Apple Coding Questions 2025 - Interview Problems with Solutions

Apple Coding Questions - Complete Problem Set

Practice 25+ Apple coding interview questions with solutions in multiple languages. Apple focuses on clean code, optimal solutions, and edge case handling.

Apple Interview Focus

Apple values: Clean, production-quality code, Optimal time/space complexity, Thorough edge case handling, Clear communication of thought process.

Arrays & Strings

1. Two Sum

Problem

Find indices of two numbers that add up to target.

Input: nums = [2,7,11,15], target = 9
Output: [0,1]

Python

def two_sum(nums, target):
    seen = {}
    for i, num in enumerate(nums):
        complement = target - num
        if complement in seen:
            return [seen[complement], i]
        seen[num] = i
    return []

Time: O(n), Space: O(n)

Java

public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement)) {
            return new int[]{map.get(complement), i};
        }
        map.put(nums[i], i);
    }
    return new int[]{};
}

2. Container With Most Water

Problem

Find two lines that form container with most water.

Input: [1,8,6,2,5,4,8,3,7]
Output: 49

Python

def max_area(height):
    left, right = 0, len(height) - 1
    max_water = 0
    while left < right:
        width = right - left
        h = min(height[left], height[right])
        max_water = max(max_water, width * h)
        if height[left] < height[right]:
            left += 1
        else:
            right -= 1
    return max_water

Time: O(n), Space: O(1)

Java

public int maxArea(int[] height) {
    int left = 0, right = height.length - 1;
    int maxWater = 0;
    while (left < right) {
        int width = right - left;
        int h = Math.min(height[left], height[right]);
        maxWater = Math.max(maxWater, width * h);
        if (height[left] < height[right]) left++;
        else right--;
    }
    return maxWater;
}

3. Longest Substring Without Repeating Characters

Problem

Find length of longest substring without repeating characters.

Input: “abcabcbb”
Output: 3 (“abc”)

Python

def length_of_longest_substring(s):
    char_index = {}
    max_length = start = 0
    for i, char in enumerate(s):
        if char in char_index and char_index[char] >= start:
            start = char_index[char] + 1
        char_index[char] = i
        max_length = max(max_length, i - start + 1)
    return max_length

Time: O(n), Space: O(min(n, 26))

Java

public int lengthOfLongestSubstring(String s) {
    Map<Character, Integer> map = new HashMap<>();
    int maxLen = 0, start = 0;
    for (int i = 0; i < s.length(); i++) {
        char c = s.charAt(i);
        if (map.containsKey(c) && map.get(c) >= start) {
            start = map.get(c) + 1;
        }
        map.put(c, i);
        maxLen = Math.max(maxLen, i - start + 1);
    }
    return maxLen;
}

4. Product of Array Except Self

Problem

Return array where each element is product of all other elements.

Input: [1,2,3,4]
Output: [24,12,8,6]

Python

def product_except_self(nums):
    n = len(nums)
    result = [1] * n

    # Left pass
    left_product = 1
    for i in range(n):
        result[i] = left_product
        left_product *= nums[i]

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

    return result

Time: O(n), Space: O(1) excluding output

Java

public int[] productExceptSelf(int[] nums) {
    int n = nums.length;
    int[] result = new int[n];

    int leftProduct = 1;
    for (int i = 0; i < n; i++) {
        result[i] = leftProduct;
        leftProduct *= nums[i];
    }

    int rightProduct = 1;
    for (int i = n - 1; i >= 0; i--) {
        result[i] *= rightProduct;
        rightProduct *= nums[i];
    }

    return result;
}

5. Trapping Rain Water

Problem

Calculate water trapped after raining.

Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6

Python

def trap(height):
    if not height:
        return 0

    left, right = 0, len(height) - 1
    left_max = right_max = 0
    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

Time: O(n), Space: O(1)

Trees & Graphs

6. Validate Binary Search Tree

Problem

Check if binary tree is a valid BST.

Python

def is_valid_bst(root, min_val=float('-inf'), max_val=float('inf')):
    if not root:
        return True
    if root.val <= min_val or root.val >= max_val:
        return False
    return (is_valid_bst(root.left, min_val, root.val) and
            is_valid_bst(root.right, root.val, max_val))

Java

public boolean isValidBST(TreeNode root) {
    return isValidBST(root, null, null);
}

private boolean isValidBST(TreeNode node, Integer min, Integer max) {
    if (node == null) return true;
    if (min != null && node.val <= min) return false;
    if (max != null && node.val >= max) return false;
    return isValidBST(node.left, min, node.val) &&
           isValidBST(node.right, node.val, max);
}

7. Binary Tree Maximum Path Sum

Problem

Find maximum path sum in binary tree. Path can start and end at any node.

Input: [-10,9,20,null,null,15,7]
Output: 42 (15 + 20 + 7)

Python

def max_path_sum(root):
    max_sum = float('-inf')

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

        left = max(0, dfs(node.left))
        right = max(0, dfs(node.right))

        max_sum = max(max_sum, left + right + node.val)
        return max(left, right) + node.val

    dfs(root)
    return max_sum

8. Number of Islands

Problem

Count number of islands in 2D grid.

Python

def num_islands(grid):
    if not grid:
        return 0

    count = 0
    rows, cols = len(grid), len(grid[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'
        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':
                count += 1
                dfs(r, c)

    return count

Time: O(m×n), Space: O(m×n)

9. Lowest Common Ancestor

Problem

Find LCA of two nodes in binary tree.

