Smallest Residue From Freely Chosen Stone Smashes

### Problem You may choose any two remaining stones to smash. Equal weights disappear; weights x<y become a single stone of weight y-x. Continue until at most one remains. Return the smallest final weight achievable, counting an empty result as 0. ```cpp class Solution { public: int lastStoneWeightII(vector<int>& stones); }; ``` ### Examples **Example 1:** ```text Input: stones = [1,2,3,6] Output: 0 Explanation: Smash6 with3 to get3, then3 with2 to get1, then the two1s vanish. ``` **Example 2:** ```text Input: stones = [2,2,5] Output: 1 Explanation:5-2=3 and3-2=1. The total is odd, so residue0 is impossible. ``` ### Constraints - The input contains 1 to 30 stones. - Each weight is an integer between 1 and 100, inclusive. ### Contract considerations Any pair can be chosen, but the returned optimum weight is unique. ```hint Read the boundary condition Is choosing the heaviest pair required here, or may each turn choose any remaining pair? ``` ### Boundary reminders Any pair can be chosen, but the returned optimum weight is unique.