What is the time complexity of Sliding Window Maximum?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Sliding Window Maximum?

This problem uses Array, Queue, Sliding Window, Heap (Priority Queue), Monotonic Queue. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Sliding Window Maximum?

Insight 1: We need an efficient way to track the potential maximums within the current window. A naive approach of iterating through each window to find the maximum would be inefficient. Insight 2: A monotonic decreasing queue (deque) is the ideal data structure. It allows us to maintain a queue where the elements are sorted in decreasing order, ensuring the maximum element is always at the front. The queue stores indices rather than values, enabling easy out-of-window checks.

Which companies ask LeetCode 239. Sliding Window Maximum in interviews?

This problem is frequently asked at Autodesk, Booking.com, Cisco, Citadel, Coupang and 25 more companies.

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Sliding Window Maximum - LeetCode 239 Solution

Sliding Window Maximum - Complete Solution Guide

Sliding Window Maximum is LeetCode problem 239, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an array of integers nums , there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position. Return the max sliding window . Example 1: Input: nums = [1,3,-1,-3,5,3,6,7], k = 3 Output: [3,3,5,5,6,7] Explanation: Window position Max --------------- ----- [1 3 -1] -3 5 3 6 7 3 1 [3 -1 -3] 5 3 6 7 3 1 3 [-1 -3 5] 3 6 7 5 1 3 -1 [-3 5 3] 6 7 5 1

Detailed Explanation

The problem asks us to find the maximum value within a sliding window of size 'k' as it moves from left to right across an array of integers 'nums'. The window encompasses 'k' consecutive elements, and we must determine the maximum element within that window for each position as it slides. The output should be an array containing these maximum values.

Solution Approach

The solution uses a monotonic decreasing deque to keep track of the indices of potential maximum elements within the current sliding window. The deque stores indices, not the values themselves. For each element in the input array, we first remove any indices from the front of the deque that are outside the current window. Then, we remove any indices from the back of the deque whose corresponding values are smaller than the current element, as they can no longer be the maximum. Finally, we add the index of the current element to the back of the deque. The maximum element for the current window is always the element at the front of the deque.

Step-by-Step Algorithm

Step 1: Initialize an empty deque 'q' and an empty output array 'output'.
Step 2: Iterate through the input array 'nums' using an index 'i'.
Step 3: Check if the deque is not empty and if the element at the front of the deque (q[0]) is outside the current window (i.e., q[0] == i - k). If so, remove it from the front.
Step 4: While the deque is not empty and the value at the back of the deque (nums[q[-1]]) is smaller than the current value (nums[i]), remove elements from the back of the deque. This maintains the monotonic decreasing property.
Step 5: Append the current index 'i' to the back of the deque.
Step 6: If the current index 'i' is greater than or equal to k - 1 (meaning the window is full), add the value at the front of the deque (nums[q[0]]) to the 'output' array, as it represents the maximum value in the current window.
Step 7: Return the 'output' array.

Key Insights

Insight 1: We need an efficient way to track the potential maximums within the current window. A naive approach of iterating through each window to find the maximum would be inefficient.
Insight 2: A monotonic decreasing queue (deque) is the ideal data structure. It allows us to maintain a queue where the elements are sorted in decreasing order, ensuring the maximum element is always at the front. The queue stores indices rather than values, enabling easy out-of-window checks.
Insight 3: When a new element is encountered, we need to remove smaller elements from the back of the queue to maintain the monotonic property. We also need to remove elements from the front of the queue if they are no longer within the current window.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(k)

Topics

This problem involves: Array, Queue, Sliding Window, Heap (Priority Queue), Monotonic Queue.

Companies

Asked at: Autodesk, Booking.com, Cisco, Citadel, Coupang, DE Shaw, DoorDash, Flipkart, Gameskraft, Gojek, Goldman Sachs, Juspay, LINE, MakeMyTrip, Media.net, MongoDB, Nuro, Nutanix, Oracle, Palo Alto Networks, PhonePe, Roku, Rubrik, ServiceNow, Visa, Wayfair, Zenefits, Zepto, oyo, tcs.

Sliding Window Maximum | LeetCode 239

Hard

Array Queue Sliding Window Heap (Priority Queue)Monotonic Queue

Time: O(n)Space: O(k)

Solution Code

Available in:

import java.util.ArrayDeque;
import java.util.Deque;
import java.util.ArrayList;
import java.util.List;

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        List<Integer> output = new ArrayList<>();
        Deque<Integer> q = new ArrayDeque<>();

        for (int i = 0; i < nums.length; i++) {
            if (!q.isEmpty() && q.peekFirst() == i - k) {
                q.pollFirst();
            }

            while (!q.isEmpty() && nums[q.peekLast()] < nums[i]) {
                q.pollLast();
            }

            q.offerLast(i);

            if (i >= k - 1) {
                output.add(nums[q.peekFirst()]);
            }
        }

        int[] result = new int[output.size()];
        for (int i = 0; i < output.size(); i++) {
            result[i] = output.get(i);
        }
        return result;
    }
}

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Problem Info

Description

Problem 239: Sliding Window Maximum - Detailed explanation and solution approaches.

