<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 {
public:
   int maxOperations(std::vector<int> &nums, int k)
  {
    std::vector<int> selected_nums;
    int min = INT_MAX;
    int max = 0;
    for (const int num : nums)
    {
      if (num < k)
        selected_nums.push_back(num);
    }
    std::sort(selected_nums.begin(), selected_nums.end());

    int left = 0;
    int right = selected_nums.size() - 1;
    int count_matching_sum = 0;
    while (left < right)
    {
      int sum_tmp = selected_nums.at(left) + selected_nums.at(right);
      if (sum_tmp == k)
      {
        count_matching_sum++;
        right--;
        left++;
      }
      else
      {
        int left_tmp = left + 1;
        while (left_tmp < right)
        {
          if (selected_nums.at(left_tmp) + selected_nums.at(right) == k)
          {
            left = left_tmp + 1;
            count_matching_sum++;
            break;
          }
          else
          {
            left_tmp++;
          }
        }
        right--;
      }
    }
   
    return count_matching_sum;
  }
    
};
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