function myPromiseAll(promises) {
  return new Promise((resolve, reject) => {
    const results = [];
    let completed = 0;
    if (promises.length === 0) resolve([]);
    promises.forEach((promise, index) => {
      Promise.resolve(promise).then(value => {
        results[index] = value;
        completed++;
        if (completed === promises.length) resolve(results);
      }).catch(error => reject(error));
    });
  });
}

def merge_sorted_arrays_no_duplicates(arr1, arr2):
    p1, p2 = 0, 0
    result = []
    while p1 < len(arr1) or p2 < len(arr2):
        val1 = arr1[p1] if p1 < len(arr1) else float('inf')
        val2 = arr2[p2] if p2 < len(arr2) else float('inf')
        if val1 < val2:
            current_val = val1
            p1 += 1
        else:
            current_val = val2
            p2 += 1
        if not result or result[-1] != current_val:
            result.append(current_val)
    return result

import java.util.Stack;
class Solution {
    public boolean isValid(String s) {
        Stack<character> stack = new Stack<>();
        for (char c : s.toCharArray()) {
            if (c == '(' || c == '[' || c == '{') {
                stack.push(c);
            } else {
                if (stack.isEmpty()) return false;
                char top = stack.pop();
                if ((c == ')' && top != '(') || (c == ']' && top != '[') || (c == '}' && top != '{')) {
                    return false;
                }
            }
        }
        return stack.isEmpty();
    }
}<

import collections
def validSquare(p1, p2, p3, p4):
    points = [p1, p2, p3, p4]
    def distSq(a, b):
        return (a[0] - b[0])**2 + (a[1] - b[1])**2
    
<pre><code># 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>
</code></pre>
<h2>5. Implement a dropdown component.</h2>
<h4>Why you might get asked this:</h4>
<p>Assesses front-end development skills: HTML structure, CSS for styling/animation, and JavaScript for interactivity and accessibility.</p>
<h4>How to answer:</h4>
<p>Discuss HTML structure for button and list, CSS for showing/hiding and basic styling, and JS for click handlers and state management.</p>
<h4>Example answer:</h4>
<p>A dropdown needs a trigger element (button) and a hidden list. Use JS to toggle <code>display: block</code> or <code>opacity: 1</code> on click. CSS transitions provide animation. Ensure keyboard navigation with <code>tabindex</code> and <code>aria-*</code> attributes for accessibility. Handle clicks outside the dropdown to close it.</p>
<h2>6. SQL query to find duplicates or optimize joins.</h2>
<h4>Why you might get asked this:</h4>
<p>Tests database understanding, SQL proficiency, and ability to write efficient queries for data manipulation or retrieval.</p>
<h4>How to answer:</h4>
<p>For duplicates, use <code>GROUP BY</code> and <code>HAVING COUNT(*) > 1</code>. For joins, discuss <code>INNER JOIN</code>, <code>LEFT JOIN</code>, and consider indexing.</p>
<h4>Example answer:</h4>
<p>To find duplicate emails in an <code>Users</code> table: <code>SELECT email, COUNT(email) FROM Users GROUP BY email HAVING COUNT(email) > 1;</code>. To optimize a join, ensure indexed columns are used in <code>ON</code> clauses: <code>SELECT * FROM Orders o INNER JOIN Customers c ON o.customer<em>id = c.id;</em></code><em> (assuming <code>customer</code></em><code>id</code> and <code>id</code> are indexed).</p>
<h2>7. Implement a function that extends <code>Array.prototype</code>.</h2>
<h4>Why you might get asked this:</h4>
<p>Probes JavaScript prototype chain understanding, <code>this</code> context, and how to safely extend built-in objects.</p>
<h4>How to answer:</h4>
<p>Define the function directly on <code>Array.prototype</code>. Explain how <code>this</code> inside the function refers to the array instance.</p>
<h4>Example answer:</h4>
<pre><code>Array.prototype.myFilter = function(callback) {
  const filteredArr = [];
  for (let i = 0; i < this.length; i++) {
    if (callback(this[i], i, this)) {
      filteredArr.push(this[i]);
    }
  }
  return filteredArr;
};
// Usage: [1, 2, 3].myFilter(num => num > 1); // [2, 3]</code></pre>
<h2>8. Rotate an image 180 degrees with animation on mouse hover.</h2>
<h4>Why you might get asked this:</h4>
<p>Tests CSS <code>transform</code> and <code>transition</code> properties, along with JavaScript event handling for interactive animations.</p>
<h4>How to answer:</h4>
<p>Use CSS <code>transform: rotateZ(180deg)</code> and <code>transition: transform 0.5s ease-in-out</code> on the image. Apply the <code>rotateZ</code> on <code>:hover</code> pseudo-class.</p>
<h4>Example answer:</h4>
<pre><code><img src="image.jpg" class="rotatable-image">
<style>
.rotatable-image {
  transition: transform 0.5s ease-in-out;
}
.rotatable-image:hover {
  transform: rotateZ(180deg);
}
</style></code></pre>
<h2>9. Find islands in a grid map.</h2>
<h4>Why you might get asked this:</h4>
<p>Evaluates graph traversal algorithms (BFS/DFS), grid manipulation, and handling visited states to prevent infinite loops.</p>
<h4>How to answer:</h4>
<p>Iterate through the grid. If an 'island' (1) is found, increment count, then use DFS/BFS to mark all connected land cells as 'visited' (0).</p>
<h4>Example answer:</h4>
<pre><code>def numIslands(grid):
    rows, cols = len(grid), len(grid[0])
    count = 0
    def dfs(r, c):
        if r < 0 or r >= rows or c < 0 or c >= cols or grid[r][c] == '0':
            return
        grid[r][c] = '0' # Mark as visited
        dfs(r + 1, c)
        dfs(r - 1, c)
        dfs(r, c + 1)
        dfs(r, c - 1)
    
