Equalize All Mirrored Pair Sums

### Problem The array `nums` has even length. One move replaces one entry by an integer in `[1,limit]`. Make every mirrored pair `(i,n-1-i)` have the same sum, and return the fewest moves required. ```cpp class Solution { public: int minMoves(vector<int>& nums, int limit); }; ``` ### Examples **Example 1:** ```text nums = [1,2,3,3], limit = 4 Output: 1 Explanation: Change the 3 at index 2 to 2. Both distinct mirrored pairs then sum to 4. ``` **Example 2:** ```text nums = [2,4,4,2], limit = 4 Output: 2 Explanation: The pair sums are 4 and 8. They cannot match after one change; two changes can make both sums 6. ``` ### Constraints - `n == nums.length`, with even `n` in `[2,10^5]`. - `1 <= nums[i] <= limit <= 10^5`. ### Clarifying Questions Each mirrored pair is considered once when computing costs. The target sum is a choice shared by every pair; only the minimum move count is returned. ```hint Inspect the structure What must be common to every mirrored pair, and what counts as one replacement? ``` ### Reasoning checkpoints - Ensure that every mirrored pair reaches one shared sum using only permitted replacement values. - Prove that the total move count is minimal, counting a changed occurrence once. ### Optional extensions How would the method change if replacement values came from an arbitrary sparse set?