Most Negative Number in Each Window

Given an integer array nums and an integer window length k, return one value for every contiguous subarray of length k, in left-to-right window order.
For a window that contains at least one negative value, return its most negative value: the numerically smallest value in that window. If a window contains no negative value, return 0.

Function
mostNegativeInWindows(nums: int[], k: int) → int[]

Examples
Example 1
nums = [12,-1,-7,8,-15,30,16,28]
k = 3
return = [-7,-7,-15,-15,-15,0]
The first window [12,-1,-7] contributes -7. The last window [30,16,28] has no negative value, so it contributes 0.
Example 2
nums = [5,4,3]
k = 2
return = [0,0]
Neither length-2 window contains a negative value.
Example 3
nums = [-2,-2,-1]
k = 2
return = [-2,-2]
Repeated values are retained normally. The smallest negative value in each window is -2.

Constraints
1 <= nums.length <= 200000.
1 <= k <= nums.length.
-10^9 <= nums[i] <= 10^9.

The trick: the answer for each window is min(window minimum, 0). If the smallest value is negative, return it. Otherwise return 0. Zero and positives both collapse to 0, so you never need to special-case them. Compute the window minimum with a monotonic deque of indices. Keep values increasing from front to back. Before pushing index i, pop from the back while nums[back] >= nums[i]. Pop from the front when its index is at or before i-k. Once i >= k-1, the front is the minimum, so append min(nums[front], 0). Common pitfalls: storing values instead of indices so you can't expire them, using a heap without lazy deletion, and forgetting that a window of all positives returns 0, not the smallest positive. It's O(n) time. If the deque logic slips under pressure, StealthCoder is the live-OA hedge that gives you a clean version.

If this hits your live OA and you blank, StealthCoder solves it in seconds, invisible to the proctor.