2870. Minimum Number of Operations to Make Array Empty

Problem Link

Description

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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

 

Note: This question is the same as 2244: Minimum Rounds to Complete All Tasks.

Solution

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Python3

class Solution:
    def minOperations(self, nums: List[int]) -> int:
        N = len(nums)
        counter = Counter(nums)
        res = 0
 
        # v % 3 == 0, v // 3
        # v % 3 == 1, (v - 4) // 3 + 2
        # v % 3 == 2, (v - 2) // 3 + 1
 
        for v in counter.values():
            if v < 2: return -1
 
            if v % 3 == 0: 
                res += v // 3
            elif v % 3 == 1:
                res += (v - 4) // 3 + 2
            elif v % 3 == 2:
                res += (v - 2) // 3 + 1
 
        return res