1217. Play with Chips

Here is the post.

It is the first algorithm in weekly contest 157. I spend almost 20 minutes to read and think, and then finally I learn that odd or even position should be counted, and sum of odd position and sum of even position can be used to calculate the minimum cost.

time complexity: O(N), N is the size of the array

public class Solution {
    public int MinCostToMoveChips(int[] chips) {
        var even = 0; 
        var odd = 0;
        
        foreach(var item in chips)
        {
            if(item % 2 == 0)
                even++;
            else
                odd++;             
        }
        
        return even >= odd? odd : even;
    }
}