Minimum number of operations to make array empty

class Solution:

    def minOperations(self, nums: List[int]) -> int:
        def getminans(num):
            for i in range(num, -1, -1):
                comp = num - i
                if i % 3 == 0 and comp % 2 == 0:
                    return i // 3 + comp // 2
            return -1

        counts = Counter(nums)
        ops = 0
        for i in counts.keys():
            count = counts[i]
            ans = getminans(count)
            if ans == -1:
                return -1
            else:
                ops += ans

        return ops

Minimum Number of Operations to Make Array Empty

---

You are given a 0-indexed array nums consisting of positive integers.

There are two types of operations that you can apply on the array any number of times:

• Choose two elements with equal values and delete them from the array.
• Choose three elements with equal values and delete them from the array.

Return the minimum number of operations required to make the array empty, or -1 if it is not possible.

 

Example 1:

Input: nums = [2,3,3,2,2,4,2,3,4]
Output: 4
Explanation: We can apply the following operations to make the array empty:
- Apply the first operation on the elements at indices 0 and 3. The resulting array is nums = [3,3,2,4,2,3,4].
- Apply the first operation on the elements at indices 2 and 4. The resulting array is nums = [3,3,4,3,4].
- Apply the second operation on the elements at indices 0, 1, and 3. The resulting array is nums = [4,4].
- Apply the first operation on the elements at indices 0 and 1. The resulting array is nums = [].
It can be shown that we cannot make the array empty in less than 4 operations.

Example 2:

Input: nums = [2,1,2,2,3,3]
Output: -1
Explanation: It is impossible to empty the array.

 

Constraints:

• 2 <= nums.length <= 105
• 1 <= nums[i] <= 106