<pre><code>for r in range(rows):
    for c in range(cols):
        if grid[r][c] == '1':
            count += 1
            dfs(r, c)
return count</code></pre>
</code></pre>
<h2>10. Tell me about a product you launched from start to finish.</h2>
<h4>Why you might get asked this:</h4>
<p>Behavioral question assessing project management, end-to-end thinking, leadership, and ability to deliver.</p>
<h4>How to answer:</h4>
<p>STAR method: Situation, Task, Action, Result. Focus on your role, challenges, decisions, and quantifiable outcomes.</p>
<h4>Example answer:</h4>
<p>"S: As lead developer, I was tasked with building a new internal analytics dashboard. T: My goal was to deliver a user-friendly tool to track key performance metrics. A: I designed the architecture, led a small team, implemented core features, and collaborated with stakeholders. R: The product launched on time, reduced manual reporting by 30%, and significantly improved data-driven decision-making."</p>
<h2>11. What’s been your biggest achievement?</h2>
<h4>Why you might get asked this:</h4>
<p>Gauges your impact, what you value, and your ability to articulate successes and lessons learned.</p>
<h4>How to answer:</h4>
<p>Choose a specific achievement, describe the context, your unique contribution, and its positive outcome or learning.</p>
<h4>Example answer:</h4>
<p>"My biggest achievement was refactoring a legacy system that was causing frequent outages. I designed and implemented a modular solution, reducing downtime by 90% and improving system reliability. This not only prevented revenue loss but also allowed the team to deliver new features faster."</p>
<h2>12. Describe a failure and what you learned.</h2>
<h4>Why you might get asked this:</h4>
<p>Tests self-awareness, resilience, ability to learn from mistakes, and problem-solving under adverse conditions.</p>
<h4>How to answer:</h4>
<p>Pick a genuine failure, take ownership, explain the root cause, and detail specific, actionable lessons applied since.</p>
<h4>Example answer:</h4>
<p>"Early in my career, I once underestimated the complexity of integrating a third-party API, leading to a missed deadline. I learned the crucial importance of thorough upfront research, breaking down tasks more granularly, and proactive communication with stakeholders about potential risks and delays."</p>
<h2>13. Why ByteDance?</h2>
<h4>Why you might get asked this:</h4>
<p>Assesses your motivation, research into the company, and how your career goals align with their mission and culture.</p>
<h4>How to answer:</h4>
<p>Connect ByteDance's products, innovation, or culture to your aspirations and skills. Be specific and genuine.</p>
<h4>Example answer:</h4>
<p>"I'm drawn to ByteDance's innovative approach to content creation and distribution, particularly how products like TikTok leverage AI to personalize user experiences. My passion for building scalable, user-centric applications aligns perfectly with your mission, and I'm eager to contribute to such a dynamic global platform."</p>
<h2>14. What excites you about this role?</h2>
<h4>Why you might get asked this:</h4>
<p>Evaluates your understanding of the role, your enthusiasm, and whether your skills and interests are a good match.</p>
<h4>How to answer:</h4>
<p>Highlight specific responsibilities, technologies, or challenges mentioned in the job description that align with your strengths and interests.</p>
<h4>Example answer:</h4>
<p>"The opportunity to work on cutting-edge machine learning infrastructure truly excites me. Specifically, the challenge of optimizing data pipelines for petabytes of data, as mentioned in the job description, perfectly matches my experience in distributed systems and my drive to build highly efficient, impactful solutions."</p>
<h2>15. Reverse a singly linked list.</h2>
<h4>Why you might get asked this:</h4>
<p>Fundamental linked list manipulation. Tests pointer handling, iterative vs. recursive thinking, and edge cases.</p>
<h4>How to answer:</h4>
<p>Use three pointers: <code>prev</code>, <code>curr</code>, <code>next_node</code>. Iterate, re-pointing <code>curr.next</code> to <code>prev</code>, then advancing all pointers.</p>
<h4>Example answer:</h4>
<pre><code>class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

