# 1679. Max Number of K-Sum Pairs

#### 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`

#### Solution 1:

```python
class Solution:
    def maxOperations(self, nums: List[int], k: int) -> int:
        d = defaultdict(int)
        for num in nums:
            d[num] += 1
        count = 0
        for index, num in enumerate(nums):
            diff = k - nums[index]
            if (num == diff and d[num] >= 2) or (num != diff and d[num] > 0 and d[diff] > 0):
                d[diff] -= 1
                d[num] -= 1
                count += 1
                
        return count
            
```

#### Solution 2 (Using Hint3) :

```python
class Solution:
    def maxOperations(self, nums: List[int], k: int) -> int:
        counter = Counter(nums)
        result = 0
        for num in counter:
            result += min(counter[num], counter[k-num])
        return result//2
            
```