Preparing for software engineering interviews often feels like navigating a dense forest of algorithms and data structures. At the heart of this preparation are LeetCode questions, which serve as a critical benchmark for assessing a candidate's problem-solving skills and coding proficiency. Many aspiring engineers and seasoned developers alike turn to community-curated lists, with the "Blind 75" being a particularly renowned collection. This list, frequently referenced on Reddit and in technical interview prep communities, offers comprehensive coverage of the most common problem patterns. Mastering these questions provides a robust foundation, significantly increasing your readiness for technical interviews at top tech companies. This article compiles 30 such essential LeetCode questions, guiding you on why they're asked, how to approach them, and what an effective answer entails.
What Are LeetCode Questions?
LeetCode questions are coding challenges designed to test a candidate's understanding of data structures, algorithms, and general problem-solving abilities. These problems range in difficulty from easy to hard and cover a wide array of topics, including arrays, strings, linked lists, trees, graphs, dynamic programming, and more. They simulate the types of challenges that software engineers might encounter in real-world scenarios or, more directly, during a technical interview. The platform provides an online judge that compiles and runs your code against various test cases, giving instant feedback on correctness and performance. This iterative feedback loop is invaluable for refining solutions and understanding subtle edge cases. Practicing LeetCode questions is essential for building algorithmic intuition and efficient coding habits.
Why Do Interviewers Ask LeetCode Questions?
Interviewers use LeetCode-style questions to evaluate several key competencies crucial for a software engineer. Firstly, they assess your problem-solving approach: can you break down a complex problem, identify relevant data structures, and devise an efficient algorithm? Secondly, these questions gauge your coding proficiency, including syntax, logic, debugging, and writing clean, maintainable code. Thirdly, they reveal your ability to analyze time and space complexity, demonstrating an understanding of how your solution scales with different input sizes. Beyond technical skills, LeetCode problems also test your communication. Explaining your thought process, discussing trade-offs, and handling edge cases aloud are all vital aspects of a successful technical interview. These questions serve as a standardized way to compare candidates' analytical and practical coding skills.
Two Sum
Add Two Numbers
Longest Substring Without Repeating Characters
Median of Two Sorted Arrays
Longest Palindromic Substring
Container With Most Water
Merge Two Sorted Lists
Valid Parentheses
Merge Intervals
Reverse Linked List
Number of Islands
Climbing Stairs
Word Break
Binary Tree Inorder Traversal
Symmetric Tree
Maximum Subarray
House Robber
Linked List Cycle
Minimum Depth of Binary Tree
Validate Binary Search Tree
Merge k Sorted Lists
Serialize and Deserialize Binary Tree
Binary Tree Level Order Traversal
Maximum Product Subarray
Course Schedule
Word Search
Find Minimum in Rotated Sorted Array
Lowest Common Ancestor of BST
LRU Cache
Sliding Window Maximum
Preview List
1. Two Sum
Why you might get asked this:
This classic problem tests your understanding of hash maps (or hash tables) for efficient lookups. It's often an introductory problem to gauge basic data structure usage and time complexity awareness. It's foundational for many other problems.
How to answer:
Iterate through the array. For each number, calculate its "complement" (target minus the current number). Check if this complement already exists in your hash map. If not, add the current number and its index to the map.
Example answer:
The optimal approach involves using a hash map to store numbers and their indices. As you iterate through the array, for each number num, compute complement = target - num. If complement is present in the hash map, you've found the pair, returning their indices. Otherwise, add num and its index to the map. This achieves an O(n) time complexity and O(n) space complexity.
2. Add Two Numbers
Why you might get asked this:
This problem assesses your ability to manipulate linked lists, handle carry-overs in arithmetic operations, and manage pointer logic correctly. It's a good test of iterative linked list construction.
How to answer:
Create a dummy head for the result list. Iterate through both input lists simultaneously, adding corresponding digits and any carry from the previous sum. Update the carry and create new nodes for the sum's digit.
