Grokking the Top 5 Most Challenging Coding Interview Questions for Google, Facebook, Amazon, Microsoft, and Apple.

Master the Most Challenging Coding Interview Questions from Top Tech Companies and Boost Your Problem-Solving Skills.

Landing a job at a top tech company like Google, Facebook, or Amazon is a dream come true for many software engineers. These companies are known for their innovative work culture, excellent compensation packages, and opportunities for growth. However, getting through their notoriously challenging coding interviews can be a daunting task.

The key to cracking these tough coding interview questions lies in understanding and mastering various coding patterns. Patterns like Sliding Window, Two-Pointers, Island, and Merge Interval can be applied to a wide range of problems, allowing you to tackle even the most complex questions with confidence. By learning to recognize and apply these patterns, you can significantly increase your chances of success in coding interviews.

In this article, I’ll walk you through the top 5 most difficult coding interview questions from major tech companies like Google, Facebook, and Amazon. We’ll dive deep into the problem statements, explore the optimal solutions using the aforementioned coding patterns, and provide step-by-step explanations to ensure you grasp the concepts.

So, whether you’re a seasoned software engineer looking to level up your skills or a newcomer preparing for your first big interview, this article will be your go-to resource for conquering these challenging coding questions. Don’t forget to follow me on Medium and join my mailing list for more expert advice and resources to help you land your dream job in the tech industry.

Question 1: LRU Cache (Google)

Problem Statement: Design and implement a data structure for a Least Recently Used (LRU) Cache. The cache should support the following operations: get and put. When the cache reaches its specified capacity, it should remove the least recently used item before adding a new item. If an item is not found in the cache, return -1 for the get operation.

Company: This question is often associated with interviews at Google.

Data structure and coding pattern used: The LRU Cache problem can be efficiently solved using a combination of a doubly-linked list and a hash map.

Step-by-step solution

1. Initialize a doubly-linked list and a hash map.
2. For the get operation, check if the key is in the hash map.
   a. If it is, move the corresponding node in the doubly-linked list to the front and return its value.
   b. If not, return -1.
3. For the put operation, check if the key is in the hash map.
   a. If it is, update the value and move the corresponding node in the doubly-linked list to the front.
   b If not, create a new node and add it to the front of the doubly-linked list and the hash map.
   c. If the cache is now over capacity, remove the least recently used item, which is at the end of the doubly-linked list, and delete its entry in the hash map.

Code

class LRUCache:
    class Node:
        def __init__(self, key, value):
            self.key = key
            self.value = value
            self.prev = None
            self.next = None

    def __init__(self, capacity: int):
        self.capacity = capacity
        self.cache = {}
        self.head = self.Node(0, 0)
        self.tail = self.Node(0, 0)
        self.head.next = self.tail
        self.tail.prev = self.head

    def _remove(self, node):
        prev_node = node.prev
        next_node = node.next
        prev_node.next = next_node
        next_node.prev = prev_node

    def _add(self, node):
        node.prev = self.head
        node.next = self.head.next
        self.head.next.prev = node
        self.head.next = node

    def get(self, key: int) -> int:
        if key in self.cache:
            node = self.cache[key]
            self._remove(node)
            self._add(node)
            return node.value
        return -1

    def put(self, key: int, value: int) -> None:
        if key in self.cache:
            self._remove(self.cache[key])
        node = self.Node(key, value)
        self._add(node)
        self.cache[key] = node
        if len(self.cache) > self.capacity:
            lru_node = self.tail.prev
            self._remove(lru_node)
            del self.cache[lru_node.key]

Question 2: Merge Intervals (Facebook)

Problem Statment: Given a collection of intervals, merge any overlapping intervals. The input list is not necessarily ordered by start time. The output should be a list of merged intervals sorted by start time.

Example: 
Input: intervals = [[1, 3], [2, 6], [8, 10], [15, 18]] 
Output: [[1, 6], [8, 10], [15, 18]]

Company: This question is often associated with interviews at Facebook.

Data structure and coding pattern used: The Merge Intervals problem can be efficiently solved using the sorting technique combined with iteration.

