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1679. Max Number of K-Sum Pairs

Description

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

Solutions

Solution 1: Sorting

We sort $nums$. Then $l$ and $r$ point to the first and last elements of $nums$ respectively, and we compare the sum $s$ of the two integers with $k$.

• If $s = k$, it means that we have found two integers whose sum is $k$. We increment the answer and then move $l$ and $r$ towards the middle;
• If $s > k$, then we move the $r$ pointer to the left;
• If $s < k$, then we move the $l$ pointer to the right;
• We continue the loop until $l \geq r$.

After the loop ends, we return the answer.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of $nums$.

Solution 2: Hash Table

We use a hash table $cnt$ to record the current remaining integers and their occurrence counts.

We iterate over $nums$. For the current integer $x$, we check if $k - x$ is in $cnt$. If it exists, it means that we have found two integers whose sum is $k$. We increment the answer and then decrement the occurrence count of $k - x$; otherwise, we increment the occurrence count of $x$.

After the iteration ends, we return the answer.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of $nums$.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• class Solution {
      public int maxOperations(int[] nums, int k) {
          Arrays.sort(nums);
          int l = 0, r = nums.length - 1;
          int ans = 0;
          while (l < r) {
              int s = nums[l] + nums[r];
              if (s == k) {
                  ++ans;
                  ++l;
                  --r;
              } else if (s > k) {
                  --r;
              } else {
                  ++l;
              }
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      int maxOperations(vector<int>& nums, int k) {
          sort(nums.begin(), nums.end());
          int cnt = 0;
          int i = 0, j = nums.size() - 1;
          while (i < j) {
              if (nums[i] + nums[j] == k) {
                  i++;
                  j--;
                  cnt++;
              } else if (nums[i] + nums[j] > k) {
                  j--;
              } else {
                  i++;
              }
          }
          return cnt;
      }
  };
  

• class Solution:
      def maxOperations(self, nums: List[int], k: int) -> int:
          nums.sort()
          l, r, ans = 0, len(nums) - 1, 0
          while l < r:
              s = nums[l] + nums[r]
              if s == k:
                  ans += 1
                  l, r = l + 1, r - 1
              elif s > k:
                  r -= 1
              else:
                  l += 1
          return ans
  
  

• func maxOperations(nums []int, k int) int {
  	sort.Ints(nums)
  	l, r, ans := 0, len(nums)-1, 0
  	for l < r {
  		s := nums[l] + nums[r]
  		if s == k {
  			ans++
  			l++
  			r--
  		} else if s > k {
  			r--
  		} else {
  			l++
  		}
  	}
  	return ans
  }
  

• function maxOperations(nums: number[], k: number): number {
      const cnt = new Map();
      let ans = 0;
      for (const x of nums) {
          if (cnt.get(k - x)) {
              cnt.set(k - x, cnt.get(k - x) - 1);
              ++ans;
          } else {
              cnt.set(x, (cnt.get(x) | 0) + 1);
          }
      }
      return ans;
  }
  
  

• impl Solution {
      pub fn max_operations(nums: Vec<i32>, k: i32) -> i32 {
          let mut nums = nums.clone();
          nums.sort();
          let (mut l, mut r, mut ans) = (0, nums.len() - 1, 0);
          while l < r {
              match nums[l] + nums[r] {
                  sum if sum == k => {
                      ans += 1;
                      l += 1;
                      r -= 1;
                  }
                  sum if sum > k => {
                      r -= 1;
                  }
                  _ => {
                      l += 1;
                  }
              }
          }
          ans
      }
  }
  
  

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