Place Even Values at Even Indices and Odd Values at Odd Indices

### Problem Exactly half the values in `nums` are even and half are odd. Rearrange the array so every index has a value of matching parity: even indices contain even values and odd indices contain odd values. Return any valid arrangement, preserving all occurrences. ```cpp class Solution { public: vector<int> sortArrayByParityII(vector<int>& nums); }; ``` ### Examples **Example 1:** ```text Input: nums = [3,2,1,4] Output: [2,3,4,1] Explanation: Indices0 and2 contain even values; indices1 and3 contain odd values. ``` **Example 2:** ```text Input: nums = [1,0] Output: [0,1] Explanation: Swapping the two values gives the required index parities. ``` ### Constraints - `2 <= nums.length <= 2*10^4`, and length is even. - Exactly half the values are even. - `0 <= nums[i] <= 1000`. ### Contract considerations The input length is even and parity counts are balanced, so a valid rearrangement always exists. ```hint Read the boundary condition Pair an odd value at an even index with an even value at an odd index. ``` ### What a Strong Answer Covers The input length is even and parity counts are balanced, so a valid rearrangement always exists. ### Optional extensions Original follow-up: perform the rearrangement in place. The reference uses only two indices and swaps. ### Follow-up Questions Optional practice extension, separate from the original task: How would you handle an input with unequal counts of odd and even values?