Example answer:
Initialize a dummy head node and a current pointer for the resulting linked list. Iterate while either input list has nodes or there's a carry value. Sum the digits from the current nodes (treating null as zero) and the carry. The new node's value is sum % 10, and the new carry is sum / 10. Advance pointers and the current node in the result list. This process is O(max(m, n)) time and space.
3. Longest Substring Without Repeating Characters
Why you might get asked this:
This problem is a prime example for evaluating sliding window technique and efficient character tracking, typically with a hash set or hash map. It tests string manipulation and optimization skills.
How to answer:
Use a sliding window approach with two pointers (left and right) and a hash set to store characters within the current window. Expand the right pointer, adding characters to the set. If a duplicate is found, shrink the window from the left.
Example answer:
Employ a sliding window defined by two pointers, left and right, and a hash set to track unique characters. As right advances, add s[right] to the set. If s[right] is already in the set (a duplicate), remove s[left] from the set and increment left until the duplicate is gone. Update the maximum length of the substring at each step. This yields O(n) time complexity.
4. Median of Two Sorted Arrays
Why you might get asked this:
This is a challenging problem that tests advanced binary search techniques and understanding of partitioning arrays. It requires a deeper grasp of logarithmic time complexity solutions.
How to answer:
Perform a binary search on the partitions of the two arrays such that the combined left halves contain (m+n+1)/2 elements and all elements in the left halves are less than or equal to elements in the right halves.
Example answer:
The solution requires finding a partition in both arrays such that elements to the left are less than or equal to elements to the right, and the number of elements in the left halves combined is (m+n+1)/2. This involves a binary search on the shorter array to find the correct partitionX. Calculate partitionY based on partitionX. Adjust high or low in binary search until the conditions are met. Time complexity is O(log(min(m, n))).
5. Longest Palindromic Substring
Why you might get asked this:
This problem explores string manipulation, dynamic programming, or the expand-around-center technique. It assesses your ability to handle nested loops or memoization for substring analysis.
How to answer:
The "expand around center" approach is often simplest. Iterate through each character (and between characters) as a potential center of a palindrome. Expand outwards, checking for matching characters.
Example answer:
The most intuitive approach is "expand around center." For each character in the string, consider it as the center of a potential palindrome (for odd-length palindromes). Also, consider the space between adjacent characters as a center (for even-length palindromes). Expand outwards from these centers, checking for character equality. Keep track of the longest palindrome found. This results in O(n^2) time complexity.
6. Container With Most Water
Why you might get asked this:
This problem is an excellent test of the two-pointer technique. It requires understanding how to optimize a search space by strategically moving pointers to find the maximum area.
How to answer:
Initialize two pointers, one at the beginning and one at the end of the height array. In a loop, calculate the area formed by these pointers. Move the pointer pointing to the shorter line inward, as moving the taller one won't improve the area.
Example answer:
Initialize two pointers, left at the start and right at the end of the array. Maintain a variable maxarea. In each step, calculate the current area: min(height[left], height[right]) * (right - left). Update maxarea. Move the pointer associated with the shorter height inward (left++ if height[left] < height[right], else right--). This ensures you always try to potentially increase height. The time complexity is O(n).
7. Merge Two Sorted Lists
Why you might get asked this:
A fundamental linked list problem that tests recursive or iterative merging logic. It's essential for understanding how to build new lists by combining elements from existing ones.
How to answer:
Create a dummy head for the merged list. Iterate, comparing the current nodes of both lists. Append the smaller node to the merged list and advance its pointer. Handle remaining nodes after one list is exhausted.
Example answer:
Initialize a dummy node and a current pointer to it. Iterate while both input lists have elements. Compare list1.val and list2.val, appending the smaller node to current.next and advancing the corresponding input list's pointer. After the loop, append any remaining nodes from the non-empty list. Return dummy.next. This operation takes O(m+n) time and O(1) space (if not counting output list).
8. Valid Parentheses
Why you might get asked this:
This problem introduces the classic use case for a stack data structure. It assesses your ability to match opening and closing symbols in a specific order, which is common in parsing or compiler design.