Topics

Array Queue Sliding Window Heap (Priority Queue)Monotonic Queue

Companies

AutodeskBooking.comCiscoCitadelCoupangDE ShawDoorDashFlipkart

Stats

Difficulty:Hard

Languages:4

Companies:30

Problem #:239

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize an empty deque 'q' and an empty output array 'output'.
2
Step 2: Iterate through the input array 'nums' using an index 'i'.
3
Step 3: Check if the deque is not empty and if the element at the front of the deque (q[0]) is outside the current window (i.e., q[0] == i - k). If so, remove it from the front.
4
Step 4: While the deque is not empty and the value at the back of the deque (nums[q[-1]]) is smaller than the current value (nums[i]), remove elements from the back of the deque. This maintains the monotonic decreasing property.
5
Step 5: Append the current index 'i' to the back of the deque.
6
Step 6: If the current index 'i' is greater than or equal to k - 1 (meaning the window is full), add the value at the front of the deque (nums[q[0]]) to the 'output' array, as it represents the maximum value in the current window.
7
Step 7: Return the 'output' array.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(k)

📝

Why?

The algorithm iterates through the input array once, resulting in O(n) complexity. The deque stores indices of elements within the current window, and the maximum size of the window is 'k', leading to O(k) space complexity. The monotonic queue ensures that we don't have to revisit or re-evaluate elements already present in the queue.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Priority Queue (Heap)

Use a max-heap (priority queue) to store elements within the current window. The maximum element can be retrieved in O(1) time. However, removing elements that are no longer in the window requires O(k) time or lazy deletion techniques.

Time: O(n log k), where n is the length of nums and k is the window size. Adding and removing from a heap of size k takes O(log k) time, and these operations are performed at most n times.Space: O(k), the space required to store k elements in the priority queue.

Trade-offs: While simpler to implement than the monotonic queue approach, it has a higher time complexity due to the O(log k) heap operations. Use this when ease of implementation is more important than absolute performance.

Brute Force

For each window, iterate through all 'k' elements and find the maximum. This is the most basic approach.

Time: O(n*k), where n is the length of nums and k is the window size.Space: O(1) excluding the space for output.

Trade-offs: Extremely inefficient, should only be considered for very small input sizes due to its high time complexity.

✨

Key Insights

1Insight 1: We need an efficient way to track the potential maximums within the current window. A naive approach of iterating through each window to find the maximum would be inefficient.
2Insight 2: A monotonic decreasing queue (deque) is the ideal data structure. It allows us to maintain a queue where the elements are sorted in decreasing order, ensuring the maximum element is always at the front. The queue stores indices rather than values, enabling easy out-of-window checks.
3Insight 3: When a new element is encountered, we need to remove smaller elements from the back of the queue to maintain the monotonic property. We also need to remove elements from the front of the queue if they are no longer within the current window.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Queue

FIFO structure - First In, First Out. Like a line at a store!

Sliding Window

A moving "window" over data to process subsets efficiently.

Edge Cases

Edge case 1: Empty input array: The code will correctly return an empty output array because the loop condition `i < nums.length` will be false from the start.
Edge case 2: k = 1: The code handles this case correctly. The window size is 1, so the maximum will always be the element itself, which the algorithm accurately captures.
Edge case 3: k = nums.length: The window spans the entire array. The code will correctly identify and return the maximum element in the whole array as the first element in the result.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to remove elements from the front of the deque when they fall out of the sliding window. This leads to incorrect maximum values being reported.

✗Common mistake 2: Not maintaining the monotonic property of the deque, which leads to inefficiency. The while loop that removes smaller elements from the back of the deque is crucial.

✗Common mistake 3: Storing the values themselves in the queue rather than indices. If values are stored, it is impossible to determine if the maximum is outside of the window.

✗Common mistake 4: Not handling edge cases such as an empty input array or a window size of 1.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Stock price analysis: Finding the maximum stock price within a rolling window of a certain number of days can help identify trends and potential buy/sell opportunities.

🏦Real-world scenario 2: Signal processing: Smoothing a noisy signal by finding the maximum value within a moving window. This can filter out spurious noise peaks.

🎮Real-world scenario 3: Network traffic monitoring: Identifying peak traffic periods by analyzing the maximum data flow within sliding time windows.

📱Real-world scenario 4: Weather data analysis: Finding the highest temperature within a sliding window of days to identify heatwaves.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 30, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Sliding Window Maximum"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether queue techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?