code"] → True
- Time: O(n³), Space: O(n) - Amazon Context: Text processing, parsing validation - Manager Insight: Substring validation with memoization - Similar: Word Break II, Concatenated Words

72. Unique Paths (LeetCode #62)

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def unique_paths(m, n):
    """Number of unique paths in grid."""
    dp = [[1] * n for _ in range(m)]

    for i in range(1, m):
        for j in range(1, n):
            dp[i][j] = dp[i-1][j] + dp[i][j-1]

    return dp[m-1][n-1]

# Space optimized version:
def unique_paths_optimized(m, n):
    dp = [1] * n

    for i in range(1, m):
        for j in range(1, n):
            dp[j] += dp[j-1]

    return dp[n-1]

# Example: m = 3, n = 7 → 28
- Time: O(m*n), Space: O(m*n) or O(n) - Amazon Context: Path planning, routing optimization - Manager Insight: Grid-based DP patterns - Similar: Unique Paths II, Minimum Path Sum

73. Decode Ways (LeetCode #91)

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def num_decodings(s):
    """Number of ways to decode string."""
    if not s or s[0] == '0':
        return 0

    prev2, prev1 = 1, 1

    for i in range(1, len(s)):
        current = 0

        # Single digit
        if s[i] != '0':
            current += prev1

        # Two digits
        two_digit = int(s[i-1:i+1])
        if 10 <= two_digit <= 26:
            current += prev2

        prev2, prev1 = prev1, current

    return prev1

# Example: "12" → 2 ("AB" or "L"), "226" → 3
- Time: O(n), Space: O(1) - Amazon Context: Encoding/decoding systems, format validation - Manager Insight: Multiple choice DP optimization - Similar: Decode Ways II, Number of Ways to Stay in Same Place

74. Edit Distance (LeetCode #72)

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def min_distance(word1, word2):
    """Minimum operations to convert word1 to word2."""
    m, n = len(word1), len(word2)
    dp = [[0] * (n + 1) for _ in range(m + 1)]

    # Initialize base cases
    for i in range(m + 1):
        dp[i][0] = i
    for j in range(n + 1):
        dp[0][j] = j

    # Fill DP table
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if word1[i-1] == word2[j-1]:
                dp[i][j] = dp[i-1][j-1]
            else:
                dp[i][j] = 1 + min(
                    dp[i-1][j],    # Delete
                    dp[i][j-1],    # Insert
                    dp[i-1][j-1]   # Replace
                )

    return dp[m][n]

# Example: word1 = "horse", word2 = "ros" → 3
- Time: O(m*n), Space: O(m*n) - Amazon Context: Spell check, fuzzy matching systems - Manager Insight: String distance algorithms - Similar: One Edit Distance, Delete Operation for Two Strings

Hard Difficulty (1 problem)

75. Regular Expression Matching (LeetCode #10)

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def is_match(s, p):
    """Check if string matches pattern with . and *."""
    m, n = len(s), len(p)
    dp = [[False] * (n + 1) for _ in range(m + 1)]

    # Base case
    dp[0][0] = True

    # Handle patterns like a*, a*b*, a*b*c*
    for j in range(2, n + 1):
        if p[j-1] == '*':
            dp[0][j] = dp[0][j-2]

    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if p[j-1] == '*':
                # Zero occurrences
                dp[i][j] = dp[i][j-2]

                # One or more occurrences
                if p[j-2] == '.' or p[j-2] == s[i-1]:
                    dp[i][j] |= dp[i-1][j]
            else:
                # Exact match or wildcard
                if p[j-1] == '.' or p[j-1] == s[i-1]:
                    dp[i][j] = dp[i-1][j-1]

    return dp[m][n]

# Example: s = "aa", p = "a*" → True
- Time: O(m*n), Space: O(m*n) - Amazon Context: Pattern matching, regex validation - Manager Insight: Complex state transitions in DP - Similar: Wildcard Matching, Expression Add Operators

Category 7: Stack & Queue (10 Problems)

Easy Difficulty (4 problems)

