Today I have done leetcode's January LeetCoding Challenge with cpp.

January LeetCoding Challenge 4

Description

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

Solution

class Solution {
public:
  int minOperations(vector<int>& nums) {
    map<int, int> count;
    for(auto n : nums) {
      count[n] += 1;
    }
    int answer = 0;
    for(const auto &[_, c] : count) {
      if(c < 2) return -1;
      if(c % 3 == 1) {
        answer += (c - 4) / 3 + 2;
      } else if(c % 3 == 2) {
        answer += c / 3 + 1;
      } else {
        answer += c / 3;
      }
    }
    return answer;
  }
};

// Accepted
// 747/747 cases passed (129 ms)
// Your runtime beats 36.65 % of cpp submissions
// Your memory usage beats 19.34 % of cpp submissions (85.6 MB)