Sliding Window Maximum using JavaScript

Given an array of integers and a window size, the task is to find the maximum element in each window of the array as it slides from left to right.

Below are the approaches to solve the Sliding Window Maximum problem using JavaScript:

Brute Force Approach

The brute force approach involves iterating through all possible windows of size k and finding the maximum element in each window. A function "maxSlidingWindowBruteForce" takes an array "nums" and a window size "k", iterating through "nums" to find maximum elements in each window using "Math.max". These maximums are stored in the "result" array and then returned.

Example: The example below shows Sliding Window Maximum using JavaScript using the Brute Force Approach.

function maxSlideWinBF(nums, k) {
    const result = [];
    for (let i = 0; i <= nums.length - k; i++) {
        let max = Number.MIN_SAFE_INTEGER;
        for (let j = i; j < i + k; j++) {
            max = Math.max(max, nums[j]);
        }
        result.push(max);
    }
    return result;
}

const nums = [1, 3, -1, -3, 5, 3, 6, 7];
const k = 3;
console.log(maxSlideWinBF(nums, k));

[ 3, 3, 5, 5, 6, 7 ]

Optimal Approach

The optimal approach utilizes a deque (double-ended queue) to efficiently find the maximum element in each window as it slides from left to right. Define a function "maxSlidingWindowOptimal" with parameters "nums" and "k". Initialize an empty "result" array to store maximum elements. Use a deque to maintain indices of elements in decreasing order. Iterate through "nums", updating the deque according to the algorithm. Store max element based on deque's first element, push to "result".

Example: The example below shows Sliding Window Maximum using JavaScript using the Optimal Approach.

function maxSlideWinOpt(nums, k) {
    const result = [];
    const deque = [];
    for (let i = 0; i < nums.length; i++) {
        while (deque.length && nums[i] >= nums[deque[deque.length - 1]]) {
            deque.pop();
        }
        deque.push(i);
        if (i - deque[0] + 1 > k) {
            deque.shift();
        }
        if (i >= k - 1) {
            result.push(nums[deque[0]]);
        }
    }
    return result;
}

const nums = [1, 3, -1, -3, 5, 3, 6, 7];
const k = 3;
console.log(maxSlideWinOpt(nums, k));

[ 3, 3, 5, 5, 6, 7 ]

Time Complexity: O(n), where n is the length of the input array nums.

Space Complexity: O(k), where k is the size of the sliding window.