Problem

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.

Examples

Example 1:

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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:

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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.

Solution

Method 1 – Greedy Grouping

Intuition

The key idea is to always group tasks of the same difficulty into as many groups of 3 as possible, since 3 is more efficient than 2. If a count cannot be grouped into 2 or 3, it’s impossible. For example, if there is only 1 task of a difficulty, we cannot form a valid group.

Approach

  1. Count the frequency of each task difficulty.
  2. For each frequency:
  • If the count is 1, return -1 (impossible).
  • Otherwise, greedily use as many groups of 3 as possible, and use groups of 2 for the remainder.
  1. Sum the rounds for all difficulties.

Code

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class Solution {
public:
   int minimumRounds(vector<int>& tasks) {
      unordered_map<int, int> cnt;
      for (int t : tasks) cnt[t]++;
      int ans = 0;
      for (auto& [k, v] : cnt) {
        if (v == 1) return -1;
        ans += (v + 2) / 3;
      }
      return ans;
   }
};
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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 + 2) / 3
   }
   return ans
}
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class Solution {
   public int minimumRounds(int[] tasks) {
      Map<Integer, Integer> cnt = new HashMap<>();
      for (int t : tasks) cnt.put(t, cnt.getOrDefault(t, 0) + 1);
      int ans = 0;
      for (int v : cnt.values()) {
        if (v == 1) return -1;
        ans += (v + 2) / 3;
      }
      return ans;
   }
}
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class Solution {
   fun minimumRounds(tasks: IntArray): Int {
      val cnt = mutableMapOf<Int, Int>()
      for (t in tasks) cnt[t] = cnt.getOrDefault(t, 0) + 1
      var ans = 0
      for (v in cnt.values) {
        if (v == 1) return -1
        ans += (v + 2) / 3
      }
      return ans
   }
}
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class Solution:
   def minimumRounds(self, tasks: list[int]) -> int:
      cnt: dict[int, int] = {}
      for t in tasks:
        cnt[t] = cnt.get(t, 0) + 1
      ans = 0
      for v in cnt.values():
        if v == 1:
           return -1
        ans += (v + 2) // 3
      return ans
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impl Solution {
   pub fn minimum_rounds(tasks: Vec<i32>) -> i32 {
      use std::collections::HashMap;
      let mut cnt = HashMap::new();
      for &t in &tasks {
        *cnt.entry(t).or_insert(0) += 1;
      }
      let mut ans = 0;
      for &v in cnt.values() {
        if v == 1 {
           return -1;
        }
        ans += (v + 2) / 3;
      }
      ans
   }
}

Complexity

  • Time: O(n), where n is the number of tasks.
  • Space: O(m), where m is the number of unique task difficulties.