How to answer:
Use a stack. Iterate through the string. If an opening parenthesis is encountered, push it onto the stack. If a closing parenthesis is found, pop from the stack and check if it matches the current closing type.
Example answer:
Initialize an empty stack. Iterate through the input string. If an opening bracket (, {, [ is encountered, push it onto the stack. If a closing bracket ), }, ] is found, check if the stack is empty or if the top element of the stack does not match the corresponding opening bracket. If so, it's invalid. Otherwise, pop the stack. Finally, the string is valid if the stack is empty. Time complexity is O(n), space is O(n).
9. Merge Intervals
Why you might get asked this:
This problem tests sorting algorithms combined with a greedy approach to merge overlapping ranges. It's a common pattern in scheduling, calendar applications, and data aggregation.
How to answer:
First, sort the intervals based on their start times. Then, iterate through the sorted intervals, merging them if the current interval overlaps with the last merged interval in your result list.
Example answer:
Sort the input intervals array by their start times. Initialize a mergedintervals list with the first interval. Iterate from the second interval: if its start time is less than or equal to the end time of the lastmergedinterval, update lastmergedinterval's end time to the maximum of its current end and the current interval's end. Otherwise, add the current interval to mergedintervals. This takes O(n log n) due to sorting.
10. Reverse Linked List
Why you might get asked this:
A fundamental linked list manipulation problem. It assesses your ability to correctly re-point nodes, often using iterative or recursive approaches. Crucial for understanding pointer logic.
How to answer:
Iteratively, keep track of a previous node (initially null) and a current node (initially head). In each step, store current.next, then set current.next to previous, update previous to current, and current to temp_next.
Example answer:
Initialize previous to null and current to head. Iterate while current is not null. In each iteration, store current.next in a temporary variable. Then, set current.next to previous. Update previous to current, and current to the stored temporary next node. previous will hold the new head of the reversed list upon completion. This is an O(n) time and O(1) space solution.
11. Number of Islands
Why you might get asked this:
This problem is a classic graph traversal question using DFS or BFS on a 2D grid. It tests your ability to identify connected components and mark visited nodes to avoid re-counting.
How to answer:
Iterate through the grid. When you find a '1' (land), increment the island count. Then, perform a DFS or BFS from that cell to mark all connected land cells as '0' (visited) so they aren't counted again.
Example answer:
Initialize an islandcount to zero. Iterate through each cell (r, c) of the grid. If grid[r][c] is '1', increment islandcount and then initiate a Depth First Search (DFS) or Breadth First Search (BFS) from (r, c). The DFS/BFS function will change all connected '1's to '0's (or some other visited marker) to ensure they are part of the current island and not recounted. This is an O(rows * cols) time and space complexity solution.
12. Climbing Stairs
Why you might get asked this:
A foundational dynamic programming problem that illustrates the concept of overlapping subproblems and optimal substructure. It often introduces the Fibonacci sequence pattern.
How to answer:
Recognize that the number of ways to climb n stairs is the sum of ways to climb n-1 stairs (taking 1 step) and n-2 stairs (taking 2 steps). This forms a Fibonacci sequence. Use memoization or tabulation.
Example answer:
This problem can be solved using dynamic programming or by recognizing the Fibonacci sequence. dp[i] represents the number of ways to reach stair i. dp[i] = dp[i-1] + dp[i-2]. Base cases are dp[0]=1 (or dp[1]=1) and dp[1]=1 (or dp[2]=2). Build up the DP table iteratively to find dp[n]. This solution has O(n) time complexity and O(n) space (can be optimized to O(1) space).
13. Word Break
Why you might get asked this:
This problem effectively tests dynamic programming or recursion with memoization. It requires checking if a string can be segmented into dictionary words, which is common in parsing.
How to answer:
Use dynamic programming. Create a boolean dp array where dp[i] is true if s[0...i-1] can be segmented. Iterate i from 1 to n, and for each i, iterate j from 0 to i-1. If dp[j] is true and s[j...i-1] is in the dictionary, then dp[i] is true.
