得到新鲜甜甜圈的最多组数

/*
 * 思路：
 * 1. 计算每个group对batchSize的余数，统计每个余数的数量。
 * 2. 如果余数为0的组，可以直接使顾客开心。
 * 3. 对于其他余数，通过配对使顾客开心。
 * 4. 使用深度优先搜索（DFS）或动态规划（DP）来优化配对过程。
 */

import java.util.*;
import java.util.stream.*;

public class DonutShopStream {
    public int maxHappyGroups(int batchSize, int[] groups) {
        Map<Integer, Long> count = Arrays.stream(groups)
            .mapToObj(g -> g % batchSize)
            .collect(Collectors.groupingBy(g -> g, Collectors.counting()));

        int happyGroups = count.getOrDefault(0, 0L).intValue();
        int left = 1, right = batchSize - 1;

        while (left < right) {
            int min = Math.min(count.getOrDefault(left, 0L).intValue(), count.getOrDefault(right, 0L).intValue());
            happyGroups += min;
            count.put(left, count.getOrDefault(left, 0L) - min);
            count.put(right, count.getOrDefault(right, 0L) - min);
            left++;
            right--;
        }

        if (batchSize % 2 == 0) {
            happyGroups += count.getOrDefault(batchSize / 2, 0L).intValue() / 2;
        }

        return happyGroups;
    }

    public static void main(String[] args) {
        DonutShopStream shop = new DonutShopStream();
        int batchSize = 4;
        int[] groups = {1, 3, 2, 5, 2, 2, 1, 6};
        System.out.println(shop.maxHappyGroups(batchSize, groups)); // 输出: 4
    }
}