76. Valid Parentheses (LeetCode #20)

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def is_valid(s):
    """Check if parentheses are properly matched."""
    stack = []
    mapping = {')': '(', '}': '{', ']': '['}

    for char in s:
        if char in mapping:
            if not stack or stack.pop() != mapping[char]:
                return False
        else:
            stack.append(char)

    return not stack

# Example: "()[]{}" → True, "([)]" → False
- Time: O(n), Space: O(n) - Amazon Context: Configuration validation, bracket matching - Manager Insight: Stack for LIFO matching problems - Similar: Generate Parentheses, Remove Invalid Parentheses

77. Implement Queue using Stacks (LeetCode #232)

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class MyQueue:
    """Queue implementation using two stacks."""

    def __init__(self):
        self.input_stack = []
        self.output_stack = []

    def push(self, x):
        """Push element to back of queue."""
        self.input_stack.append(x)

    def pop(self):
        """Remove element from front of queue."""
        self.peek()
        return self.output_stack.pop()

    def peek(self):
        """Get front element."""
        if not self.output_stack:
            while self.input_stack:
                self.output_stack.append(self.input_stack.pop())
        return self.output_stack[-1]

    def empty(self):
        """Check if queue is empty."""
        return not self.input_stack and not self.output_stack

# Example: queue.push(1); queue.push(2); queue.peek() → 1; queue.pop() → 1
- Time: O(1) amortized, Space: O(n) - Amazon Context: Data structure design, system interfaces - Manager Insight: Amortized analysis importance - Similar: Implement Stack using Queues, Design Circular Queue

78. Min Stack (LeetCode #155)

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class MinStack:
    """Stack with constant time minimum element access."""

    def __init__(self):
        self.stack = []
        self.min_stack = []

    def push(self, val):
        """Push element onto stack."""
        self.stack.append(val)
        if not self.min_stack or val <= self.min_stack[-1]:
            self.min_stack.append(val)

    def pop(self):
        """Remove top element."""
        if self.stack:
            val = self.stack.pop()
            if val == self.min_stack[-1]:
                self.min_stack.pop()

    def top(self):
        """Get top element."""
        return self.stack[-1] if self.stack else None

    def get_min(self):
        """Get minimum element."""
        return self.min_stack[-1] if self.min_stack else None

# Example: minStack.push(-2); minStack.push(0); minStack.getMin() → -2
- Time: O(1) all operations, Space: O(n) - Amazon Context: Monitoring systems, metric tracking - Manager Insight: Auxiliary data structures for optimization - Similar: Max Stack, Design a Stack With Increment Operation

79. Backspace String Compare (LeetCode #844)

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def backspace_compare(s, t):
    """Compare strings after backspace operations."""
    def build_string(string):
        stack = []
        for char in string:
            if char != '#':
                stack.append(char)
            elif stack:
                stack.pop()
        return ''.join(stack)

    return build_string(s) == build_string(t)

# Space optimized two-pointer approach:
def backspace_compare_optimized(s, t):
    def get_next_char(string, index):
        backspace_count = 0
        while index >= 0:
            if string[index] == '#':
                backspace_count += 1
            elif backspace_count > 0:
                backspace_count -= 1
            else:
                break
            index -= 1
        return index

    i, j = len(s) - 1, len(t) - 1

    while i >= 0 or j >= 0:
        i = get_next_char(s, i)
        j = get_next_char(t, j)

        if i < 0 and j < 0:
            return True
        if i < 0 or j < 0:
            return False
        if s[i] != t[j]:
            return False

        i -= 1
        j -= 1

    return True

# Example: s = "ab#c", t = "ad#c" → True
- Time: O(n), Space: O(n) or O(1) - Amazon Context: Text processing, undo operations - Manager Insight: Stack simulation vs space optimization - Similar: Remove All Adjacent Duplicates, Asteroid Collision

Medium Difficulty (5 problems)

80. Daily Temperatures (LeetCode #739)

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def daily_temperatures(temperatures):
    """Days until warmer temperature for each day."""
    result = [0] * len(temperatures)
    stack = []  # Store indices

    for i, temp in enumerate(temperatures):
        while stack and temperatures[stack[-1]] < temp:
            prev_index = stack.pop()
            result[prev_index] = i - prev_index
        stack.append(i)

    return result

# Example: [73,74,75,71,69,72,76,73] → [1,1,4,2,1,1,0,0]
- Time: O(n), Space: O(n) - Amazon Context: Time series analysis, next greater element - Manager Insight: Monotonic stack applications - Similar: Next Greater Element, Next Greater Element II

