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Formatted question description: https://leetcode.ca/all/2244.html

2244. Minimum Rounds to Complete All Tasks (Medium)

You are given a 0-indexed integer array tasks, where tasks[i] represents the difficulty level of a task. In each round, you can complete either 2 or 3 tasks of the same difficulty level.

Return the minimum rounds required to complete all the tasks, or -1 if it is not possible to complete all the tasks.

 

Example 1:

Input: tasks = [2,2,3,3,2,4,4,4,4,4]
Output: 4
Explanation: To complete all the tasks, a possible plan is:
- In the first round, you complete 3 tasks of difficulty level 2. 
- In the second round, you complete 2 tasks of difficulty level 3. 
- In the third round, you complete 3 tasks of difficulty level 4. 
- In the fourth round, you complete 2 tasks of difficulty level 4.  
It can be shown that all the tasks cannot be completed in fewer than 4 rounds, so the answer is 4.

Example 2:

Input: tasks = [2,3,3]
Output: -1
Explanation: There is only 1 task of difficulty level 2, but in each round, you can only complete either 2 or 3 tasks of the same difficulty level. Hence, you cannot complete all the tasks, and the answer is -1.

 

Constraints:

  • 1 <= tasks.length <= 105
  • 1 <= tasks[i] <= 109

Similar Questions:

Solution 1.

  • class Solution {
    public int minimumRounds(int[] tasks) {
        Map<Integer, Integer> cnt = new HashMap<>();
        for (int t : tasks) {
            cnt.merge(t, 1, Integer::sum);
        }
        int ans = 0;
        for (int v : cnt.values()) {
            if (v == 1) {
                return -1;
            }
            ans += v / 3 + (v % 3 == 0 ? 0 : 1);
        }
        return ans;
    }
}

  • class Solution {
public:
    int minimumRounds(vector<int>& tasks) {
        unordered_map<int, int> cnt;
        for (auto& t : tasks) {
            ++cnt[t];
        }
        int ans = 0;
        for (auto& [_, v] : cnt) {
            if (v == 1) {
                return -1;
            }
            ans += v / 3 + (v % 3 != 0);
        }
        return ans;
    }
};

  • class Solution:
    def minimumRounds(self, tasks: List[int]) -> int:
        cnt = Counter(tasks)
        ans = 0
        for v in cnt.values():
            if v == 1:
                return -1
            ans += v // 3 + (v % 3 != 0)
        return ans


  • func minimumRounds(tasks []int) int {
	cnt := map[int]int{}
	for _, t := range tasks {
		cnt[t]++
	}
	ans := 0
	for _, v := range cnt {
		if v == 1 {
			return -1
		}
		ans += v / 3
		if v%3 != 0 {
			ans++
		}
	}
	return ans
}

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