Minimum Rounds to Complete All Tasks LeetCode Solution

Problem – Minimum Rounds to Complete All Tasks LeetCode Solution

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

Minimum Rounds to Complete All Tasks LeetCode Solution in Java

    public int minimumRounds(int[] A) {
        HashMap<Integer, Integer> count = new HashMap<>();
        for (int a : A)
            count.put(a, count.getOrDefault(a, 0) + 1);
        int res = 0;
        for (int freq : count.values()) {
            if (freq == 1) return -1;
            res += (freq + 2) / 3;
        }
        return res;
    }

Minimum Rounds to Complete All Tasks LeetCode Solution in C++

    int minimumRounds(vector<int>& A) {
        unordered_map<int, int> count;
        int res = 0, freq1;
        for (int a: A)
            ++count[a];
        for (auto& it: count) {
            if (it.second == 1) return -1;
            res += (it.second + 2) / 3;
        }
        return res;
    }

Minimum Rounds to Complete All Tasks LeetCode Solution in Python

    def minimumRounds(self, tasks):
        freq = Counter(tasks).values()
        return -1 if 1 in freq else sum((a + 2) // 3 for a in freq)

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