<p>You are given an integer array <code>nums</code> and an integer <code>k</code>.</p>

<p>In one operation, you can pick two numbers from the array whose sum equals <code>k</code> and remove them from the array.</p>

<p>Return <em>the maximum number of operations you can perform on the array</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4], k = 5
<strong>Output:</strong> 2
<strong>Explanation:</strong> 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.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,1,3,4,3], k = 6
<strong>Output:</strong> 1
<strong>Explanation:</strong> Starting with nums = [3,1,3,4,3]:
- Remove the first two 3&#39;s, then nums = [1,4,3]
There are no more pairs that sum up to 6, hence a total of 1 operation.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:

    def maxOperations(self, nums: List[int], k: int) -> int:

        result = 0
        nums.sort()

        while len(nums) > 1 and nums[0] < k:
            #print(nums)
            first = nums[0]
            complement = k - nums[0]
            #print("first == " + str(first))
            #print("complement == " + str(complement))
            nums.pop(0)
            index = self.binary_search(nums,0,len(nums) - 1,complement)
            #print("index == " + str(index))
            if index != -1:
                result +=1
                nums.pop(index)
            #print(nums)
            #print()
        
        return result

    def binary_search(self, arr: List[int], low: int, high: int, x: int) -> int:
 
        # Check base case
        if high >= low:
    
            mid = (high + low) // 2
    
            # If element is present at the middle itself
            if arr[mid] == x:
                return mid
    
            # If element is smaller than mid, then it can only
            # be present in left subarray
            elif arr[mid] > x:
                return self.binary_search(arr, low, mid - 1, x)
    
            # Else the element can only be present in right subarray
            else:
                return self.binary_search(arr, mid + 1, high, x)
    
        else:
            # Element is not present in the array
            return -1
   
    
            

        
27700