LeetCode in Kotlin
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
numsthat 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 <= 1000 <= nums[i] <= 100
Solution
class Solution {
fun numberOfPairs(nums: IntArray): IntArray {
nums.sort()
var pairs = 0
for (i in nums.indices) {
if (i > 0 && nums[i] == nums[i - 1]) {
nums[i] = -1
nums[i - 1] = -1
pairs++
}
}
return intArrayOf(pairs, nums.size - 2 * pairs)
}
}