# 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 $nums$.

• 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
}
}