21. Sliding Window Maximum
hardAsked at RappiReturn the max in each window of size k as it slides across the array — Rappi frames this as reporting the peak courier supply in a rolling time-window during dispatch matching.
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Problem
Given an array nums and a window size k, return an array of the maximum of each contiguous window of length k.
Constraints
1 <= k <= nums.length <= 10^5-10^4 <= nums[i] <= 10^4
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, scan all k elements to find the max.
- Time
- O(n*k)
- Space
- O(1)
const out = [];
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]);
out.push(m);
}
return out;Tradeoff:
2. Monotonic deque
Maintain a deque of indices whose values strictly decrease; pop the back when a larger value arrives, pop the front when it falls out of the window. Front is always the max.
- Time
- O(n)
- Space
- O(k)
function maxSlidingWindow(nums, k) {
const dq = [], out = [];
for (let i = 0; i < nums.length; i++) {
while (dq.length && dq[0] <= i - k) dq.shift();
while (dq.length && nums[dq.at(-1)] < nums[i]) dq.pop();
dq.push(i);
if (i >= k - 1) out.push(nums[dq[0]]);
}
return out;
}Tradeoff:
Rappi-specific tips
Rappi explicitly tests the monotonic-deque pattern — their rolling-supply metric uses the exact same data structure to avoid O(nk) per dispatch tick.
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