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2244. Minimum Rounds to Complete All Tasks

Description

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

Solutions

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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
  }
  

• function minimumRounds(tasks: number[]): number {
      const cnt = new Map();
      for (const t of tasks) {
          cnt.set(t, (cnt.get(t) || 0) + 1);
      }
      let ans = 0;
      for (const v of cnt.values()) {
          if (v == 1) {
              return -1;
          }
          ans += Math.floor(v / 3) + (v % 3 === 0 ? 0 : 1);
      }
      return ans;
  }
  
  

• use std::collections::HashMap;
  
  impl Solution {
      pub fn minimum_rounds(tasks: Vec<i32>) -> i32 {
          let mut cnt = HashMap::new();
          for &t in tasks.iter() {
              let count = cnt.entry(t).or_insert(0);
              *count += 1;
          }
          let mut ans = 0;
          for &v in cnt.values() {
              if v == 1 {
                  return -1;
              }
              ans += v / 3 + (if v % 3 == 0 { 0 } else { 1 });
          }
          ans
      }
  }
  
  

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