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

```
public class Solution {
public int minimumRounds(int[] tasks) {
HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();// Creating HashMap
int a = 0;
int c = 0;
for (int i = 0; i < tasks.length; i++)// Traversing array
{
// Checking if the element is present
if (hm.containsKey(tasks[i])) {
a = hm.get(tasks[i]) + 1;
hm.put(tasks[i], a);
a = 0;
} else // If element is not present then add it to the HashMap
{
hm.put(tasks[i], 1);
}
}
// Iterating over the HashMap to count the minimum number of rounds or return -1
// if it is not possible to complete all the tasks.
for (int z : hm.keySet()) {
int d = hm.get(z);
if (d == 1) {
return -1;
} else if (d == 2) {
c += 1;
} else if (d % 3 == 0) {
c += d / 3;
} else {
c += d / 3 + 1;
}
}
return c;// returning the minimum number of rounds.
// Time Complexity: O(n)
// Space Complexity: O(n)
// End of the program.
}
}
```

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