</code><p><code>def reverseList(head):<br>    prev = None<br>    curr = head<br>    while curr:<br>        next_node = curr.next<br>        curr.next = prev<br>        prev = curr<br>        curr = next_node<br>    return prev</code></p></pre><p></p>
<h2>16. Find the Kth smallest element in a BST.</h2>
<h4>Why you might get asked this:</h4>
<p>Tests understanding of Binary Search Tree properties and tree traversal algorithms (in-order traversal).</p>
<h4>How to answer:</h4>
<p>Perform an in-order traversal (Left-Root-Right). The Kth element visited will be the Kth smallest. Use a counter.</p>
<h4>Example answer:</h4>
<pre><code>class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode() {}
    TreeNode(int val) { this.val = val; }
}

</code><p><code>class Solution {<br>    int count = 0;<br>    int result = 0;<br>    public int kthSmallest(TreeNode root, int k) {<br>        inorder(root, k);<br>        return result;<br>    }<br>    private void inorder(TreeNode node, int k) {<br>        if (node == null) return;<br>        inorder(node.left, k);<br>        count++;<br>        if (count == k) {<br>            result = node.val;<br>            return;<br>        }<br>        inorder(node.right, k);<br>    }<br>}</code></p></pre><p></p>
<h2>17. Longest common subsequence.</h2>
<h4>Why you might get asked this:</h4>
<p>A classic dynamic programming problem. Tests ability to identify overlapping subproblems and optimal substructure.</p>
<h4>How to answer:</h4>
<p>Use a 2D DP table where <code>dp[i][j]</code> stores the LCS length for <code>text1[0...i-1]</code> and <code>text2[0...j-1]</code>.</p>
<h4>Example answer:</h4>
<pre><code>def longestCommonSubsequence(text1, text2):
    m, n = len(text1), len(text2)
    dp = [[0] * (n + 1) for _ in range(m + 1)]
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if text1[i - 1] == text2[j - 1]:
                dp[i][j] = 1 + dp[i - 1][j - 1]
            else:
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
    return dp[m][n]</code></pre>
<h2>18. Design a URL shortener.</h2>
<h4>Why you might get asked this:</h4>
<p>A common system design question. Tests understanding of scalability, unique ID generation, database choices, and redirects.</p>
<h4>How to answer:</h4>
<p>Discuss core components: API, hash generation (base62 encoding, custom hash, collision handling), database (NoSQL for read-heavy), and redirect logic.</p>
<h4>Example answer:</h4>
<p>Use a distributed key-value store (e.g., Redis for short-term, Cassandra for long-term) for mapping short URLs to long URLs. Generate unique short codes using Base62 encoding of an auto-incrementing ID. Implement a REST API for creation and redirection. Consider caching and rate limiting.</p>
<h2>19. Implement a LRU Cache.</h2>
<h4>Why you might get asked this:</h4>
<p>Tests data structure design, combining a hash map for O(1) lookup and a doubly linked list for O(1) eviction policy.</p>
<h4>How to answer:</h4>
<p>Maintain a <code>HashMap</code> (key to node) and a <code>DoublyLinkedList</code> (maintains LRU order). On <code>get</code>, move node to front. On <code>put</code>, add to front; if capacity exceeded, remove from tail.</p>
<h4>Example answer:</h4>
<pre><code>class LRUCache:
    class Node:
        def __init__(self, key, val):
            self.key = key
            self.val = val
            self.prev = None
            self.next = None

