Final Weight Under Heaviest-Pair Smashing

### Problem Repeatedly take the two heaviest remaining stones. Equal weights both disappear; unequal weights x<y are replaced by one stone of weight y-x. Return the last weight, or 0 when no stone remains. ```cpp class Solution { public: int lastStoneWeight(vector<int>& stones); }; ``` ### Examples **Example 1:** ```text Input: stones = [6,3,2] Output: 1 Explanation:6 and3 become3;3 and2 then become1. ``` **Example 2:** ```text Input: stones = [4,4,2,2] Output: 0 Explanation: The two4s disappear, then the two2s disappear. ``` ### Constraints - There are at least 1 and at most 30 stones. - Each weight is an integer in `[1,1000]`. ### Contract considerations The heaviest-pair choice is mandatory, rather than a decision to optimize. ```hint Read the boundary condition Does each turn use the original heaviest pair or the stones that remain after the preceding smash? ``` ### Boundary reminders The heaviest-pair choice is mandatory, rather than a decision to optimize.