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class Solution {
    public int minimumRounds(int[] tasks) {
        Map<Integer, Integer> taskCount = new HashMap<>();
        for (int task : tasks) {
            taskCount.put(task, taskCount.getOrDefault(task, 0) + 1);
        }
        int totalRound = 0;
        for (Integer count : taskCount.values()) {
            if (count == 1) {
                return -1;
            }
            if (count % 3 == 0) {
                totalRound += (count / 3);
            } else if (count % 3 == 1) {
                totalRound += ((count / 3) + 1);
            } else {
                totalRound += ((count / 3) + 1);
            }
        }
        return totalRound;
    }
}

首先统计每个task的count. 如果task的个数是1, 直接返回-1; 如果除以3能整除那就是totalRound直接加上count除以3; 如果余数是1, 那么我们需要让最后四个以两轮完成也就是(count / 3) - 1 + 2. 如果余数是2, 那么还需要额外一轮于是就是(count / 3) + 1.

时间复杂度: O(n)
空间复杂度: O(n)