LeetCode 2870: Minimum Number of Operations to Make Array Empty Solution

Master LeetCode problem 2870 (Minimum Number of Operations to Make Array Empty), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2870. Minimum Number of Operations to Make Array Empty

Medium

Array Hash Table Greedy Counting

Problem Explanation

Explanation

To solve this problem, we can use a greedy approach. We want to remove elements in pairs or triplets to empty the array. We can iterate through the array and count the occurrences of each element. Then, we can iterate through the counts and calculate the number of operations needed to remove pairs and triplets. If at any point we cannot form pairs or triplets to remove, we return -1.

The algorithmic steps are as follows:

Count the occurrences of each element in the array.
Iterate through the counts of elements and calculate the number of operations needed.
Keep track of the total number of operations required to empty the array.
Return the total number of operations or -1 if it is not possible to empty the array.

The time complexity of this approach is O(n) where n is the number of elements in the array. The space complexity is also O(n) to store the counts of elements.

Solution Code

class Solution {
    public int minOperations(int[] nums) {
        Map<Integer, Integer> countMap = new HashMap<>();
        for (int num : nums) {
            countMap.put(num, countMap.getOrDefault(num, 0) + 1);
        }
        
        int operations = 0;
        for (int count : countMap.values()) {
            if (count % 2 == 1) {
                return -1;
            }
            operations += count / 2;
        }
        
        return operations;
    }
}

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