The job hunt can feel like a marathon, with technical interviews acting as one of the toughest hurdles. You’ve put in the hours learning algorithms and data structures, but sitting across from an interviewer who’s asking you to solve a problem on the spot? That’s where many candidates freeze. I’ve coached hundreds of job seekers through this exact situation, and I can tell you that preparation makes all the difference.
With the right approach to LeetCode-style questions, you can turn these challenges into opportunities to shine. Let’s walk through 15 of the most common LeetCode interview questions and arm you with strategies that will help you stand out from other candidates.
LeetCode Interview Questions & Answers
LeetCode problems form the backbone of technical interviews at many top companies. These questions test your problem-solving abilities and coding skills in ways that directly relate to on-the-job challenges.
1. Two Sum
Interviewers love this question because it tests your ability to think about optimizing solutions. The Two Sum problem asks you to find two numbers in an array that add up to a specific target. Companies use this to assess how you approach problem-solving and your understanding of time complexity.
A naive approach would use nested loops, checking every possible pair of numbers. But this shows interviewers you’re not thinking about efficiency. Instead, consider using a hash map to store values you’ve already seen, which reduces the time complexity from O(n²) to O(n).
This question also gives you a chance to demonstrate your communication skills. Talk through your thought process, explaining why your solution is optimal and how you’re handling edge cases like duplicate numbers or an empty array.
Sample Answer: I’d approach this by using a hash map to store values I’ve already seen. For each number in the array, I’d check if the target minus the current number exists in the hash map. If it does, I’ve found my pair. If not, I add the current number to the hash map and continue. This gives us a time complexity of O(n) since we only need to go through the array once, and a space complexity of O(n) for storing the hash map. For example, with the array [2,7,11,15] and target 9, I’d first check if (9-2=7) is in my hash map. It’s not, so I add 2 to the map. Then I check if (9-7=2) is in the map, which it is, so I return the indices [0,1].
2. Valid Parentheses
This question evaluates your understanding of stack data structures. You need to determine if a string containing just the characters ‘(‘, ‘)’, ‘{‘, ‘}’, ‘[‘, ‘]’ is valid, meaning open brackets must be closed by the same type of brackets and in the correct order.
Stack data structures are perfect for this problem because they follow the Last-In-First-Out principle, which matches how nested parentheses work. You push opening brackets onto the stack and pop them off when you find matching closing brackets.
The key insight is recognizing that the most recent opening bracket needs to match the next closing bracket. If you can articulate this clearly, you’ll show that you understand not just how to use a stack, but why it’s the right tool for this job.
Sample Answer: I would use a stack data structure for this problem. I’ll iterate through each character in the string. If I encounter an opening bracket (‘(‘, ‘{‘, or ‘[‘), I’ll push it onto the stack. If I find a closing bracket (‘)’, ‘}’, or ‘]’), I’ll check if the stack is empty. If it is, that means there’s no matching opening bracket, so I return false. Otherwise, I pop the top element from the stack and check if it matches the current closing bracket. If they don’t match, I return false. After processing all characters, I check if the stack is empty – if it is, all brackets were properly matched and I return true. Otherwise, some opening brackets weren’t closed, so I return false.
3. Reverse Linked List
Reversing a linked list tests your understanding of pointer manipulation and linked data structures. You need to reverse the direction of all pointers in a singly linked list, which requires careful tracking of multiple nodes.
This question appears frequently because linked lists are common in real-world applications where elements need to be inserted or removed efficiently. It also tests your ability to visualize data structures and manipulate them step by step.
The trick is to maintain three pointers: one to the previous node, one to the current node, and one to the next node. This allows you to reverse the direction of each pointer without losing track of where you are in the list.
Sample Answer: To reverse a linked list, I need to iterate through the list while changing the direction of each pointer. I’ll use three variables: prev (initially null), current (pointing to the head), and next. For each node, I first save the next node, then change current’s next pointer to point to prev instead. Then I move prev and current forward by one step. After the loop completes, prev will point to the new head of the reversed list. This approach has O(n) time complexity and O(1) space complexity since I’m modifying the list in-place without using extra space proportional to the input size.
4. Maximum Subarray
The Maximum Subarray problem asks you to find the contiguous subarray with the largest sum. This question evaluates your ability to recognize dynamic programming opportunities and optimize for efficient solutions.
Interviewers use this to test if you can break down complex problems into simpler subproblems. A naive solution would check all possible subarrays, but that would be inefficient with a time complexity of O(n²).
