Construct an Array with Uniform Parity

# Construct an Array with Uniform Parity You are given an array `nums1` of `n` distinct positive integers. Decide whether you can construct an array `nums2` of length `n` whose elements are either all odd or all even. For every index `i`, choose exactly one of these assignments: - Keep the value: `nums2[i] = nums1[i]`. - Choose an index `j != i` and set `nums2[i] = nums1[i] - nums1[j]`, provided that the difference is at least `1`. Every assignment uses the original values in `nums1`. Choosing a value as a subtrahend does not consume or change it, so the same index `j` may be used for several other positions. You cannot apply several subtractions to one position. Return `true` if such an array exists, and `false` otherwise. You do not need to return `nums2`. ### Function `bool uniformArray(vector<int>& nums1)` ### Example 1 **Input:** `nums1 = [9, 14, 5, 8]` **Output:** `true` **Explanation:** Keep `9` and `5`, replace `14` by `14 - 5 = 9`, and replace `8` by `8 - 5 = 3`. The resulting array `[9, 9, 5, 3]` is all odd. Repeated values in the constructed array are allowed. ### Example 2 **Input:** `nums1 = [6, 11, 18]` **Output:** `false` **Explanation:** The value `6` cannot be replaced by a positive difference, so the constructed array would have to be all even. The value `11` cannot become even: keeping it or subtracting `6` gives an odd value, while subtracting `18` is not permitted. ### Constraints - `1 <= n == nums1.length <= 100000` - `1 <= nums1[i] <= 1000000000` - All values in `nums1` are distinct. ### Clarifications - The subtraction result must be strictly positive; zero and negative results are invalid. - Only `nums1` must contain distinct values. The output values may repeat. - A one-element array already has uniform parity. ### What a Strong Answer Covers - Respecting the one-choice-per-position rule and the positivity condition. - Justifying both possible and impossible outcomes. - Handling the full input size without constructing an unnecessary output array.