Python

def lowest_common_ancestor(root, p, q):
    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

10. Course Schedule (Topological Sort)

Problem

Check if all courses can be finished given prerequisites.

Python

from collections import defaultdict, deque

def can_finish(num_courses, prerequisites):
    graph = defaultdict(list)
    in_degree = [0] * num_courses

    for course, prereq in prerequisites:
        graph[prereq].append(course)
        in_degree[course] += 1

    queue = deque([i for i in range(num_courses) if in_degree[i] == 0])
    completed = 0

    while queue:
        course = queue.popleft()
        completed += 1
        for next_course in graph[course]:
            in_degree[next_course] -= 1
            if in_degree[next_course] == 0:
                queue.append(next_course)

    return completed == num_courses

Dynamic Programming

11. Longest Increasing Subsequence

Problem

Find length of longest increasing subsequence.

Input: [10,9,2,5,3,7,101,18]
Output: 4

Python

import bisect

def length_of_lis(nums):
    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)

Time: O(n log n), Space: O(n)

12. Coin Change

Problem

Find minimum coins needed to make amount.

Input: coins = [1,2,5], amount = 11
Output: 3 (5+5+1)

Python

def coin_change(coins, 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

13. Word Break

Problem

Can string be segmented into dictionary words?

Input: s = “leetcode”, wordDict = [“leet”,“code”]
Output: true

Python

def word_break(s, word_dict):
    word_set = set(word_dict)
    dp = [False] * (len(s) + 1)
    dp[0] = True

    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)]

14. House Robber

Problem

Max money without robbing adjacent houses.

Input: [2,7,9,3,1]
Output: 12 (2+9+1)

Python

def rob(nums):
    if not nums:
        return 0
    if len(nums) == 1:
        return nums[0]

    prev2, prev1 = 0, 0
    for num in nums:
        curr = max(prev1, prev2 + num)
        prev2 = prev1
        prev1 = curr

    return prev1

15. Edit Distance

Problem

Minimum operations to convert word1 to word2.

Input: word1 = “horse”, word2 = “ros”
Output: 3

Python

def min_distance(word1, word2):
    m, n = len(word1), len(word2)
    dp = [[0] * (n + 1) for _ in range(m + 1)]

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

    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], dp[i][j-1], dp[i-1][j-1])

    return dp[m][n]

System Design & Design Patterns

16. LRU Cache

Problem

Design LRU Cache with O(1) get and put.

Python

from collections import OrderedDict

class LRUCache:
    def __init__(self, capacity):
        self.cache = OrderedDict()
        self.capacity = capacity

    def get(self, key):
        if key not in self.cache:
            return -1
        self.cache.move_to_end(key)
        return self.cache[key]

    def put(self, key, value):
        if key in self.cache:
            self.cache.move_to_end(key)
        self.cache[key] = value
        if len(self.cache) > self.capacity:
            self.cache.popitem(last=False)

17. Design Twitter

Problem

Design simplified Twitter with follow, unfollow, post, getNewsFeed.

Python

import heapq
from collections import defaultdict

class Twitter:
    def __init__(self):
        self.time = 0
        self.tweets = defaultdict(list)
        self.following = defaultdict(set)

    def postTweet(self, userId, tweetId):
        self.tweets[userId].append((self.time, tweetId))
        self.time += 1

    def getNewsFeed(self, userId):
        heap = []
        users = self.following[userId] | {userId}

        for user in users:
            for tweet in self.tweets[user][-10:]:
                heapq.heappush(heap, tweet)
                if len(heap) > 10:
                    heapq.heappop(heap)

        return [t[1] for t in sorted(heap, reverse=True)]

    def follow(self, followerId, followeeId):
        self.following[followerId].add(followeeId)

    def unfollow(self, followerId, followeeId):
        self.following[followerId].discard(followeeId)

Advanced Problems

18. Merge K Sorted Lists

Python

import heapq

def merge_k_lists(lists):
    heap = []
    for i, lst in enumerate(lists):
        if lst:
            heapq.heappush(heap, (lst.val, i, lst))

    dummy = ListNode(0)
    current = dummy

    while heap:
        val, i, node = heapq.heappop(heap)
        current.next = node
        current = current.next
        if node.next:
            heapq.heappush(heap, (node.next.val, i, node.next))

    return dummy.next

19. Serialize and Deserialize Binary Tree

Python

class Codec:
    def serialize(self, root):
        def dfs(node):
            if not node:
                return ['null']
            return [str(node.val)] + dfs(node.left) + dfs(node.right)
        return ','.join(dfs(root))

    def deserialize(self, data):
        nodes = iter(data.split(','))

        def dfs():
            val = next(nodes)
            if val == 'null':
                return None
            node = TreeNode(int(val))
            node.left = dfs()
            node.right = dfs()
            return node

        return dfs()

20. Find Median from Data Stream

Python

import heapq

class MedianFinder:
    def __init__(self):
        self.small = []  # max heap (negated)
        self.large = []  # min heap

    def addNum(self, num):
        heapq.heappush(self.small, -num)
        heapq.heappush(self.large, -heapq.heappop(self.small))

        if len(self.large) > len(self.small):
            heapq.heappush(self.small, -heapq.heappop(self.large))

    def findMedian(self):
        if len(self.small) > len(self.large):
            return -self.small[0]
        return (-self.small[0] + self.large[0]) / 2

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Last updated: February 2026