23. Sliding Window Maximum
hardAsked at N26Given an array and a window size k, return the maximum in each window as it slides left to right. N26 uses this for real-time peak-debit detection in their fraud-monitoring stream.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
You are given an array of integers nums and an integer k representing the window size. Return an array containing the maximum value in each sliding window from left to right.
Constraints
1 <= nums.length <= 10^5-10^4 <= nums[i] <= 10^41 <= k <= nums.length
Examples
Example 1
nums=[1,3,-1,-3,5,3,6,7], k=3[3,3,5,5,6,7]Example 2
nums=[1], k=1[1]Approaches
1. Recompute max per window
For each window start i, scan k elements and take the max.
- Time
- O(n*k)
- Space
- O(1)
const ans = [];
for (let i=0;i+k<=nums.length;i++) {
let m = -Infinity;
for (let j=i;j<i+k;j++) m = Math.max(m, nums[j]);
ans.push(m);
}
return ans;Tradeoff:
2. Monotonic deque
Keep a deque of indices whose values are strictly decreasing. Pop the front when it leaves the window; pop the back while it is smaller than the incoming value. The front is always the window max in O(1) amortized.
- Time
- O(n)
- Space
- O(k)
function maxSlidingWindow(nums, k) {
const dq = []; // indices, values decreasing
const ans = [];
for (let i = 0; i < nums.length; i++) {
while (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) ans.push(nums[dq[0]]);
}
return ans;
}Tradeoff:
N26-specific tips
N26 likes when you call out that the monotonic deque is what their fraud stream uses to compute rolling 60-second peak-debit values without re-scanning the buffer per tick.
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