Instead, Kadane’s algorithm provides an elegant solution that runs in O(n) time. The key insight is that for any position, the maximum subarray ending at that position is either the current element alone or the maximum subarray ending at the previous position plus the current element.
Sample Answer: I would solve this using Kadane’s algorithm, which is an efficient way to find the maximum subarray sum in O(n) time. I’ll maintain two variables: currentSum (the maximum sum ending at the current position) and maxSum (the overall maximum subarray sum found so far). For each element in the array, I update currentSum to be the maximum of the current element alone or currentSum plus the current element. Then I update maxSum if currentSum is larger. This works because at each step, we’re deciding whether to start a new subarray or continue the existing one. For the array [-2,1,-3,4,-1,2,1,-5,4], we’d get a maximum subarray of [4,-1,2,1] with a sum of 6.
5. Merge Two Sorted Lists
This problem asks you to merge two sorted linked lists into a single sorted list. It tests your ability to work with linked data structures and implement a merge algorithm efficiently.
Interviewers like this question because it combines linked list manipulation with merge logic, which is fundamental to many algorithms like merge sort. The problem evaluates both your technical knowledge and your attention to detail.
The key to solving this efficiently is to maintain a pointer to the current position in each list and repeatedly choose the smaller value to add to the result list. This approach mimics the merge step in merge sort and runs in O(n+m) time where n and m are the lengths of the input lists.
Sample Answer: I’d solve this recursively or iteratively, but I’ll describe the iterative approach for clarity. First, I create a dummy head node to simplify edge cases. Then I keep two pointers, one for each list, and compare the current nodes from both lists. I take the smaller value, connect it to my result list, and advance that list’s pointer. I continue this process until I’ve exhausted one of the lists, then append any remaining nodes from the non-empty list. This gives us a time complexity of O(n+m) where n and m are the lengths of the input lists, and space complexity of O(1) since we’re reusing the original nodes.
6. Binary Tree Level Order Traversal
This question asks you to return the level order traversal of a binary tree’s values as an array of arrays, where each inner array contains the values of nodes at that level. It tests your understanding of tree data structures and breadth-first search algorithms.
Employers value this question because trees appear in many real-world scenarios, from file systems to organization charts. Your ability to traverse and process tree structures efficiently demonstrates important problem-solving skills.
A breadth-first approach using a queue is perfect for this problem. The queue helps you process nodes level by level, and you can keep track of the current level’s size to group nodes correctly.
Sample Answer: I would use a breadth-first search approach with a queue data structure. First, I check if the root is null – if so, I return an empty array. Otherwise, I initialize a queue with the root node and create a result array. For each level, I determine how many nodes are currently in the queue, which represents the number of nodes at the current level. I process exactly that many nodes, adding their values to the current level’s array and enqueueing their children for the next level. This continues until the queue is empty. The approach has a time and space complexity of O(n), where n is the number of nodes in the tree, since we visit each node exactly once and store all node values.
7. Valid Palindrome
This problem asks you to determine if a string is a palindrome, considering only alphanumeric characters and ignoring cases. It tests your string manipulation skills and attention to detail.
Interviewers use this question to see how you handle string processing and edge cases. While it seems simple, there are several approaches with different trade-offs in terms of code cleanliness and efficiency.
The two-pointer technique works well here, with one pointer starting from the beginning and another from the end, moving toward each other while skipping non-alphanumeric characters. The pointers should always find matching characters until they meet in the middle.
Sample Answer: I would use the two-pointer approach for this problem. First, I convert the string to lowercase. Then I initialize two pointers, one at the beginning and one at the end of the string. While the pointers haven’t met, I check if both characters are alphanumeric. If not, I skip that character and move the appropriate pointer. Once both pointers are at alphanumeric characters, I compare them. If they don’t match, the string isn’t a palindrome. If all comparisons pass until the pointers meet or cross, then the string is a palindrome. This approach has O(n) time complexity and O(1) space complexity since we’re just using two pointers regardless of string length.
8. Climbing Stairs
This problem asks you how many distinct ways you can climb a staircase with n steps if you can take either 1 or 2 steps at a time. It introduces you to dynamic programming concepts in a simple context.
Interviewers like this question because it tests your ability to recognize patterns and optimize recursive solutions. A naive recursive solution would have exponential time complexity, which is inefficient for larger inputs.
The key insight is that the number of ways to reach step n equals the sum of ways to reach step n-1 and step n-2. This creates a Fibonacci-like sequence that can be computed efficiently using dynamic programming.
