What are the most challenging coding interview questions at Facebook?
Facebook is known for its challenging coding interviews, which often test a candidate's problem-solving skills, knowledge of data structures and algorithms, and ability to think critically. Here are some of the most challenging coding interview questions that you might encounter at Facebook, along with brief descriptions and tips on how to approach them:
1. Clone Graph
Problem: Given a reference of a node in a connected undirected graph, return a deep copy (clone) of the graph.
Approach: Use BFS or DFS with a hash map to keep track of visited nodes and their corresponding clones.
Solution:
def cloneGraph(node): if not node: return None node_map = {} def dfs(node): if node in node_map: return node_map[node] clone = Node(node.val, []) node_map[node] = clone for neighbor in node.neighbors: clone.neighbors.append(dfs(neighbor)) return clone return dfs(node)
2. LRU Cache
Problem: Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.
Approach: Use a combination of a hash map and a doubly linked list to achieve O(1) time complexity for get and put operations.
Solution:
class LRUCache: def __init__(self, capacity: int): self.cache = {} self.capacity = capacity self.head, self.tail = ListNode(0), ListNode(0) self.head.next, self.tail.prev = self.tail, self.head def _add(self, node): p = self.head.next self.head.next = node node.prev = self.head node.next = p p.prev = node def _remove(self, node): p = node.prev n = node.next p.next, n.prev = n, p 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 = ListNode(key, value) self._add(node) self.cache[key] = node if len(self.cache) > self.capacity: lru = self.tail.prev self._remove(lru) del self.cache[lru.key] class ListNode: def __init__(self, key=0, value=0): self.key = key self.value = value self.prev = None self.next = None
3. Word Ladder
Problem: Given two words (beginWord and endWord), and a dictionary's word list, find the length of the shortest transformation sequence from beginWord to endWord.
Approach: Use BFS to explore all possible transformations level by level.
Solution:
from collections import deque def ladderLength(beginWord, endWord, wordList): if endWord not in wordList: return 0 wordList = set(wordList) queue = deque([(beginWord, 1)]) while queue: current_word, level = queue.popleft() for i in range(len(current_word)): for char in 'abcdefghijklmnopqrstuvwxyz': next_word = current_word[:i] + char + current_word[i+1:] if next_word == endWord: return level + 1 if next_word in wordList: wordList.remove(next_word) queue.append((next_word, level + 1)) return 0
4. Longest Valid Parentheses
Problem: Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring.
Approach: Use dynamic programming or stack to keep track of valid parentheses.
Solution:
def longestValidParentheses(s): stack = [-1] max_length = 0 for i, char in enumerate(s): if char == '(': stack.append(i) else: stack.pop() if stack: max_length = max(max_length, i - stack[-1]) else: stack.append(i) return max_length
5. Serialize and Deserialize Binary Tree
Problem: Design an algorithm to serialize and deserialize a binary tree.
Approach: Use BFS or DFS to convert the tree into a string and back to a tree structure.
Solution:
class Codec: def serialize(self, root): if not root: return '' queue = deque([root]) result = [] while queue: node = queue.popleft() if node: result.append(str(node.val)) queue.append(node.left) queue.append(node.right) else: result.append('null') return ','.join(result) def deserialize(self, data): if not data: return None nodes = data.split(',') root = TreeNode(int(nodes[0])) queue = deque([root]) i = 1 while queue: node = queue.popleft() if nodes[i] != 'null': node.left = TreeNode(int(nodes[i])) queue.append(node.left) i += 1 if nodes[i] != 'null': node.right = TreeNode(int(nodes[i])) queue.append(node.right) i += 1 return root class TreeNode: def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right
6. Median of Two Sorted Arrays
Problem: Given two sorted arrays nums1 and nums2, return the median of the two sorted arrays.
Approach: Use binary search to partition the arrays and find the median.
Solution:
def findMedianSortedArrays(nums1, nums2): A, B = nums1, nums2 total = len(A) + len(B) half = total // 2 if len(A) > len(B): A, B = B, A l, r = 0, len(A) - 1 while True: i = (l + r) // 2 j = half - i - 2 Aleft = A[i] if i >= 0 else float('-infinity') Aright = A[i + 1] if (i + 1) < len(A) else float('infinity') Bleft = B[j] if j >= 0 else float('-infinity') Bright = B[j + 1] if (j + 1) < len(B) else float('infinity') if Aleft <= Bright and Bleft <= Aright: if total % 2: return min(Aright, Bright) return (max(Aleft, Bleft) + min(Aright, Bright)) / 2 elif Aleft > Bright: r = i - 1 else: l = i + 1
7. Maximum Subarray
Problem: Find the contiguous subarray within an array (containing at least one number) which has the largest sum.
Approach: Use Kadane’s Algorithm.
Solution:
def maxSubArray(nums): max_current = max_global = nums[0] for num in nums[1:]: max_current = max(num, max_current + num) if max_current > max_global: max_global = max_current return max_global
8. Regular Expression Matching
Problem: Implement regular expression matching with support for '.' and '*'.
Approach: Use dynamic programming to match the patterns.
Solution:
def isMatch(s, p): dp = [[False] * (len(p) + 1) for _ in range(len(s) + 1)] dp[0][0] = True for j in range(1, len(p) + 1): if p[j-1] == '*': dp[0][j] = dp[0][j-2] for i in range(1, len(s) + 1): for j in range(1, len(p) + 1): if p[j-1] == '.' or p[j-1] == s[i-1]: dp[i][j] = dp[i-1][j-1] elif p[j-1] == '*': dp[i][j] = dp[i][j-2] or (dp[i-1][j] if p[j-2] == '.' or p[j-2] == s[i-1] else False) return dp[-1][-1]
Tips to Tackle Challenging Problems
- Understand the Problem Statement: Make sure you fully understand the problem and clarify any doubts before you start coding.
- Break Down the Problem: Divide the problem into smaller subproblems and solve each one step by step.
- Optimize: Always look for ways to optimize your solution for time and space complexity.
- Practice: Regular practice of these and other challenging problems will help you become proficient and confident.
By mastering these challenging coding problems, you'll be well-prepared for your interview at
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