Example answer:
Employ dynamic programming where dp[i] is a boolean indicating if s[0...i-1] can be segmented into words from the dictionary. Initialize dp[0] to true (empty string can be segmented). Iterate i from 1 to n (string length). For each i, iterate j from 0 to i-1. If dp[j] is true and the substring s[j...i-1] is found in the wordDict (use a hash set for quick lookups), set dp[i] to true and break the inner loop. Return dp[n]. Time complexity is O(n^2 * L) where L is max word length.
14. Binary Tree Inorder Traversal
Why you might get asked this:
A fundamental tree traversal algorithm, essential for understanding tree structures. It tests recursive thinking or iterative traversal using a stack.
How to answer:
For inorder traversal (Left-Root-Right): Recursively traverse the left subtree, then visit the current node, then recursively traverse the right subtree. Iteratively, use a stack to keep track of nodes to visit.
Example answer:
For recursive inorder traversal, define a helper function that takes a node. First, call itself on node.left. Then, add node.val to the result list. Finally, call itself on node.right. The iterative approach uses a stack: push nodes and their left children until null, then pop, process, and go right. Both methods correctly visit nodes in sorted order for a BST. Time complexity is O(n), and space complexity is O(h) for recursion (stack space) or O(n) for iterative (worst-case for skew tree).
15. Symmetric Tree
Why you might get asked this:
This problem tests recursive problem-solving and understanding of tree symmetry. It requires comparing two subtrees as mirror images of each other.
How to answer:
Define a helper function isMirror(t1, t2) that checks if two subtrees are mirror images. It returns true if both are null, or if their values match AND isMirror(t1.left, t2.right) AND isMirror(t1.right, t2.left) are true.
Example answer:
Implement a helper function, say checkSymmetry(node1, node2), which recursively compares node1 and node2. The base cases are: if both are null, return true; if only one is null, return false; if their values differ, return false. Otherwise, return checkSymmetry(node1.left, node2.right) AND checkSymmetry(node1.right, node2.left). The main function calls checkSymmetry(root.left, root.right). Time and space complexity are O(n).
16. Maximum Subarray
Why you might get asked this:
A classic dynamic programming problem solved efficiently with Kadane's algorithm. It assesses your ability to optimize for contiguous subarray sums.
How to answer:
Use Kadane's algorithm: maintain currentmax (max sum ending at current position) and globalmax (overall max sum). At each element, currentmax = max(element, element + currentmax). Update globalmax with currentmax.
Example answer:
Apply Kadane's algorithm. Initialize maxsofar and currentmax to the first element. Iterate from the second element: currentmax = max(nums[i], currentmax + nums[i]). Then, maxsofar = max(maxsofar, currentmax). This ensures that current_max always holds the maximum sum ending at the current index, effectively handling negative numbers. This is an efficient O(n) time and O(1) space solution.
17. House Robber
Why you might get asked this:
Another excellent dynamic programming problem. It tests your ability to make optimal decisions based on previous choices, specifically avoiding adjacent elements.
How to answer:
Use dynamic programming. For each house i, the maximum money you can rob is either money[i] + maxmoneyfromhousesbeforei-2 OR maxmoneyfromhousesbeforei-1 (if you skip current house).
Example answer:
This is a dynamic programming problem. Let dp[i] be the maximum amount of money that can be robbed from houses up to i. For house i, you have two choices: rob it (nums[i] + dp[i-2]) or don't rob it (dp[i-1]). So, dp[i] = max(nums[i] + dp[i-2], dp[i-1]). Handle base cases for i=0 and i=1. This iterative approach achieves O(n) time and O(1) space complexity by only tracking the previous two maximums.
18. Linked List Cycle
Why you might get asked this:
This problem introduces Floyd's Cycle Detection Algorithm (Tortoise and Hare). It's a classic for testing linked list traversal and detecting cycles efficiently.
How to answer:
Use two pointers, a slow pointer moving one step at a time and a fast pointer moving two steps. If there's a cycle, they will eventually meet. If the fast pointer reaches the end (null), there's no cycle.
