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22. Sliding Window Maximum

hardAsked at Byju's

Return the max of every contiguous window of size k as it slides across the array.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Problem

You are given an array of integers nums, and there is a sliding window of size k which is moving from the very left to the very right of the array. 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.

Constraints

  • 1 <= nums.length <= 10^5
  • 1 <= k <= nums.length
  • -10^4 <= nums[i] <= 10^4

Examples

Example 1

Input
nums = [1,3,-1,-3,5,3,6,7], k = 3
Output
[3,3,5,5,6,7]

Example 2

Input
nums = [1], k = 1
Output
[1]

Approaches

1. Recompute max per window

Slide and call Math.max on each window slice.

Time
O(n*k)
Space
O(n)
const res=[];
for(let i=0;i<=nums.length-k;i++) res.push(Math.max(...nums.slice(i,i+k)));
return res;

Tradeoff:

2. Monotonic deque of indices

Maintain a deque of indices whose values are strictly decreasing. The front is always the current window's max; pop expired indices off the front and smaller values off the back.

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

Tradeoff:

Byju's-specific tips

Byju's adaptive-learning telemetry aggregates rolling-window engagement signals, so the deque pattern maps directly onto their streaming analytics work.

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Output

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