動機

在相等時不用煩惱該縮哪邊的two sum

Problem

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 = 5Output: 2Explanation: 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 = 6Output: 1Explanation: 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

Sol

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