2870 Minimum Number of Operations to Make Array Empty - Medium

Problem:

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

Problem Analysis:

High Level Strategy:

• As long as no element appears only once, there is always a possible solution.
• Simply consider the case of i%3
  • if the remainder is 2, it is straightforward, we can just add 1 more operations to account for the pair
  • if the remainder is 1, eg. 10%3 = 1,
    • 10 can be 5 pairs -> 5
    • 3 triplets and 1 standalone (not possible)
    • or it can be 2 triplets and 2 pairs -> 4. This is also equal to 10%3 + 1
  • Hence both conditions can be simplified to + 1.

Complexity:

• Time Complexity: The solution iterates through the counts of each number in the count_dict, which takes O(N) time, where N is the length of the input array.
• Space Complexity: Additional space is used for the count_dict, which takes O(N) space, where N is the length of the input array.

Solutions:

class Solution:
    def minOperations(self, nums: List[int]) -> int:
        count_dict = Counter(nums)
        res = 0
        
        for i in count_dict.values():
            if i == 1:
                return -1 
            res += i//3
            if i % 3:
                res += 1
        return res

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