81. Evaluate Reverse Polish Notation (LeetCode #150)

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def eval_rpn(tokens):
    """Evaluate expression in Reverse Polish Notation."""
    stack = []

    for token in tokens:
        if token in ['+', '-', '*', '/']:
            b = stack.pop()
            a = stack.pop()

            if token == '+':
                stack.append(a + b)
            elif token == '-':
                stack.append(a - b)
            elif token == '*':
                stack.append(a * b)
            else:  # token == '/'
                # Truncate towards zero
                stack.append(int(a / b))
        else:
            stack.append(int(token))

    return stack[0]

# Example: ["2","1","+","3","*"] → 9 ((2+1)*3)
- Time: O(n), Space: O(n) - Amazon Context: Expression evaluation, calculator implementation - Manager Insight: Stack-based expression parsing - Similar: Basic Calculator, Basic Calculator II

82. Decode String (LeetCode #394)

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def decode_string(s):
    """Decode string with nested brackets."""
    stack = []
    current_string = ""
    current_num = 0

    for char in s:
        if char.isdigit():
            current_num = current_num * 10 + int(char)
        elif char == '[':
            stack.append((current_string, current_num))
            current_string = ""
            current_num = 0
        elif char == ']':
            prev_string, num = stack.pop()
            current_string = prev_string + current_string * num
        else:
            current_string += char

    return current_string

# Example: "3[a2[c]]" → "accaccacc"
- Time: O(n), Space: O(n) - Amazon Context: String processing, nested structure parsing - Manager Insight: Stack for nested operations - Similar: Number of Atoms, Parse Lisp Expression

83. Asteroid Collision (LeetCode #735)

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def asteroid_collision(asteroids):
    """Simulate asteroid collisions."""
    stack = []

    for asteroid in asteroids:
        while stack and asteroid < 0 < stack[-1]:
            if stack[-1] < -asteroid:
                stack.pop()
                continue
            elif stack[-1] == -asteroid:
                stack.pop()
            break
        else:
            stack.append(asteroid)

    return stack

# Example: [5,10,-5] → [5,10], [8,-8] → [], [10,2,-5] → [10]
- Time: O(n), Space: O(n) - Amazon Context: Simulation, collision detection - Manager Insight: State machine with stack - Similar: Car Fleet, Remove All Adjacent Duplicates

84. Sliding Window Maximum (LeetCode #239)

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def max_sliding_window(nums, k):
    """Maximum in each sliding window of size k."""
    from collections import deque

    dq = deque()
    result = []

    for i, num in enumerate(nums):
        # Remove elements outside window
        while dq and dq[0] < i - k + 1:
            dq.popleft()

        # Remove smaller elements
        while dq and nums[dq[-1]] < num:
            dq.pop()

        dq.append(i)

        # Add to result if window is complete
        if i >= k - 1:
            result.append(nums[dq[0]])

    return result

# Example: nums = [1,3,-1,-3,5,3,6,7], k = 3 → [3,3,5,5,6,7]
- Time: O(n), Space: O(k) - Amazon Context: Real-time monitoring, window-based analytics - Manager Insight: Monotonic deque for range queries - Similar: Sliding Window Minimum, Constrained Subsequence Sum

Hard Difficulty (1 problem)

85. Largest Rectangle in Histogram (LeetCode #84)

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def largest_rectangle_area(heights):
    """Find largest rectangle area in histogram."""
    stack = []
    max_area = 0

    for i, height in enumerate(heights):
        while stack and heights[stack[-1]] > height:
            h = heights[stack.pop()]
            w = i if not stack else i - stack[-1] - 1
            max_area = max(max_area, h * w)
        stack.append(i)

    while stack:
        h = heights[stack.pop()]
        w = len(heights) if not stack else len(heights) - stack[-1] - 1
        max_area = max(max_area, h * w)

    return max_area

# Example: [2,1,5,6,2,3] → 10
- Time: O(n), Space: O(n) - Amazon Context: Resource utilization, capacity optimization - Manager Insight: Stack for range optimization problems - Similar: Maximal Rectangle, Remove K Digits

