Max number of k sum pairs

class Solution:
    def maxOperations(self, nums: List[int], k: int) -> int:
        op=0
        left=0
        right=len(nums)-1
        nums.sort()
        while left<right:
            if nums[left]+nums[right]==k and left<right:
                op+=1
                left+=1
                right-=1
            elif nums[left]+nums[right]>k:
                right-=1
            else:
                left+=1
        return op

Max Number of K-Sum Pairs

Difficulty: Medium

You are given an integer array nums and an integer k.

In one operation, you can pick two numbers from the array whose sum equals k and remove them from the array.

Return the maximum number of operations you can perform on the array.

 

Example 1:

Input: nums = [1,2,3,4], k = 5
Output: 2
Explanation: Starting with nums = [1,2,3,4]:
- Remove numbers 1 and 4, then nums = [2,3]
- Remove numbers 2 and 3, then nums = []
There are no more pairs that sum up to 5, hence a total of 2 operations.

Example 2:

Input: nums = [3,1,3,4,3], k = 6
Output: 1
Explanation: Starting with nums = [3,1,3,4,3]:
- Remove the first two 3's, then nums = [1,4,3]
There are no more pairs that sum up to 6, hence a total of 1 operation.

 

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 109
  • 1 <= k <= 109