Example answer:
Employ the "Floyd's Cycle Detection" algorithm with two pointers: slow and fast. Initialize slow to head and fast to head.next (or both to head for slight variation). In each step, slow moves one node forward, and fast moves two nodes forward. If slow and fast meet at any point, a cycle exists. If fast or fast.next becomes null, there is no cycle. Time complexity is O(n), space is O(1).
19. Minimum Depth of Binary Tree
Why you might get asked this:
Tests tree traversal (BFS or DFS) with an emphasis on finding the shortest path. BFS is often preferred for minimum path problems as it explores level by level.
How to answer:
Use Breadth-First Search (BFS). Start with the root and a depth of 1. Use a queue to store nodes. The first time you encounter a leaf node (no children), its depth is the minimum depth.
Example answer:
The most efficient way is often Breadth-First Search (BFS). Initialize a queue with the root and its depth (1). While the queue is not empty, dequeue a node and its depth. If this node is a leaf (both left and right children are null), return its depth. Otherwise, enqueue its non-null children with depth + 1. If the tree is empty, return 0. This guarantees finding the shortest path first, resulting in O(n) time complexity and O(W) space (W is max width of tree).
20. Validate Binary Search Tree
Why you might get asked this:
This problem requires a deep understanding of BST properties and how to apply them recursively, typically using a min/max range check to ensure validity.
How to answer:
Implement a recursive helper function that takes node, minval, and maxval. For each node, check if node.val is within (minval, maxval). Recurse on left with (minval, node.val) and on right with (node.val, maxval).
Example answer:
Use a recursive approach with a helper function isValid(node, minbound, maxbound). For each node, verify that its val is strictly greater than minbound and strictly less than maxbound. Then, recursively call isValid for the left child with (minbound, node.val) and for the right child with (node.val, maxbound). Initialize the call with (root, -infinity, +infinity). This ensures all nodes conform to BST properties. Time and space complexity are O(n).
21. Merge k Sorted Lists
Why you might get asked this:
This problem combines linked lists with efficient data structures like a min-heap (priority queue). It tests your ability to manage multiple sorted streams of data.
How to answer:
Use a min-heap (priority queue). Add the head of each list to the heap. While the heap is not empty, extract the minimum element, add it to your result list, and then add its next element (if it exists) back to the heap.
Example answer:
Initialize a min-heap (priority queue). Add the head node of each non-empty list to the heap. Create a dummy head for the result list. While the heap is not empty, extract the node with the smallest value. Append this node to the result list. If the extracted node has a next node, add node.next to the heap. This leverages the heap's O(log k) extraction and insertion for k lists. Total time complexity is O(N log k), where N is total nodes, k is number of lists.
22. Serialize and Deserialize Binary Tree
Why you might get asked this:
This is a design problem that tests your ability to convert a complex data structure (tree) into a string representation and then reconstruct it. DFS (preorder) or BFS are common serialization methods.
How to answer:
For serialization, use a preorder traversal, appending node values (and a marker for null nodes) to a string, separated by delimiters. For deserialization, parse the string and reconstruct the tree recursively based on the preorder sequence.
Example answer:
For serialization, perform a preorder traversal (Root, Left, Right). Append each node's value to a string, using a special marker (e.g., "#") for null children and a delimiter (e.g., ",") to separate values. For deserialization, split the string by the delimiter into a list/queue. Recursively build the tree: if the current value is the null marker, return null; otherwise, create a node with the value and recursively build its left and right children from the remaining list. Time and space are O(n).
23. Binary Tree Level Order Traversal
Why you might get asked this:
A fundamental tree traversal, primarily solved using Breadth-First Search (BFS). It assesses your ability to process nodes level by level, often using a queue.
How to answer:
Use Breadth-First Search (BFS) with a queue. In each iteration, process all nodes currently in the queue (which represent a single level), then add their children to the queue for the next level.
Example answer:
Initialize an empty list of lists for the results and a queue with the root node. While the queue is not empty, get the number of nodes at the current level (levelsize). Create a list for the current level's values. For levelsize times, dequeue a node, add its value to the current level list, and enqueue its non-null children. After the loop, add the current level list to the results. This is O(n) time and O(W) space (max width of tree).