Category 8: Binary Search (5 Problems)

Easy Difficulty (2 problems)

86. Binary Search (LeetCode #704)

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def search(nums, target):
    """Binary search in sorted array."""
    left, right = 0, len(nums) - 1

    while left <= right:
        mid = (left + right) // 2

        if nums[mid] == target:
            return mid
        elif nums[mid] < target:
            left = mid + 1
        else:
            right = mid - 1

    return -1

# Example: nums = [-1,0,3,5,9,12], target = 9 → 4
- Time: O(log n), Space: O(1) - Amazon Context: Efficient lookups, data retrieval - Manager Insight: Logarithmic search fundamentals - Similar: Search Insert Position, Find Peak Element

87. First Bad Version (LeetCode #278)

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def first_bad_version(n):
    """Find first bad version using isBadVersion API."""
    left, right = 1, n

    while left < right:
        mid = (left + right) // 2

        if isBadVersion(mid):
            right = mid
        else:
            left = mid + 1

    return left

# Example: n = 5, bad = 4 → 4
- Time: O(log n), Space: O(1) - Amazon Context: Bug detection, version control - Manager Insight: API-based binary search - Similar: Search Insert Position, Find Peak Element

Medium Difficulty (2 problems)

88. Search in Rotated Sorted Array (LeetCode #33)

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def search_rotated(nums, target):
    """Search in rotated sorted array."""
    left, right = 0, len(nums) - 1

    while left <= right:
        mid = (left + right) // 2

        if nums[mid] == target:
            return mid

        # Left half is sorted
        if nums[left] <= nums[mid]:
            if nums[left] <= target < nums[mid]:
                right = mid - 1
            else:
                left = mid + 1
        # Right half is sorted
        else:
            if nums[mid] < target <= nums[right]:
                left = mid + 1
            else:
                right = mid - 1

    return -1

# Example: nums = [4,5,6,7,0,1,2], target = 0 → 4
- Time: O(log n), Space: O(1) - Amazon Context: Distributed system searches, circular buffers - Manager Insight: Modified binary search for rotated data - Similar: Search in Rotated Sorted Array II, Find Minimum in Rotated Sorted Array

89. Find Peak Element (LeetCode #162)

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def find_peak_element(nums):
    """Find peak element in array."""
    left, right = 0, len(nums) - 1

    while left < right:
        mid = (left + right) // 2

        if nums[mid] > nums[mid + 1]:
            right = mid
        else:
            left = mid + 1

    return left

# Example: nums = [1,2,3,1] → 2
- Time: O(log n), Space: O(1) - Amazon Context: Performance analysis, local maxima detection - Manager Insight: Binary search on unimodal functions - Similar: Peak Index in Mountain Array, Find Peak Element II

Hard Difficulty (1 problem)

90. Median of Two Sorted Arrays (LeetCode #4)

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def find_median_sorted_arrays(nums1, nums2):
    """Find median of two sorted arrays."""
    if len(nums1) > len(nums2):
        nums1, nums2 = nums2, nums1

    m, n = len(nums1), len(nums2)
    left, right = 0, m

    while left <= right:
        partition_x = (left + right) // 2
        partition_y = (m + n + 1) // 2 - partition_x

        max_left_x = float('-inf') if partition_x == 0 else nums1[partition_x - 1]
        min_right_x = float('inf') if partition_x == m else nums1[partition_x]

        max_left_y = float('-inf') if partition_y == 0 else nums2[partition_y - 1]
        min_right_y = float('inf') if partition_y == n else nums2[partition_y]

        if max_left_x <= min_right_y and max_left_y <= min_right_x:
            if (m + n) % 2 == 0:
                return (max(max_left_x, max_left_y) + min(min_right_x, min_right_y)) / 2
            else:
                return max(max_left_x, max_left_y)
        elif max_left_x > min_right_y:
            right = partition_x - 1
        else:
            left = partition_x + 1

# Example: nums1 = [1,3], nums2 = [2] → 2.0
- Time: O(log(min(m,n))), Space: O(1) - Amazon Context: Data analysis, statistical computation - Manager Insight: Binary search on answer space - Similar: Kth Smallest Element in Sorted Matrix

