玩筹码

/*
 * Problem: There are n chips. The i-th chip is at position[i].
 * We can move a chip from position[i] to position[i] + 2 or position[i] - 2 with cost = 0,
 * or to position[i] + 1 or position[i] - 1 with cost = 1.
 * Our goal is to move all chips to the same position with the minimum cost.
 * 
 * Solution:
 * - Chips on even positions can be moved to other even positions with no cost.
 * - Chips on odd positions can be moved to other odd positions with no cost.
 * - Thus, the optimal strategy is to move all chips to the even or odd position with the maximum count,
 *   and then move chips from the other group to this position with cost = 1 per chip.
 */
import java.util.Arrays;

public class Solution {
    public int minCostToMoveChips(int[] position) {
        long evenCount = Arrays.stream(position)
                               .filter(pos -> pos % 2 == 0)
                               .count();
        long oddCount = position.length - evenCount;
        return (int) Math.min(evenCount, oddCount);
    }
}