Implement sliding window maximum (using a deque)?

Short answer: for (int i = 0; i < n; i++) { // Remove indices out of window if (deque.Count > 0 && deque.First.Value <= i - k) Follow on: deque.RemoveFirst(); // Remove smaller values from the back while (deque.Count > 0 && nums[deque.Last.Value] < nums[i]) deque.RemoveLast(); deque.AddLast(i); if (i >= k - 1)

Example code

public int[] MaxSlidingWindow(int[] nums, int k) {
if (nums == null || k <= 0) return new int[0];
int n = nums.Length;
int[] result = new int[n - k + 1];
LinkedList<int> deque = new LinkedList<int>(); // store indices
result[i - k + 1] = nums[deque.First.Value];
}
return result;
} Explanation: Use a deque to keep indexes of useful elements in current window, ensuring the front is always max.

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.