Category 9: Design Problems (5 Problems)

Medium Difficulty (4 problems)

91. LRU Cache (LeetCode #146)

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class LRUCache:
    """Least Recently Used Cache implementation."""

    class Node:
        def __init__(self, key=0, val=0):
            self.key = key
            self.val = val
            self.prev = None
            self.next = None

    def __init__(self, capacity):
        self.capacity = capacity
        self.cache = {}

        # Create dummy head and tail
        self.head = self.Node()
        self.tail = self.Node()
        self.head.next = self.tail
        self.tail.prev = self.head

    def _add_node(self, node):
        """Add node right after head."""
        node.prev = self.head
        node.next = self.head.next

        self.head.next.prev = node
        self.head.next = node

    def _remove_node(self, node):
        """Remove an existing node."""
        prev_node = node.prev
        next_node = node.next

        prev_node.next = next_node
        next_node.prev = prev_node

    def _move_to_head(self, node):
        """Move node to head."""
        self._remove_node(node)
        self._add_node(node)

    def _pop_tail(self):
        """Remove last node."""
        last_node = self.tail.prev
        self._remove_node(last_node)
        return last_node

    def get(self, key):
        """Get value and mark as recently used."""
        node = self.cache.get(key)

        if not node:
            return -1

        self._move_to_head(node)
        return node.val

    def put(self, key, value):
        """Put key-value pair."""
        node = self.cache.get(key)

        if not node:
            new_node = self.Node(key, value)

            self.cache[key] = new_node
            self._add_node(new_node)

            if len(self.cache) > self.capacity:
                tail = self._pop_tail()
                del self.cache[tail.key]
        else:
            node.val = value
            self._move_to_head(node)

# Example: cache = LRUCache(2); cache.put(1,1); cache.get(1) → 1
- Time: O(1) for both get and put, Space: O(capacity) - Amazon Context: Caching systems, memory management - Manager Insight: Combining hash map with doubly linked list - Similar: LFU Cache, Design Twitter

92. Design Add and Search Words Data Structure (LeetCode #211)

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class WordDictionary:
    """Add and search words with wildcard support."""

    class TrieNode:
        def __init__(self):
            self.children = {}
            self.is_word = False

    def __init__(self):
        self.root = self.TrieNode()

    def add_word(self, word):
        """Add word to data structure."""
        node = self.root
        for char in word:
            if char not in node.children:
                node.children[char] = self.TrieNode()
            node = node.children[char]
        node.is_word = True

    def search(self, word):
        """Search word with wildcard '.' support."""
        def dfs(node, index):
            if index == len(word):
                return node.is_word

            char = word[index]
            if char == '.':
                for child in node.children.values():
                    if dfs(child, index + 1):
                        return True
                return False
            else:
                if char not in node.children:
                    return False
                return dfs(node.children[char], index + 1)

        return dfs(self.root, 0)

# Example: wd.addWord("bad"); wd.search("pad") → False; wd.search(".ad") → True
- Time: O(m) for add, O(m*26^k) for search worst case, Space: O(ALPHABET_SIZE * N * M) - Amazon Context: Search systems, pattern matching - Manager Insight: Trie with backtracking for wildcards - Similar: Implement Trie, Word Search II

93. Design Twitter (LeetCode #355)

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class Twitter:
    """Simple Twitter-like social media system."""

    def __init__(self):
        self.timestamp = 0
        self.user_tweets = {}  # user_id -> list of (timestamp, tweet_id)
        self.user_follows = {}  # user_id -> set of followed user_ids

    def post_tweet(self, user_id, tweet_id):
        """User posts a tweet."""
        if user_id not in self.user_tweets:
            self.user_tweets[user_id] = []

        self.user_tweets[user_id].append((self.timestamp, tweet_id))
        self.timestamp += 1

    def get_news_feed(self, user_id):
        """Get 10 most recent tweets from user and followees."""
        import heapq

        # Collect all relevant tweets
        tweets = []

        # User's own tweets
        if user_id in self.user_tweets:
            tweets.extend(self.user_tweets[user_id])

        # Followees' tweets
        if user_id in self.user_follows:
            for followee_id in self.user_follows[user_id]:
                if followee_id in self.user_tweets:
                    tweets.extend(self.user_tweets[followee_id])

