Last updated on October 5th, 2024 at 09:25 pm

Here, We see Sliding Window Maximum LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Sliding Window Maximum LeetCode Solution

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:

Example 2:
Input: nums = [1], k = 1
Output: [1]

1. Sliding Window Maximum LeetCode Solution C++

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        vector<int> result;
        if (k == 0) return result;
        multiset<int> w;
        for (int i = 0, n = (int)nums.size(); i < n; i++) {
            if (i >= k)
                w.erase(w.find(nums[i-k]));
            w.insert(nums[i]);
            if (i >= k-1)
                result.push_back(*w.rbegin());
        }
        return result;
    }
};

2. Sliding Window Maximum LeetCode Solution Java

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
		if (nums == null || k <= 0) {
			return new int[0];
		}
		int n = nums.length;
		int[] r = new int[n-k+1];
		int ri = 0;
		Deque<Integer> q = new ArrayDeque<>();
		for (int i = 0; i < nums.length; i++) {
			while (!q.isEmpty() && q.peek() < i - k + 1) {
				q.poll();
			}
			while (!q.isEmpty() && nums[q.peekLast()] < nums[i]) {
				q.pollLast();
			}
			q.offer(i);
			if (i >= k - 1) {
				r[ri++] = nums[q.peek()];
			}
		}
		return r;        
    }
}

3. Sliding Window Maximum LeetCode Solution JavaScript

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

4. Sliding Window Maximum Solution Python

class Solution(object):
    def maxSlidingWindow(self, nums, k):
        d = collections.deque()
        out = []
        for i, n in enumerate(nums):
            while d and nums[d[-1]] < n:
                d.pop()
            d += i,
            if d[0] == i - k:
                d.popleft()
            if i >= k - 1:
                out += nums[d[0]],
        return out