24. Maximum Product Subarray
Why you might get asked this:
This dynamic programming problem is similar to Maximum Subarray but complicated by negative numbers (a negative times a negative becomes positive). It requires tracking both max and min products.
How to answer:
Iterate through the array, maintaining currentmaxproduct and currentminproduct ending at the current position. Update these by considering nums[i], nums[i] current_max_product, and nums[i] currentminproduct. Keep track of the globalmaxproduct.
Example answer:
Iterate through the array, maintaining two variables: maxsofar and minsofar for the maximum and minimum products ending at the current position i. At each nums[i], the new maxsofar could be nums[i], nums[i] max_so_far, or nums[i] minsofar (if nums[i] is negative). Similarly for minsofar. Update result with the overall maximum. This handles negative numbers correctly and achieves O(n) time and O(1) space.
25. Course Schedule
Why you might get asked this:
A classic graph problem involving topological sort and cycle detection. It tests your understanding of directed acyclic graphs (DAGs) and their properties.
How to answer:
Represent courses and prerequisites as a directed graph. Use either Kahn's algorithm (BFS with in-degrees) or DFS to detect cycles. If a cycle exists, a valid course order is impossible.
Example answer:
Model the courses as a directed graph where an edge u -> v means u is a prerequisite for v. Calculate the in-degree for each course. Initialize a queue with all courses that have an in-degree of zero. While the queue is not empty, dequeue a course. For each of its neighbors, decrement their in-degree. If a neighbor's in-degree becomes zero, enqueue it. If the total number of processed courses equals numCourses, a valid order exists (no cycle); otherwise, a cycle exists. Time complexity is O(V+E).
26. Word Search
Why you might get asked this:
This problem is a typical application of backtracking and Depth-First Search (DFS) on a grid. It tests your ability to explore paths and revert choices.
How to answer:
Iterate through each cell of the board as a starting point. From each cell, initiate a DFS. In the DFS, mark the current cell as visited, then recursively explore its unvisited neighbors to match the next character of the word. Backtrack if a path doesn't lead to the word.
Example answer:
Iterate through each cell (r, c) of the board. If board[r][c] matches the first character of the word, initiate a Depth First Search (DFS) from that cell. The DFS function recursively tries to match subsequent characters by exploring adjacent cells (up, down, left, right). Before recursion, mark the current cell as visited (e.g., change its value temporarily). After the recursive call, backtrack by restoring the cell's original value. If the entire word is matched, return true. Time complexity can be roughly O(mn4^L) where L is word length.
27. Find Minimum in Rotated Sorted Array
Why you might get asked this:
A common binary search variant. It tests your ability to adapt binary search to arrays with a rotated pivot point.
How to answer:
Use a modified binary search. In a rotated sorted array, one half will always be sorted. Compare mid with right. If nums[mid] > nums[right], the minimum is in the right half. Otherwise, it's in the left half or mid.
Example answer:
Apply a modified binary search. Initialize left to 0 and right to n-1. While left < right: calculate mid. If nums[mid] > nums[right], the pivot (and minimum) must be in the right half, so left = mid + 1. Otherwise, the right half is sorted, meaning nums[mid] could be the minimum, or the minimum is in the left half including mid, so right = mid. The loop terminates when left == right, and nums[left] is the minimum. Time complexity is O(log n).
28. Lowest Common Ancestor of BST
Why you might get asked this:
This problem leverages the properties of a Binary Search Tree (BST) to efficiently find the LCA. It's a good test of recursive traversal and BST understanding.
How to answer:
Recursively traverse the BST. If both p and q are on the left side of the current node, move left. If both are on the right, move right. If one is on the left and the other on the right (or one is the current node), the current node is the LCA.
Example answer:
Start from the root. If both p and q are smaller than the root.val, the LCA must be in the left subtree, so recurse on root.left. If both p and q are larger than the root.val, the LCA must be in the right subtree, so recurse on root.right. If root.val is between p.val and q.val (inclusive), or if root is p or q, then root is the LCA. This approach is O(h) time, where h is tree height (O(log n) for balanced BST).
