Decide the Winner of Optimal End-Pile Stone Selection

### Problem Two players alternately take an entire pile from either end of a row, with the first player moving first. Both maximize their final stone total. The number of piles is even, every pile is positive, and the total number of stones is odd. Return whether the first player wins under optimal play. ```cpp class Solution { public: bool stoneGame(vector<int>& piles); }; ``` ### Examples **Example 1:** ```text Input: piles = [2,7,4,6] Output: true Explanation: Original odd-index piles total13 and even-index piles total6. The first player can secure all piles of the larger parity. ``` **Example 2:** ```text Input: piles = [9,2] Output: true Explanation: The first player chooses9; the second receives2. ``` ### Constraints - `2 <= piles.length <= 500`, and the length is even. - `1 <= piles[i] <= 500`. - The sum of all pile values is odd. ### Contract considerations Players take whole endpoint piles; they do not remove an arbitrary number of stones or an interior pile. ```hint Read the boundary condition Consider the sums of the original even and odd positions and which player can force one parity. ``` ### Boundary reminders Players take whole endpoint piles; they do not remove an arbitrary number of stones or an interior pile.