Sliding Window

A sliding window is a contiguous slice of an array that glides across it one step at a time. The trick is that when the window moves, almost everything inside it stays the same — so instead of recomputing the slice from scratch, you add the value entering on the right and subtract the value leaving on the left. That turns what looks like nested work into a single left-to-right pass.

The naive version wastes work

Suppose you want the largest sum of any 3 neighbouring numbers in [2, 1, 5, 1, 3, 2]. The brute-force way re-adds three numbers for every position — and if the window were width k over n elements that's O(n·k) work. But consecutive windows overlap almost entirely, so re-summing throws away answers you already had.

The window trick, step by step

1. 1Sum the first window directly: arr[0..k-1].
2. 2To slide right one step, add the entering element arr[r] and subtract the leaving element arr[r-k].
3. 3That single add-and-subtract gives the new window's sum in O(1) — no re-scan.
4. 4Track the best sum seen as the window sweeps to the end.

Worked trace — max sum of 3 in [2, 1, 5, 1, 3, 2]

1. 1First window [2, 1, 5] → sum 8. best = 8.
2. 2Slide: +1 (enters), -2 (leaves) → 8 - 2 + 1 = 7. Window [1, 5, 1].
3. 3Slide: +3, -1 → 7 - 1 + 3 = 9. Window [5, 1, 3]. best = 9.
4. 4Slide: +2, -5 → 9 - 5 + 2 = 6. Window [1, 3, 2].
5. 5No more elements → answer is 9.

function maxSumK(arr, k) {
  let sum = 0;
  for (let i = 0; i < k; i++) sum += arr[i]; // first window
  let best = sum;
  for (let r = k; r < arr.length; r++) {
    sum += arr[r] - arr[r - k];  // add entering, drop leaving
    best = Math.max(best, sum);
  }
  return best;
}
// maxSumK([2, 1, 5, 1, 3, 2], 3) === 9