<pre><code>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>
</code></pre>
<h2>20. Find all permutations of a string.</h2>
<h4>Why you might get asked this:</h4>
<p>A classic recursion/backtracking problem. Tests understanding of state management and exploration of all possibilities.</p>
<h4>How to answer:</h4>
<p>Use recursion. For each character, place it at the current position and recursively find permutations for the remaining characters.</p>
<h4>Example answer:</h4>
<pre><code>def permute(s):
    res = []
    def backtrack(current_perm, remaining_chars):
        if not remaining_chars:
            res.append("".join(current_perm))
            return
        for i in range(len(remaining_chars)):
            char = remaining_chars[i]
            backtrack(current_perm + [char], remaining_chars[:i] + remaining_chars[i+1:])
    
<pre><code>backtrack([], list(s))
return res</code></pre>
</code></pre>
<h2>21. Implement <code>debounce</code> function.</h2>
<h4>Why you might get asked this:</h4>
<p>Common JavaScript utility function. Tests understanding of closures, <code>setTimeout</code>, <code>clearTimeout</code>, and event optimization.</p>
<h4>How to answer:</h4>
<p>Return a new function. Inside, use <code>clearTimeout</code> to cancel previous calls and <code>setTimeout</code> to delay execution.</p>
<h4>Example answer:</h4>
<pre><code>function debounce(func, delay) {
  let timeout;
  return function(...args) {
    const context = this;
    clearTimeout(timeout);
    timeout = setTimeout(() => func.apply(context, args), delay);
  };
}
// Usage: const debouncedSearch = debounce(searchFunction, 300);
// inputElement.addEventListener('keyup', debouncedSearch);</code></pre>
<h2>22. Deep copy a linked list with random pointers.</h2>
<h4>Why you might get asked this:</h4>
<p>A challenging linked list problem requiring careful handling of pointers (next and random) and avoiding cycles.</p>
<h4>How to answer:</h4>
<p>First pass: create copy nodes and interleave them (e.g., <code>A -> A' -> B -> B'</code>). Second pass: assign random pointers for copy nodes. Third pass: separate original and copy lists.</p>
<h4>Example answer:</h4>
<pre><code>class Node:
    def __init__(self, x: int, next: 'Node' = None, random: 'Node' = None):
        self.val = int(x)
        self.next = next
        self.random = random

<p>def copyRandomList(head: 'Node') -> 'Node':<br>    if not head: return None<br>    curr = head<br>    # 1. Create interleaved nodes<br>    while curr:<br>        new_node = Node(curr.val, curr.next)<br>        curr.next = new_node<br>        curr = new_node.next</p>
<pre><code>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>
</code></pre>
<h2>23. Find the median of two sorted arrays.</h2>
<h4>Why you might get asked this:</h4>
<p>A classic problem testing binary search and understanding of partitioning, typically with O(log(min(m,n))) complexity.</p>
<h4>How to answer:</h4>
<p>Use a binary search approach to find the partition point in the smaller array such that elements are correctly split for median calculation.</p>
<h4>Example answer:</h4>
<pre><code>def findMedianSortedArrays(nums1, nums2):
    A, B = nums1, nums2
    m, n = len(A), len(B)
    if m > n: # Ensure A is the shorter array
        A, B = B, A
        m, n = n, m
    
<pre><code>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>
</code></pre>
<h2>24. Binary tree level order traversal.</h2>
<h4>Why you might get asked this:</h4>
<p>Tests BFS (Breadth-First Search) algorithm using a queue, a fundamental tree traversal method.</p>
<h4>How to answer:</h4>
<p>Use a queue. Add the root. While the queue is not empty, dequeue a node, add its children to the queue, and record the node's value.</p>
<h4>Example answer:</h4>
<pre><code>import java.util.*;

