Reorder One Array to Win as Many Position Comparisons as Possible

### Problem Rearrange `nums1` to maximize the number of indices i where the rearranged value is strictly greater than `nums2[i]`. Return any maximizing permutation of nums1; nums2 stays in its original order. ```cpp class Solution { public: vector<int> advantageCount(vector<int>& nums1, vector<int>& nums2); }; ``` ### Examples **Example 1:** ```text Input: nums1 = [3,5,8], nums2 = [4,7,2] Output: [5,8,3] Explanation: All three comparisons are wins, so no permutation can improve the count. ``` **Example 2:** ```text Input: nums1 = [1,2,2], nums2 = [2,1,3] Output: [1,2,2] Explanation: The only beatable target position contains 1; either copy of 2 can beat it. No source value exceeds the targets 2 or 3, so at most one position can be won. The returned permutation achieves that one win. ``` ### Constraints - The arrays have a common length n, with `1 <= n <= 100000`. - Each element in either array is an integer between0 and1000000000, inclusive. ### Contract considerations Advantage counts strict wins, not the sum of numeric differences. Any maximum-win permutation is accepted. ```hint Read the boundary condition Which target should receive a source value that cannot beat even the smallest unmatched target? ``` ### What a Strong Answer Covers Advantage counts strict wins, not the sum of numeric differences. Any maximum-win permutation is accepted. ### Follow-up Questions Optional practice extension, separate from the original task: How could you validate optimality independently of the particular returned permutation?