        # Get 10 most recent
        tweets.sort(reverse=True)  # Sort by timestamp descending
        return [tweet_id for _, tweet_id in tweets[:10]]

    def follow(self, follower_id, followee_id):
        """Follower follows followee."""
        if follower_id not in self.user_follows:
            self.user_follows[follower_id] = set()
        self.user_follows[follower_id].add(followee_id)

    def unfollow(self, follower_id, followee_id):
        """Follower unfollows followee."""
        if (follower_id in self.user_follows and 
            followee_id in self.user_follows[follower_id]):
            self.user_follows[follower_id].remove(followee_id)

# Example: twitter.postTweet(1, 5); twitter.getNewsFeed(1) → [5]
- Time: O(n log n) for news feed, O(1) for others, Space: O(n) - Amazon Context: Social media systems, feed algorithms - Manager Insight: System design with multiple operations - Similar: Design Chat System, Design Instagram

94. Insert Delete GetRandom O(1) (LeetCode #380)

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class RandomizedSet:
    """Data structure with O(1) insert, delete, and getRandom."""

    def __init__(self):
        self.vals = []
        self.val_to_index = {}

    def insert(self, val):
        """Insert value, return False if already exists."""
        if val in self.val_to_index:
            return False

        self.vals.append(val)
        self.val_to_index[val] = len(self.vals) - 1
        return True

    def remove(self, val):
        """Remove value, return False if doesn't exist."""
        if val not in self.val_to_index:
            return False

        # Move last element to position of element to delete
        last_val = self.vals[-1]
        index_to_remove = self.val_to_index[val]

        self.vals[index_to_remove] = last_val
        self.val_to_index[last_val] = index_to_remove

        # Remove last element
        self.vals.pop()
        del self.val_to_index[val]

        return True

    def get_random(self):
        """Get random element."""
        import random
        return random.choice(self.vals)

# Example: rs.insert(1) → True; rs.remove(2) → False; rs.getRandom() → 1
- Time: O(1) for all operations, Space: O(n) - Amazon Context: Random sampling, A/B testing systems - Manager Insight: Array and hash map combination for O(1) operations - Similar: Insert Delete GetRandom O(1) - Duplicates allowed

Hard Difficulty (1 problem)

95. Design In-Memory File System (LeetCode #588)

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class FileSystem:
    """In-memory file system with directories and files."""

    class Node:
        def __init__(self):
            self.is_file = False
            self.content = ""
            self.children = {}

    def __init__(self):
        self.root = self.Node()

    def _get_node(self, path):
        """Get node at given path."""
        node = self.root
        if path == "/":
            return node

        parts = path.split("/")[1:]  # Skip empty first part
        for part in parts:
            if part not in node.children:
                return None
            node = node.children[part]
        return node

    def _create_path(self, path):
        """Create path if it doesn't exist."""
        node = self.root
        if path == "/":
            return node

        parts = path.split("/")[1:]
        for part in parts:
            if part not in node.children:
                node.children[part] = self.Node()
            node = node.children[part]
        return node

    def ls(self, path):
        """List contents of directory or return file name."""
        node = self._get_node(path)
        if not node:
            return []

        if node.is_file:
            return [path.split("/")[-1]]

        return sorted(node.children.keys())

    def mkdir(self, path):
        """Create directory."""
        self._create_path(path)

    def add_content_to_file(self, file_path, content):
        """Add content to file."""
        node = self._create_path(file_path)
        node.is_file = True
        node.content += content

    def read_content_from_file(self, file_path):
        """Read content from file."""
        node = self._get_node(file_path)
        return node.content if node and node.is_file else ""

# Example: fs.mkdir("/a/b/c"); fs.addContentToFile("/a/b/c/d", "hello"); fs.readContentFromFile("/a/b/c/d") → "hello"
- Time: O(m) where m is path length, Space: O(total content) - Amazon Context: File storage systems, directory management - Manager Insight: Tree-based file system representation - Similar: Design File System, Design Log Storage System

Category 10: Mathematical (5 Problems)

Easy Difficulty (2 problems)

