Easy
You are given a 0-indexed integer array nums
. In one operation, you may do the following:
nums
that are equal.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
import java.util.Arrays;
public 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)};
}
}