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Top 30 Most Common ByteDance Coding Interview Questions You Should Prepare For

Top 30 Most Common ByteDance Coding Interview Questions You Should Prepare For

Top 30 Most Common ByteDance Coding Interview Questions You Should Prepare For

Top 30 Most Common ByteDance Coding Interview Questions You Should Prepare For

Top 30 Most Common ByteDance Coding Interview Questions You Should Prepare For

Top 30 Most Common ByteDance Coding Interview Questions You Should Prepare For

Written by

Written by

Written by

Kent McAllister, Career Advisor

Kent McAllister, Career Advisor

Kent McAllister, Career Advisor

💡Even the best candidates blank under pressure. AI Interview Copilot helps you stay calm and confident with real-time cues and phrasing support when it matters most. Let’s dive in.

💡Even the best candidates blank under pressure. AI Interview Copilot helps you stay calm and confident with real-time cues and phrasing support when it matters most. Let’s dive in.

💡Even the best candidates blank under pressure. AI Interview Copilot helps you stay calm and confident with real-time cues and phrasing support when it matters most. Let’s dive in.

# Calculate all 6 pairwise distances
distances = []
for i in range(4):
    for j in range(i + 1, 4):
        distances.append(distSq(points[i], points[j]))

# Count frequency of distances
counts = collections.Counter(distances)

# A square must have 4 equal side lengths and 2 equal diagonal lengths.
# The side length squared should be non-zero.
if len(counts) == 2 and 0 not in counts:
    sides = min(counts.keys())
    diagonals = max(counts.keys())
    if counts[sides] == 4 and counts[diagonals] == 2:
        return True
return False</code></pre

for r in range(rows):
    for c in range(cols):
        if grid[r][c] == '1':
            count += 1
            dfs(r, c)
return count</code></pre

def reverseList(head):
prev = None
curr = head
while curr:
next_node = curr.next
curr.next = prev
prev = curr
curr = next_node
return prev

class Solution {
int count = 0;
int result = 0;
public int kthSmallest(TreeNode root, int k) {
inorder(root, k);
return result;
}
private void inorder(TreeNode node, int k) {
if (node == null) return;
inorder(node.left, k);
count++;
if (count == k) {
result = node.val;
return;
}
inorder(node.right, k);
}
}

def __init__(self, capacity: int):
    self.cache = {} # map key to node
    self.capacity = capacity
    self.head = self.Node(0,0) # dummy head
    self.tail = self.Node(0,0) # dummy tail
    self.head.next = self.tail
    self.tail.prev = self.head

def _remove_node(self, node):
    node.prev.next = node.next
    node.next.prev = node.prev

def _add_to_front(self, node):
    node.next = self.head.next
    node.prev = self.head
    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(node)
        self._add_to_front(node)
        return node.val
    return -1

def put(self, key: int, value: int) -> None:
    if key in self.cache:
        self._remove_node(self.cache[key])
    new_node = self.Node(key, value)
    self.cache[key] = new_node
    self._add_to_front(new_node)

    if len(self.cache) > self.capacity:
        lru_node = self.tail.prev
        self._remove_node(lru_node)
        del self.cache[lru_node.key]</code></pre

backtrack([], list(s))
return res</code></pre

def copyRandomList(head: 'Node') -> 'Node':
if not head: return None
curr = head
# 1. Create interleaved nodes
while curr:
new_node = Node(curr.val, curr.next)
curr.next = new_node
curr = new_node.next

curr = head
# 2. Assign random pointers for new nodes
while curr:
    if curr.random:
        curr.next.random = curr.random.next
    curr = curr.next.next

curr = head
new_head = head.next
# 3. Separate original and new lists
while curr:
    copy = curr.next
    curr.next = copy.next
    if copy.next:
        copy.next = copy.next.next
    curr = curr.next
return new_head</code></pre

half_len = (m + n + 1) // 2
low, high = 0, m

while low <= high:
    i = (low + high) // 2 # Partition point for A
    j = half_len - i     # Partition point for B

    if i < m and B[j-1] > A[i]:
        low = i + 1
    elif i > 0 and A[i-1] > B[j]:
        high = i - 1
    else: # Partition found
        max_left = 0
        if i == 0: max_left = B[j-1]
        elif j == 0: max_left = A[i-1]
        else: max_left = max(A[i-1], B[j-1])

        if (m + n) % 2 == 1:
            return float(max_left)
        
        min_right = 0
        if i == m: min_right = B[j]
        elif j == n: min_right = A[i]
        else: min_right = min(A[i], B[j])
        
        return (max_left + min_right) / 2.0</code></pre

class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode() {}
TreeNode(int val) { this.val = val; }
}

class Solution {
public List<List> levelOrder(TreeNode root) {
List<List> result = new ArrayList<>();
if (root == null) return result;
Queue queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
int levelSize = queue.size();
List currentLevel = new ArrayList<>();
for (int i = 0; i < levelSize; i++) {
TreeNode node = queue.poll();
currentLevel.add(node.val);
if (node.left != null) queue.offer(node.left);
if (node.right != null) queue.offer(node.right);
}
result.add(currentLevel);
}
return result;
}
}

def shortestPathBinaryMatrix(grid):
n = len(grid)
if grid[0][0] == 1 or grid[n-1][n-1] == 1:
return -1

q = collections.deque([(0, 0, 1)]) # (r, c, dist)
grid[0][0] = 1 # Mark as visited

directions = [
    (1, 0), (-1, 0), (0, 1), (0, -1),
    (1, 1), (1, -1), (-1, 1), (-1, -1)
]

while q:
    r, c, dist = q.popleft()
    if r == n - 1 and c == n - 1:
        return dist
    
    for dr, dc in directions:
        nr, nc = r + dr, c + dc
        if 0 <= nr < n and 0 <= nc < n and grid[nr][nc] == 0:
            grid[nr][nc] = 1 # Mark as visited
            q.append((nr, nc, dist + 1))

return -1</code></pre

class Solution {
public boolean isValidBST(TreeNode root) {
return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
}

private boolean isValidBST(TreeNode node, long minVal, long maxVal) {
    if (node == null) return true;
    if (node.val <= minVal || node.val >= maxVal) return false;
    return isValidBST(node.left, minVal, node.val) && isValidBST(node.right, node.val, maxVal);
}

}

class Trie:
def init(self):
self.root = TrieNode()

def insert(self, word: str) -> None:
    node = self.root
    for char in word:
        if char not in node.children:
            node.children[char] = TrieNode()
        node = node.children[char]
        node.words.append(word) # Store word at each node for easy retrieval
    node.is_end_of_word = True

def search_prefix(self, prefix: str) -> list[str]:
    node = self.root
    for char in prefix:
        if char not in node.children:
            return []
        node = node.children[char]
    return node.words # All words passing through this prefix node</code></pre

class Solution {
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> numMap = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
int complement = target - nums[i];
if (numMap.containsKey(complement)) {
return new int[]{numMap.get(complement), i};
}
numMap.put(nums[i], i);
}
return new int[]{}; // Should not reach here for valid inputs
}
}

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