96. Palindrome Number (LeetCode #9)

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def is_palindrome(x):
    """Check if integer is palindrome."""
    if x < 0:
        return False

    original = x
    reversed_num = 0

    while x > 0:
        reversed_num = reversed_num * 10 + x % 10
        x //= 10

    return original == reversed_num

# Alternative without converting to string:
def is_palindrome_half(x):
    if x < 0 or (x % 10 == 0 and x != 0):
        return False

    reversed_half = 0
    while x > reversed_half:
        reversed_half = reversed_half * 10 + x % 10
        x //= 10

    return x == reversed_half or x == reversed_half // 10

# Example: 121 → True, -121 → False
- Time: O(log n), Space: O(1) - Amazon Context: Data validation, number processing - Manager Insight: Mathematical manipulation vs string conversion - Similar: Reverse Integer, Valid Perfect Square

97. Fizz Buzz (LeetCode #412)

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def fizz_buzz(n):
    """Return fizz buzz sequence from 1 to n."""
    result = []

    for i in range(1, n + 1):
        if i % 15 == 0:
            result.append("FizzBuzz")
        elif i % 3 == 0:
            result.append("Fizz")
        elif i % 5 == 0:
            result.append("Buzz")
        else:
            result.append(str(i))

    return result

# More extensible approach:
def fizz_buzz_extensible(n):
    result = []

    for i in range(1, n + 1):
        output = ""
        if i % 3 == 0:
            output += "Fizz"
        if i % 5 == 0:
            output += "Buzz"

        result.append(output or str(i))

    return result

# Example: n = 15 → ["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14","FizzBuzz"]
- Time: O(n), Space: O(n) - Amazon Context: Interview warm-up, basic logic - Manager Insight: Code clarity and extensibility - Similar: Count and Say, Print Zero Even Odd

Medium Difficulty (2 problems)

98. Pow(x, n) (LeetCode #50)

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def my_pow(x, n):
    """Calculate x raised to power n."""
    if n < 0:
        x = 1 / x
        n = -n

    result = 1
    current_power = x

    while n > 0:
        if n % 2 == 1:
            result *= current_power
        current_power *= current_power
        n //= 2

    return result

# Recursive approach:
def my_pow_recursive(x, n):
    def helper(x, n):
        if n == 0:
            return 1

        half = helper(x, n // 2)
        if n % 2 == 0:
            return half * half
        else:
            return half * half * x

    if n < 0:
        return 1 / helper(x, -n)
    return helper(x, n)

# Example: x = 2.0, n = 10 → 1024.0
- Time: O(log n), Space: O(1) or O(log n) for recursive - Amazon Context: Mathematical computations, algorithms optimization - Manager Insight: Exponentiation by squaring - Similar: Super Pow, Sqrt(x)

99. Sqrt(x) (LeetCode #69)

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def my_sqrt(x):
    """Find integer square root."""
    if x < 2:
        return x

    left, right = 1, x // 2

    while left <= right:
        mid = (left + right) // 2
        square = mid * mid

        if square == x:
            return mid
        elif square < x:
            left = mid + 1
        else:
            right = mid - 1

    return right  # Return floor of square root

# Newton's method approach:
def my_sqrt_newton(x):
    if x < 2:
        return x

    x0 = x
    x1 = (x0 + x / x0) / 2

    while abs(x0 - x1) >= 1:
        x0 = x1
        x1 = (x0 + x / x0) / 2

    return int(x1)

# Example: x = 8 → 2
- Time: O(log n), Space: O(1) - Amazon Context: Mathematical computations, approximation algorithms - Manager Insight: Binary search vs Newton's method - Similar: Valid Perfect Square, Pow(x, n)

Hard Difficulty (1 problem)

100. Basic Calculator (LeetCode #224)

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def calculate(s):
    """Evaluate basic mathematical expression with parentheses."""
    stack = []
    result = 0
    number = 0
    sign = 1

    for char in s:
        if char.isdigit():
            number = number * 10 + int(char)
        elif char == '+':
            result += sign * number
            number = 0
            sign = 1
        elif char == '-':
            result += sign * number
            number = 0
            sign = -1
        elif char == '(':
            # Push current result and sign to stack
            stack.append(result)
            stack.append(sign)
            # Reset for new sub-expression
            result = 0
            sign = 1
        elif char == ')':
            # Complete current number
            result += sign * number
            number = 0

