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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 $nums$.

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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

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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;
    }
}

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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;
    }
};

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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
}

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function maxOperations(nums: number[], k: number): number {
    const cnt = new Map();
    let ans = 0;
    for (const x of nums) {
        if (cnt.get(k - x))