Step-by-step solution

1. Sort the input list of intervals based on the start time of each interval.
2. Initialize an empty list called ‘merged’ to store the merged intervals.
3. Iterate through the sorted list of intervals:
   a. If the ‘merged’ list is empty or the current interval’s start time is greater than the end time of the last interval in the ‘merged’ list, append the current interval to the ‘merged’ list.
   b. If the current interval overlaps with the last interval in the ‘merged’ list, update the end time of the last interval in the ‘merged’ list to be the maximum of the end times of the two overlapping intervals.
4. Return the ‘merged’ list.

Merge Intervals

Code

def merge(intervals):
    if not intervals:
        return []

    # Sort the input list of intervals based on the start time of each interval
    intervals.sort(key=lambda x: x[0])

    # Initialize an empty list called 'merged' to store the merged intervals
    merged = [intervals[0]]

    # Iterate through the sorted list of intervals, starting from the second interval
    for interval in intervals[1:]:
        # If the current interval's start time is greater than the end time of the last interval in the 'merged' list
        if interval[0] > merged[-1][1]:
            # Append the current interval to the 'merged' list, as it doesn't overlap with the previous interval
            merged.append(interval)
        else:
            # If the current interval overlaps with the last interval in the 'merged' list,
            # update the end time of the last interval in the 'merged' list to be the maximum
            # of the end times of the two overlapping intervals
            merged[-1][1] = max(merged[-1][1], interval[1])

    return merged

Question 3: Serialize and Deserialize Binary Tree (Amazon)

Problem Statment: Design an algorithm to serialize and deserialize a binary tree. Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer or transmitted across a network connection link to be reconstructed later in the same or another computer environment. Deserialization is the reverse process of recreating the binary tree from the serialized format.

Company: This question is often associated with interviews at Amazon.

Data structure and coding pattern used: The Serialize and Deserialize Binary Tree problem can be efficiently solved using Depth-First Search (DFS) in combination with pre-order traversal.

Step-by-step solution

1. For serialization, perform a pre-order traversal of the tree. a. If a node is not null, add its value to the serialized string, followed by a delimiter (e.g., ‘,’). b. If a node is null, add a placeholder (e.g., ‘#’) to the serialized string, followed by a delimiter.
2. For deserialization, use a queue to store the serialized nodes.
   a. Pop the first element from the queue.
   b. If the element is not a null placeholder, create a new TreeNode with the element’s value.
   c. Recursively build the left and right subtrees by repeating steps 2a to 2c.
3. Return the root of the deserialized tree.

Code

class Codec:
    def serialize(self, root):
        # Base case: If the current node is null, return a placeholder followed by a delimiter
        if not root:
            return '#,'
        
        # Pre-order traversal: Serialize the current node's value, followed by its left and right subtrees
        return str(root.val) + ',' + self.serialize(root.left) + self.serialize(root.right)

    def deserialize(self, data):
        def helper(queue):
            # Pop the first element from the queue
            value = queue.pop(0)
            
            # If the element is a null placeholder, return None
            if value == '#':
                return None
            
            # Create a new TreeNode with the element's value
            node = TreeNode(int(value))
            
            # Recursively build the left and right subtrees
            node.left = helper(queue)
            node.right = helper(queue)
            
            return node

        # Split the serialized data string into a queue of elements
        queue = data.split(',')
        
        # Call the helper function to deserialize the binary tree and return the root node
        return helper(queue[:-1])

Question 4: Longest Increasing Path in a Matrix (Microsoft)

Problem Statment: Given an m x n integers matrix, return the length of the longest increasing path in the matrix. From each cell, you can either move in four directions: left, right, up, or down. You may not move diagonally or move outside the matrix boundaries (i.e., wrap-around is not allowed).

Example: 
Input: matrix = [ [9, 9, 4], 
                  [6, 6, 8], 
                  [2, 1, 1] ] 
Output: 4 
Explanation: The longest increasing path is [1, 2, 6, 9].

Company: This question is often associated with interviews at Microsoft.

Data structure and coding pattern used: The Longest Increasing Path in a Matrix problem can be efficiently solved using Depth-First Search (DFS) with memoization.

Step-by-step solution

1. Create a memoization matrix of the same size as the input matrix, initialized with zeros.
2. Define a helper function for DFS traversal.
   a. If the current cell has already been visited, return its value from the memoization matrix.
   b. Otherwise, visit all valid neighboring cells that have a value greater than the current cell’s value, and find the maximum increasing path length among these neighbors.
   c. Store the result in the memoization matrix and return it.
3. Iterate through each cell in the input matrix and call the helper function to calculate the longest increasing path.
4. Return the maximum value found in the memoization matrix.

