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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

  • 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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