Sample Answer: This is a classic dynamic programming problem that follows the Fibonacci sequence. The number of ways to reach the nth step equals the number of ways to reach the (n-1)th step plus the number of ways to reach the (n-2)th step. This is because you can reach the nth step by either taking one step from the (n-1)th step or two steps from the (n-2)th step. I’d use a bottom-up approach with an array to store the number of ways to reach each step. After initializing dp[1] = 1 and dp[2] = 2, I’d compute each subsequent value as dp[i] = dp[i-1] + dp[i-2]. The final value dp[n] gives us our answer. This approach has O(n) time complexity and O(n) space complexity, though we can optimize space to O(1) by just keeping track of the two most recent values.
9. Find First and Last Position of Element in Sorted Array
This problem asks you to find the starting and ending position of a given target value in a sorted array. If the target isn’t found, you should return [-1, -1]. The challenge is to achieve O(log n) time complexity.
Interviewers use this to test your understanding of binary search and your ability to modify standard algorithms for specific requirements. Many candidates can implement basic binary search, but adapting it to find a range requires deeper understanding.
The solution involves performing two binary searches: one to find the leftmost occurrence and another to find the rightmost occurrence of the target. The key is in how you handle the case when you find the target value.
Sample Answer: I would approach this by performing two binary searches: one to find the first occurrence and another to find the last occurrence of the target. For the first occurrence, when I find the target, I continue searching on the left half to see if there are earlier occurrences. Similarly, for the last occurrence, I continue searching on the right half after finding the target. Both searches follow the binary search pattern of dividing the search space in half at each step, which gives us a time complexity of O(log n). If the target isn’t found during the first search, we can immediately return [-1, -1]. This approach is efficient and handles all edge cases, including when the target appears multiple times or not at all in the array.
10. Container With Most Water
This problem gives you an array where each element represents the height of a vertical line. You need to find two lines that, together with the x-axis, form a container that holds the maximum amount of water.
Interviewers like this question because it tests your ability to think spatially and recognize that a brute force approach (checking all pairs of lines) would be inefficient at O(n²).
The optimal solution uses the two-pointer technique, starting with the widest possible container and moving inward strategically. This gives us an O(n) time complexity solution, which is much more efficient.
Sample Answer: I would use the two-pointer technique for this problem. I start with two pointers, one at the beginning and one at the end of the array. These represent the leftmost and rightmost lines of my container. I calculate the area as the width (distance between pointers) multiplied by the height (the minimum of the two heights at those positions). Then I move the pointer that points to the shorter line inward, because moving the pointer with the taller line would only decrease the area. I continue this process, keeping track of the maximum area seen so far, until the pointers meet. This approach works in O(n) time complexity and O(1) space complexity. It’s optimal because we strategically eliminate containers that couldn’t possibly have a larger area than what we’ve already found.
11. Search in Rotated Sorted Array
This problem presents you with a sorted array that has been rotated at some unknown pivot, and asks you to find a target value in O(log n) time. It tests your ability to adapt the binary search algorithm to a more complex scenario.
Interviewers use this question to evaluate your problem-solving skills when standard algorithms need modification. Many candidates immediately recognize binary search as the approach, but struggle with the rotation aspect.
The key insight is that despite the rotation, at least one half of the array must still be sorted. By checking which half is sorted and whether the target falls within that range, you can determine which half to search next.
Sample Answer: I would modify the standard binary search algorithm for this problem. At each step, I determine the middle element and check if it equals the target. If not, I need to figure out which half of the array to search next. I can determine which half is sorted by comparing the middle element with the leftmost element. If nums[mid] >= nums[left], the left half is sorted. Otherwise, the right half is sorted. Once I know which half is sorted, I check if the target falls within that sorted range. If it does, I search that half; otherwise, I search the other half. This continues until I find the target or determine it’s not present. The approach maintains the O(log n) time complexity of binary search while handling the rotation complication.
12. Longest Substring Without Repeating Characters
This problem asks you to find the length of the longest substring without repeating characters. It tests your ability to work with strings and use sliding window techniques efficiently.
Interviewers value this question because it evaluates your understanding of string manipulation and optimization techniques. A naive approach would check all possible substrings, leading to O(n³) time complexity, which is highly inefficient.
The sliding window approach provides an elegant O(n) solution. You maintain a window of unique characters and expand it as long as adding the next character doesn’t create a duplicate. When a duplicate would be created, you shrink the window from the left until the duplicate is removed.