Code

def longestIncreasingPath(matrix):
    # Return 0 if the input matrix is empty
    if not matrix:
        return 0

    def dfs(i, j):
        # If the current cell has already been visited, return its value from the memoization matrix
        if not memo[i][j]:
            val = matrix[i][j]
            # Perform DFS on all valid neighboring cells with a greater value and find the maximum path length
            memo[i][j] = 1 + max(
                dfs(i - 1, j) if i > 0 and val > matrix[i - 1][j] else 0,
                dfs(i + 1, j) if i < m - 1 and val > matrix[i + 1][j] else 0,
                dfs(i, j - 1) if j > 0 and val > matrix[i][j - 1] else 0,
                dfs(i, j + 1) if j < n - 1 and val > matrix[i][j + 1] else 0
            )
        # Return the longest increasing path length for the current cell
        return memo[i][j]

    # Determine the dimensions of the input matrix
    m, n = len(matrix), len(matrix[0])

    # Initialize a memoization matrix of the same size as the input matrix, filled with zeros
    memo = [[0] * n for _ in range(m)]

    # Iterate through each cell in the input matrix and call the DFS function to calculate the longest increasing path
    # Return the maximum value found in the memoization matrix
    return max(dfs(i, j) for i in range(m) for j in range(n))

Question 5: Maximum Sum of Non-adjacent Elements (Apple)

Problem Statment: Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum of non-adjacent elements, and return its sum.

Example: 
Input: nums = [1, 2, 3, 1] 
Output: 4 
Explanation: The optimal solution is to pick elements 1 and 3, which are 
non-adjacent, with a sum of 4.

Company: This question is often associated with interviews at Apple.

Data structure and coding pattern used: The Maximum Sum of Non-adjacent Elements problem can be efficiently solved using Dynamic Programming.

Step-by-step solution

1. Initialize two variables, prev_sum and curr_sum, to store the previous maximum sum and current maximum sum, respectively.
2. Iterate through the input array.
   a. Calculate the new maximum sum by adding the current element and the previous maximum sum.
   b. Update the previous maximum sum to be the maximum of the previous and current maximum sums.
   c. Update the current maximum sum to be the new maximum sum.
3. Return the maximum of the previous and current maximum sums.

Code

def max_sum_non_adjacent_elements(nums):
    # Initialize prev_sum and curr_sum to store the previous and current maximum sums, respectively
    prev_sum = 0
    curr_sum = 0

    # Iterate through the input array
    for num in nums:
        # Calculate the new maximum sum by adding the current element and the previous maximum sum
        new_sum = curr_sum + num
        
        # Update the previous maximum sum to be the maximum of the previous and current maximum sums
        curr_sum, prev_sum = max(curr_sum, prev_sum), curr_sum

    # Return the maximum of the previous and current maximum sums
    return max(prev_sum, curr_sum)

Conclusion

In this blog post, we discussed five of the most difficult coding interview questions from major tech companies, such as Google, Facebook, Amazon, Microsoft, and Apple. By working through these challenging problems, you can better prepare yourself for technical interviews and sharpen your problem-solving skills. Furthermore, understanding the coding patterns and optimization techniques presented in this post will enable you to tackle a wide range of coding challenges with confidence.

It is essential to remember that practice makes perfect. By continually working on complex coding problems, you can hone your skills and develop a deep understanding of various algorithms, data structures, and coding patterns. This, in turn, will make you a more competent and competitive candidate during the interview process.

The journey to mastering coding interviews is challenging but rewarding. By studying the problems presented in this blog post and practicing other coding problems, you will be well-prepared to tackle even the most difficult interview questions from top tech companies. Keep pushing yourself, stay curious, and never stop learning. Good luck on your journey to becoming a successful software engineer!

Learn more about the coding patterns in Grokking the Coding Interview.

Don’t Just LeetCode; Follow the Coding Patterns Instead

System Design Interview Survival Guide (2023): Preparation Strategies and Practical Tips

Thanks for reading

• 📰 View my content on Coding and System Design Interviews
• 🔔 Follow me: LinkedIn | Twitter | Newsletter