Find the Smallest Stable Index I

# Find the Smallest Stable Index I You are given an integer array `nums` of length `n` and an integer `k`. For each zero-based index `i`, define its instability score as: `max(nums[0..i]) - min(nums[i..n - 1])`. The first range contains every element from index `0` through `i`, and the second contains every element from `i` through `n - 1`. Both ranges include `nums[i]`. An index is stable when its score is at most `k`. Return the smallest stable index, or `-1` if no index is stable. ### Function `int firstStableIndex(vector<int>& nums, int k)` ### Example 1 **Input:** `nums = [8, 2, 5, 7], k = 3` **Output:** `2` **Explanation:** The instability scores at indices `0`, `1`, `2` and `3` are `6`, `6`, `3` and `1`. Index `2` is the first whose score is at most `3`. ### Example 2 **Input:** `nums = [9, 3, 1], k = 7` **Output:** `-1` **Explanation:** Every prefix maximum is `9` and every suffix minimum is `1`, so all three scores equal `8`. None is at most `7`. ### Constraints - `1 <= nums.length <= 100` - `0 <= nums[i] <= 1000000000` - `0 <= k <= 1000000000` ### Clarifications - Indices are zero-based, and a score equal to `k` qualifies. - Values may repeat. - A one-element array has score `0` at its only index. ### What a Strong Answer Covers - Including the current index in both ranges. - Returning the first qualifying index rather than an arbitrary qualifying index. - Handling equality and the case where no index qualifies.