            # Pop sign and previous result
            result *= stack.pop()  # sign before parentheses
            result += stack.pop()  # result before parentheses

    return result + sign * number

# Example: " 2-1 + 2 " → 3, "(1+(4+5+2)-3)+(6+8)" → 23
- Time: O(n), Space: O(n) - Amazon Context: Expression evaluation, formula processing - Manager Insight: Stack-based parsing with state management - Similar: Basic Calculator II, Basic Calculator III

Interview Guidelines for Engineering Managers

L6 Level Expectations

Technical Competency: - Solve Easy and Medium problems efficiently - Explain time/space complexity clearly - Handle edge cases systematically - Write clean, readable code

Communication Skills: - Start with clarifying questions - Think out loud during problem solving - Explain trade-offs between solutions - Connect algorithms to business context

Leadership Context: - Discuss how you'd guide team members - Explain architectural considerations - Consider operational implications - Address scalability concerns

L7 Level Expectations

Advanced Problem Solving: - Handle Hard problems with multiple approaches - Optimize for both time and space - Design scalable solutions - Consider system-wide implications

Technical Leadership: - Explain complex algorithms simply - Discuss performance at scale - Connect to system design concepts - Evaluate different architectural approaches

Management Perspective: - How would you mentor junior engineers? - Code review and quality standards - Technical decision making process - Risk assessment and mitigation

Time Management Strategy

Problem Difficulty Time Allocation Focus Areas
Easy 15-20 minutes Clean implementation, edge cases
Medium 25-35 minutes Optimization, multiple approaches
Hard 35-45 minutes Complex logic, scalability discussion

Common Follow-up Questions

  1. Optimization: "Can you improve the complexity?"
  2. Scale: "How would this work with billions of records?"
  3. Variations: "What if constraints changed?"
  4. Production: "How would you deploy this?"
  5. Testing: "How would you validate this solution?"
  6. Maintenance: "How would you make this more robust?"

Practice Schedule for Managers

Phase 1: Foundation (Weeks 1-2)

  • Goal: Refresh algorithmic thinking
  • Focus: Easy problems (30+ problems)
  • Time: 1-2 hours daily
  • Emphasis: Pattern recognition, clean code

Phase 2: Skill Building (Weeks 3-4)

  • Goal: Master core patterns
  • Focus: Medium problems (50+ problems)
  • Time: 1.5-2 hours daily
  • Emphasis: Optimization, multiple solutions

Phase 3: Advanced Preparation (Weeks 5-6)

  • Goal: Handle complex scenarios
  • Focus: Hard problems + system design
  • Time: 2+ hours daily
  • Emphasis: Leadership explanations, scalability

Phase 4: Mock Interviews (Weeks 7-8)

  • Goal: Interview simulation
  • Focus: Timed practice, communication
  • Time: 2+ hours daily
  • Emphasis: Explaining thought process, handling pressure

Tips for Managers Returning to Coding

  1. Start Simple: Begin with Easy problems to build confidence
  2. Pattern Recognition: Focus on common patterns rather than memorizing solutions
  3. Business Context: Always connect problems to real-world scenarios
  4. Communication: Practice explaining complex concepts simply
  5. Leadership Angle: Discuss how you'd guide team members through similar problems
  6. Time Management: Practice with actual timers to build interview stamina
  7. Mock Interviews: Schedule practice sessions with peers or mentors
  8. System Design: Connect coding problems to larger system architecture
  9. Code Quality: Emphasize maintainable, readable code over clever tricks
  10. Stay Current: Review recent Amazon engineering blog posts and case studies

Key Success Factors

  • Preparation: Consistent daily practice over cramming
  • Understanding: Focus on principles over memorization
  • Communication: Clear explanation of thought process
  • Leadership: Demonstrate technical mentoring capabilities
  • Adaptability: Handle unexpected variations gracefully
  • Business Acumen: Connect technical solutions to business value

Remember: Amazon values customer obsession, ownership, and long-term thinking. Demonstrate these principles through your problem-solving approach and communication style.