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:
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<strong>Input:</strong> tasks = [2,2,3,3,2,4,4,4,4,4] <strong>Output:</strong> 4 <strong>Explanation:</strong> 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 |
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https://leetcode.com/problems/minimum-rounds-to-complete-all-tasks/description/ |
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class Solution { public int minimumRounds(int[] tasks) { //if length is less than or equals to 1 return -1 if(tasks.length<=1) return -1; Map<Integer, Integer> map = new HashMap<>(); //create a frequency map for(int task: tasks){ map.put(task, map.getOrDefault(task,0)+1); } int sum = 0; for(int count: map.values()){ //if frequenc is 1 return -1 if(count==1) return -1; //if the frequency is divisible by 3 if(count%3==0){ sum = sum + count/3; } //example count = 22 else if(count%3==1){ sum = sum + (count-4)/3+2; } //example count = 17 else if(count%3==2){ sum = sum + (count-2)/3+1; } } return sum; } } |