LeetCode-in-Java
2244. Minimum Rounds to Complete All Tasks
Medium
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 <= 1051 <= tasks[i] <= 109
Solution
import java.util.Arrays;
public class Solution {
public int minimumRounds(int[] tasks) {
Arrays.sort(tasks);
int rounds = 0;
int c = 1;
for (int i = 0; i < tasks.length - 1; i++) {
if (tasks[i] == tasks[i + 1]) {
c++;
} else {
if (c == 1) {
return -1;
}
if (c % 3 == 0) {
rounds += c / 3;
}
if (c % 3 >= 1) {
rounds += c / 3 + 1;
}
c = 1;
}
}
if (c == 1) {
return -1;
}
if (c % 3 == 0) {
rounds += c / 3;
}
if (c % 3 >= 1) {
rounds += c / 3 + 1;
}
return rounds;
}
}