2870

2870. Minimum Number of Operations to Make Array Empty (Medium)

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

Solution 1.

Find pattern

cnt	#	operations
1	-1
2	1	2
3	1	3
4	2	22
5	2	23
6	2	33
7	3	223
8	3	233
9	3	333
10	4	2233

So, if the frequency cnt of a unique number is:

• 1, return -1
• cnt % 3 == 0, add answer by cnt / 3
• otherwise, add answer by cnt / 3 + 1

// OJ: https://leetcode.com/problems/minimum-number-of-operations-to-make-array-empty
// Author: github.com/lzl124631x
// Time: O()
// Space: O()
class Solution {
public:
    int minOperations(vector<int>& A) {
        unordered_map<int, int> m;
        for (int n : A) m[n]++;
        int ans = 0;
        for (auto &[n, cnt] : m) {
            if (cnt == 1) return -1;
            if (cnt % 3 == 0) ans += cnt / 3;
            else ans += cnt / 3 + 1;
        }
        return ans;
    }
};