Minimum of Fixed-Window Maxima

You are given an integer array nums and an integer k. Consider every contiguous subarray of exactly k elements.
Compute the maximum value in each such window, then return the minimum among all of those window maxima.

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

Examples
Example 1
nums = [1,3,2,5,1,4]
k = 3
return = 3
The four window maxima are 3, 5, 5, and 5. Their minimum is 3.
Example 2
nums = [-4,-2,-7,-3]
k = 2
return = -3
The window maxima are -2, -2, and -3. The minimum is -3.
Example 3
nums = [5,1,5]
k = 1
return = 1
With k = 1, every element is its own window maximum. The smallest of 5, 1, and 5 is 1.

Constraints
1 <= nums.length.
1 <= k <= nums.length.
Every value in nums is a signed 32-bit integer.

The core is a sliding window maximum. Keep a deque of indices whose values are in decreasing order. For each new element, pop from the back while the back value is less than or equal to the new one, push the index, and pop from the front if it's outside the window. Once you've seen k elements, the front holds the window max. Track a running minimum of those maxima and return it. That's O(n) time and O(k) space. The common pitfall is the brute force that rescans every window, which is O(n*k) and dies on large arrays. Another is initializing the answer to 0 instead of the max int, which breaks on negative values like Example 2. Also check k = 1 and k = n. If the deque logic slips under pressure, StealthCoder is the hedge on the live OA, reading the problem and handing you the working version.

The honest play: practice the pattern, and have StealthCoder ready for the one you didn't see coming.