Check Whether the Middle Value Is Unique

# Check Whether the Middle Value Is Unique You are given an integer array `nums` of odd length. Return `true` if the value at its middle position appears exactly once in the entire array. Otherwise, return `false`. The middle position is the zero-based index `nums.length / 2`, using integer division. ### Function `bool isMiddleElementUnique(vector<int>& nums)` ### Example 1 **Input:** `nums = [9, 4, 6, 4, 9]` **Output:** `true` **Explanation:** The middle value is `6`, at index `2`, and it appears nowhere else. Repeated values `4` and `9` do not affect the answer. ### Example 2 **Input:** `nums = [7, 2, 5, 7, 3, 1, 7]` **Output:** `false` **Explanation:** The middle position is index `3`, whose value is `7`. That value also occurs at indices `0` and `6`, so it is not unique. ### Constraints - `1 <= nums.length <= 100` - `nums.length` is odd. - `1 <= nums[i] <= 100` ### Clarifications - Use the middle position in the given array, not the median after sorting. - Other values may repeat; only the middle value's frequency matters. - A one-element array always returns `true`. ### What a Strong Answer Covers - Identifying the middle position from the odd length. - Checking the middle value against every other position, including both ends. - Distinguishing repeated middle values from repetitions of unrelated values.