<p>class TreeNode {<br>    int val;<br>    TreeNode left;<br>    TreeNode right;<br>    TreeNode() {}<br>    TreeNode(int val) { this.val = val; }<br>}</p>
</code><p><code>class Solution {<br>    public List<List<integer>> levelOrder(TreeNode root) {<br>        List<List<integer>> result = new ArrayList<>();<br>        if (root == null) return result;<br>        Queue<treenode> queue = new LinkedList<>();<br>        queue.offer(root);<br>        while (!queue.isEmpty()) {<br>            int levelSize = queue.size();<br>            List<integer> currentLevel = new ArrayList<>();<br>            for (int i = 0; i < levelSize; i++) {<br>                TreeNode node = queue.poll();<br>                currentLevel.add(node.val);<br>                if (node.left != null) queue.offer(node.left);<br>                if (node.right != null) queue.offer(node.right);<br>            }<br>            result.add(currentLevel);<br>        }<br>        return result;<br>    }<br>}</integer></treenode></integer></integer></code></p></pre><p></p>
<h2>25. Merge all overlapping intervals.</h2>
<h4>Why you might get asked this:</h4>
<p>Common array manipulation problem. Tests sorting and iterative merging logic.</p>
<h4>How to answer:</h4>
<p>Sort intervals by start time. Iterate and merge if current interval overlaps with the last merged one.</p>
<h4>Example answer:</h4>
<pre><code>def merge(intervals):
    intervals.sort(key=lambda x: x[0])
    merged = []
    for interval in intervals:
        if not merged or interval[0] > merged[-1][1]:
            merged.append(interval)
        else:
            merged[-1][1] = max(merged[-1][1], interval[1])
    return merged</code></pre>
<h2>26. Find the shortest path in a binary matrix.</h2>
<h4>Why you might get asked this:</h4>
<p>Tests BFS on a grid, often used for pathfinding. Assesses understanding of graph representation and visited states.</p>
<h4>How to answer:</h4>
<p>Use BFS starting from (0,0). Queue stores <code>(row, col, distance)</code>. Explore 8-directional neighbors, marking visited cells.</p>
<h4>Example answer:</h4>
<pre><code>import collections

<p>def shortestPathBinaryMatrix(grid):<br>    n = len(grid)<br>    if grid[0][0] == 1 or grid[n-1][n-1] == 1:<br>        return -1</p>
<pre><code>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>
</code></pre>
<h2>27. Validate a Binary Search Tree.</h2>
<h4>Why you might get asked this:</h4>
<p>Tests tree traversal logic and understanding of BST properties (left child < parent < right child, recursively).</p>
<h4>How to answer:</h4>
<p>Use an in-order traversal; elements must be strictly increasing. Or, use a recursive helper with <code>min<em>val</em></code><em> and <code>max</code></em><code>val</code> bounds.</p>
<h4>Example answer:</h4>
<pre><code>class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode() {}
    TreeNode(int val) { this.val = val; }
}

<p>class Solution {<br>    public boolean isValidBST(TreeNode root) {<br>        return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);<br>    }</p>
<pre><code>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);
}
</code></pre>
</code><p><code>}</code></p></pre><p></p>
<h2>28. Design a distributed key-value store.</h2>
<h4>Why you might get asked this:</h4>
<p>Advanced system design, probing knowledge of distributed systems concepts like consistency, availability, partitioning, and fault tolerance.</p>
<h4>How to answer:</h4>
<p>Discuss APIs (get/put), partitioning (consistent hashing), replication (N-way), consistency models (eventual vs. strong), conflict resolution, and fault tolerance (quorum, gossip protocol).</p>
<h4>Example answer:</h4>
<p>A distributed KVS needs: <strong>Consistent Hashing</strong> for data distribution across nodes; <strong>Replication</strong> (N copies for durability/availability); <strong>Quorum Consensus</strong> for reads/writes to ensure consistency (W+R > N); <strong>Leader Election</strong> or <strong>Gossip Protocol</strong> for fault tolerance and node discovery. Data is partitioned into ranges, and each range is managed by a set of replica nodes.</p>
<h2>29. Implement an autocomplete feature.</h2>
<h4>Why you might get asked this:</h4>
<p>Tests data structure selection (Trie/Prefix Tree) and algorithmic thinking for prefix-based searching.</p>
<h4>How to answer:</h4>
<p>Use a Trie. Each node stores characters. Traverse the Trie based on the input prefix, then collect all words from the subtree.</p>
<h4>Example answer:</h4>
<pre><code>class TrieNode:
    def __init__(self):
        self.children = {}
        self.is_end_of_word = False
        self.words = [] # To store words for autocompletion