Sample Answer: I would use the sliding window technique with a hash map or set to solve this problem. I maintain two pointers, left and right, defining the current substring without repeating characters. As I move the right pointer forward, I check if the character is already in my current window. If not, I add it and update the maximum length if needed. If it is a duplicate, I move the left pointer forward until the duplicate is no longer in my window. The hash map or set helps me check for duplicates in O(1) time. For the string “abcabcbb”, the longest substring without repeating characters is “abc” with a length of 3. This approach has O(n) time complexity where n is the length of the string, as each character is processed at most twice (once when added to the window and once when removed).
13. Number of Islands
This problem asks you to count the number of islands in a 2D grid, where 1 represents land and 0 represents water. An island is formed by connecting adjacent lands horizontally or vertically.
Interviewers use this question to assess your understanding of graph traversal algorithms. It tests your ability to explore connected components in a grid, which is applicable to many real-world problems.
The solution typically involves using depth-first search (DFS) or breadth-first search (BFS) to explore each island fully when you encounter a land cell. After exploring an island, you mark all its cells as visited to avoid counting them again.
Sample Answer: I would solve this using depth-first search (DFS). First, I iterate through each cell in the grid. When I find a ‘1’ (land), I increment my island count and start a DFS from that cell to mark all connected land cells as visited (by changing them to ‘0’ or another marker). The DFS explores all four adjacent cells (up, down, left, right) and continues recursively for any land cells found. This approach ensures each island is counted exactly once, as all cells belonging to an island are marked during the first encounter. The time complexity is O(m×n) where m and n are the grid dimensions, as each cell is visited at most once. The space complexity is O(m×n) in the worst case due to the call stack for DFS, though this can be optimized to O(min(m,n)) with careful implementation.
14. LRU Cache
This problem asks you to implement a Least Recently Used (LRU) cache with O(1) average time complexity for both get and put operations. It tests your understanding of data structures and how to combine them for specific use cases.
Interviewers like this question because it evaluates your ability to design systems with performance constraints. LRU caches are common in real-world applications where you need to cache data with limited memory.
The solution typically combines a hash map for O(1) lookups with a doubly linked list to maintain the order of elements based on their recent usage. This combination allows for efficient updates to the access order.
Sample Answer: I would implement the LRU Cache using a combination of a hash map and a doubly linked list. The hash map provides O(1) lookups by mapping keys to nodes in the linked list. The doubly linked list maintains the order of elements based on their recency of use, with the most recently used item at the head and the least recently used at the tail. When get is called, I look up the key in the hash map. If found, I move that node to the head of the list (marking it as most recently used) and return its value. If not found, I return -1. When put is called, I check if the key already exists. If it does, I update the value and move the node to the head. If not, I create a new node at the head and add it to the hash map. If this exceeds the capacity, I remove the tail node (least recently used) from both the list and the hash map. This design ensures both operations have O(1) time complexity.
15. Merge Intervals
This problem asks you to merge all overlapping intervals in an array of intervals. It tests your ability to handle ranges and implement sorting-based algorithms.
Interviewers use this question to evaluate how you approach problems that require organizing and processing data in a specific order. It’s also relevant to many real-world scheduling and resource allocation scenarios.
The solution involves sorting the intervals by their start times and then merging overlapping intervals in a single pass. This requires careful handling of edge cases and a clear understanding of how to identify and combine overlapping ranges.
Sample Answer: I would start by sorting the intervals by their start times to make overlapping intervals adjacent in the array. Then I initialize the result with the first interval and iterate through the remaining intervals. For each interval, I check if it overlaps with the last interval in my result. An overlap occurs when the current interval’s start time is less than or equal to the end time of the last interval in the result. If there’s an overlap, I merge them by updating the end time of the last interval to be the maximum of the two end times. If there’s no overlap, I simply add the current interval to the result. This approach has a time complexity of O(n log n) due to the sorting step, and a space complexity of O(n) to store the result. For example, with intervals [[1,3],[2,6],[8,10],[15,18]], the merged result would be [[1,6],[8,10],[15,18]].
Wrapping Up
Mastering these LeetCode interview questions puts you on the path to technical interview success. The key is not just knowing the solutions, but understanding the underlying patterns and strategies that make them effective.
Practice these questions regularly, but go beyond memorization. Focus on the problem-solving process, the ability to communicate your approach clearly, and the confidence to handle variations of these problems. With deliberate practice and the right mindset, you’ll be ready to tackle any technical challenge that comes your way.