Cheapest Way to Gather Chips at One Position

### Problem Each chip starts at its given integer position. Moving one chip by +2 or -2 costs 0; moving it by +1 or -1 costs 1. Return the minimum total cost to bring all chips to the same position. ```cpp class Solution { public: int minCostToMoveChips(vector<int>& position); }; ``` ### Examples **Example 1:** ```text Input: position = [2,4,7] Output: 1 Explanation: Move4 to2 for free and7 through odd positions to3 for free, then3 to2 for cost1. ``` **Example 2:** ```text Input: position = [1,3,5] Output: 0 Explanation: All positions are odd and can reach1 using free steps of size2. ``` ### Constraints - There are 1 to 100 chips. - Each original position is an integer in `[1,1000000000]`. ### Contract considerations The destination is a common integer position and need not be chosen from the original positions. ```hint Read the boundary condition Can separate chips share one position, and is the objective a move count or the total paid cost? ``` ### Boundary reminders The destination is a common integer position and need not be chosen from the original positions.