<p>class Trie:<br>    def <strong>init</strong>(self):<br>        self.root = TrieNode()</p>
<pre><code>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>
</code></pre>
<h2>30. Two Sum problem (find indices of two numbers adding to target).</h2>
<h4>Why you might get asked this:</h4>
<p>A foundational problem testing basic array manipulation and the effective use of hash maps for efficient lookups.</p>
<h4>How to answer:</h4>
<p>Iterate through the array. For each number, check if <code>target - current_number</code> exists in a hash map. If so, return indices. Otherwise, add current number to map.</p>
<h4>Example answer:</h4>
<pre><code>import java.util.HashMap;
import java.util.Map;

</code><p><code>class Solution {<br>    public int[] twoSum(int[] nums, int target) {<br>        Map<Integer, Integer> numMap = new HashMap<>();<br>        for (int i = 0; i < nums.length; i++) {<br>            int complement = target - nums[i];<br>            if (numMap.containsKey(complement)) {<br>                return new int[]{numMap.get(complement), i};<br>            }<br>            numMap.put(nums[i], i);<br>        }<br>        return new int[]{}; // Should not reach here for valid inputs<br>    }<br>}</code></p></pre><p></p>
<h2>Other Tips to Prepare for a ByteDance Coding Interview</h2>
<p>Preparing for a ByteDance coding interview requires consistent effort and a strategic approach. "The only way to do great work is to love what you do," and for technical roles, that means genuinely enjoying problem-solving. Start by mastering data structures and algorithms, as these form the bedrock of ByteDance's technical assessments. Practice coding regularly on platforms like LeetCode, focusing on medium to hard problems. Don't just solve them; understand multiple approaches, analyze their time and space complexity, and be ready to explain your thought process. As "Practice makes perfect," the more diverse problems you tackle, the better equipped you'll be.</p>
<p>When practicing, simulate interview conditions: use a whiteboard or a plain text editor, and articulate your steps aloud. This hones your communication skills, which are as crucial as your coding ability. For system design, understand core concepts like scalability, consistency, and fault tolerance, and be able to draw diagrams. Remember to practice behavioral questions too; use the STAR method to structure your answers effectively. Consider using tools like <strong>Verve AI Interview Copilot</strong> (https://vervecopilot.com) to get real-time feedback on your answers and practice explaining your code and system designs. <strong>Verve AI Interview Copilot</strong> can help refine your articulation and ensure you're hitting all key points. Using <strong>Verve AI Interview Copilot</strong> for mock interviews can significantly boost your confidence and performance.</p>
<h2>Frequently Asked Questions</h2>
<p>Q1: How long do ByteDance coding interviews typically last?<br>A1: Coding interviews at ByteDance usually last 30-45 minutes, often involving real-time coding on a shared platform or whiteboard.</p>
<p>Q2: What programming languages are preferred for ByteDance interviews?<br>A2: ByteDance accepts common languages like Java, C++, Python, and JavaScript. Choose the one you are most proficient and comfortable with.</p>
<p>Q3: Should I expect system design questions for all roles?<br>A3: System design questions are typically for more senior engineering roles, but may occasionally appear for mid-level positions as well.</p>
<p>Q4: How important are behavioral questions at ByteDance?<br>A4: Behavioral questions are highly important at ByteDance, especially in final rounds, to assess your cultural fit, motivation, and teamwork skills.</p>
<p>Q5: What difficulty level are coding problems?<br>A5: Most coding problems range from medium to hard difficulty, often found on platforms like LeetCode.</p>
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