29. LRU Cache
Why you might get asked this:
A classic design problem that requires combining a hash map for O(1) lookups and a doubly linked list for O(1) insertion/deletion at both ends (for LRU policy).
How to answer:
Use a hash map to store key -> node mappings. The node should be part of a doubly linked list. The linked list maintains the order of usage (most recently used at one end, least recently used at the other). Implement get and put methods.
Example answer:
Implement the LRU Cache using a HashMap to store key-node mappings and a custom DoublyLinkedList to manage the order of usage. Node objects store key, value, prev, and next pointers. The DoublyLinkedList has head and tail sentinels. get(key) retrieves node from map, moves it to front of list. put(key, value) adds/updates node, moves to front; if capacity exceeded, removes tail.prev node. Both operations are O(1) average time complexity.
30. Sliding Window Maximum
Why you might get asked this:
This problem is a challenging application of the sliding window technique, often solved efficiently using a deque (double-ended queue) to maintain potential maximums.
How to answer:
Use a deque to store indices of elements in the current window. Maintain the deque such that elements are in decreasing order of value. When the window slides, remove outdated indices from the front and smaller elements from the back before adding the new element.
Example answer:
Employ a deque (double-ended queue) to store indices of elements. The deque will maintain indices in decreasing order of their corresponding values. When the window slides: 1) Remove indices from the front of the deque if they are no longer within the current window. 2) Remove indices from the back of the deque whose corresponding values are less than or equal to the current element nums[i] (as they can no longer be the maximum). 3) Add i to the back of the deque. 4) If the window is fully formed, the element at deque.front() is the maximum. This offers O(n) time complexity.
Other Tips to Prepare for a LeetCode Interview
Beyond practicing these specific problems, a holistic approach is key to excelling in LeetCode interviews. Firstly, understand the underlying data structures and algorithms thoroughly, not just memorizing solutions. As "the great programmer" [fictional quote, or specific tech leader if available] might say, "Understand the problem, then understand the solution; only then can you code." Actively participate in mock interviews to simulate real-world pressure and refine your communication skills. Tools like Verve AI Interview Copilot (https://vervecopilot.com) can provide invaluable real-time feedback on your verbal responses, helping you articulate your thought process more clearly.
Secondly, don't shy away from revisiting problems. Spaced repetition helps cement concepts and patterns. When stuck, review official solutions or community explanations, then try to re-solve independently. Verve AI Interview Copilot offers AI-powered coaching, allowing you to practice specific LeetCode questions and receive detailed feedback on your approach, code explanation, and complexity analysis. Remember, "Practice makes perfect, but perfect practice makes perfection," as Vince Lombardi supposedly said. Leverage Verve AI Interview Copilot's interactive environment to refine your responses and identify areas for improvement. Consistent practice, coupled with targeted feedback from platforms like Verve AI Interview Copilot, will significantly boost your confidence and performance.
Frequently Asked Questions
Q1: How much time should I spend on each LeetCode problem?
A1: Aim for 30-45 minutes per problem. If stuck, spend 10-15 minutes, then look at hints/solutions, and try to re-solve later.
Q2: Should I focus on breadth or depth in my LeetCode prep?
A2: Initially, breadth (cover various types). Once familiar, delve deeper into common patterns like DP or graphs.
Q3: Is it okay to look at solutions?
A3: Yes, but learn actively. Understand the logic, time/space complexity, and then attempt to implement it yourself without looking.
Q4: What languages are best for LeetCode?
A4: Python, Java, and C++ are most common. Choose one you are proficient in, as clarity and speed matter.
Q5: How many LeetCode problems are enough?
A5: There's no magic number. Quality over quantity. Mastering 75-150 core problems is often sufficient for most roles.
Q6: How do I improve my time and space complexity analysis?
A6: Practice. After solving, explicitly calculate Big O for your solution. Compare with optimal. Review common data structure complexities.
