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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 {
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:
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:
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
}