Maximum Number of Pairs in Array

Problem

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 (Java)

class Solution {
    public int[] numberOfPairs(int[] nums) {
        Arrays.sort(nums);
        int pairs = 0;
        for (int i = 0; i < nums.length; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                nums[i] = -1;
                nums[i - 1] = -1;
                pairs++;
            }
        }
        return new int[] {pairs, nums.length - (2 * pairs)};
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).