README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/2200-2299/2244.Minimum%20Rounds%20to%20Complete%20All%20Tasks/README_EN.md	1371	Weekly Contest 289 Q2	Greedy Array Hash Table Counting

2244. Minimum Rounds to Complete All Tasks

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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

 

Note: This question is the same as 2870: Minimum Number of Operations to Make Array Empty.

Solutions

Solution 1: Hash Table

We use a hash table to count the number of tasks for each difficulty level. Then we traverse the hash table. For each difficulty level, if the number of tasks is $1$, then it is impossible to complete all tasks, so we return $-1$. Otherwise, we calculate the number of rounds needed to complete tasks of this difficulty level and add it to the answer.

Finally, we return the answer.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the tasks array.

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}

Rust

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
    }
}