Description Link to heading

2341. Maximum Number of Pairs in Array (Easy)

You are given a 0-indexed integer array nums. In one operation, you may do the following:

  • Choose two integers in nums that are equal.
  • Remove both integers from nums, forming a pair. The operation is done on nums as many times as possible. Return a 0-indexed integer array answer of size 2 where answer[0] is the number of pairs that are formed and answer[1] is the number of leftover integers in nums after doing the operation as many times as possible. Example 1:
Input: nums = [1,3,2,1,3,2,2]
Output: [3,1]
Explanation:
Form a pair with nums[0] and nums[3] and remove them from nums. Now, nums = [3,2,3,2,2].
Form a pair with nums[0] and nums[2] and remove them from nums. Now, nums = [2,2,2].
Form a pair with nums[0] and nums[1] and remove them from nums. Now, nums = [2].
No more pairs can be formed. A total of 3 pairs have been formed, and there is 1 number leftover in
nums.

Example 2:

Input: nums = [1,1]
Output: [1,0]
Explanation: Form a pair with nums[0] and nums[1] and remove them from nums. Now, nums = [].
No more pairs can be formed. A total of 1 pair has been formed, and there are 0 numbers leftover in
nums.

Example 3:

Input: nums = [0]
Output: [0,1]
Explanation: No pairs can be formed, and there is 1 number leftover in nums.

Constraints:

  • 1 <= nums.length <= 100
  • 0 <= nums[i] <= 100

Solution Link to heading

We can use a hashtable to record the times of occurrence of each element.

Code Link to heading 

class Solution {
public:
    vector<int> numberOfPairs(vector<int>& nums) {
        int arr[101] = {0};
        vector<int> res(2, 0);
        for (int i = 0; i < nums.size(); i++) {
            if (arr[nums[i]] == 0) {
                arr[nums[i]]++;
            } else {
                arr[nums[i]] = 0;
                res[0]++;
            }
        }
        for (int i = 0; i <= 100; i++) {
            if (arr[i] > 0) {
                res[1]++;
            }
        }
        return res;
    }
};