Return the Negative x-th Order Statistic of Every Fixed Window
Company: Infosys
Role: Software Engineer
Category: Coding & Algorithms
Difficulty: medium
Interview Round: Other
### Problem
For each contiguous length-k window in nums, its beauty is its x-th smallest value if that value is negative; otherwise its beauty is zero. Return beauties from leftmost window to rightmost, counting duplicate entries as separate positions.
```cpp
class Solution {
public:
vector<int> getSubarrayBeauty(vector<int>& nums, int k, int x);
};
```
### Examples
**Example 1:**
```text
Input: nums = [-2,4,-1,-3], k = 3, x = 2
Output: [-1,-1]
Explanation: The sorted windows are [-2,-1,4] and [-3,-1,4]; each has -1 in position two.
```
**Example 2:**
```text
Input: nums = [0,-1,2], k = 2, x = 2
Output: [0,0]
Explanation: Each window contains fewer than two negative entries.
```
### Constraints
- `n == nums.length`; `1 <= n <= 100000`.
- `1 <= k <= n`; `1 <= x <= k`.
- `-50 <= nums[i] <= 50`.
### Clarifying Questions
- Is x one-based? Yes; `x = 1` asks for the smallest value.
- Do duplicates occupy separate sorted positions? Yes.
- Is zero negative? No.
- What if there are fewer than x negatives? Return zero for that window.
- What order is required? Leftmost window first.
```hint Check the boundary
Check a window whose x-th smallest value is zero.
```
Overview: For each length-k window, return its x-th smallest value when negative and zero otherwise, preserving window order and duplicate multiplicities.