<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(vector<int>& nums, int k) {
        int n = nums.size();
        int res = 0;
        std::map<int, int> sp;
        vector<int> num;
        vector<int> count;

        for(int i:nums)
        {
            sp[i]++;
        }

        for(auto& i:sp)
        {
            num.push_back(i.first);
            count.push_back(i.second);
        }

        int nn = num.size();
        for(int i = 0;i < nn;i++)
        {
            if(count[i] == 0)continue;
            if(k%2 == 0 && num[i] == k/2)
            {
                res += floor(count[i]/2.0);
                continue;
            }
            for(int j = i+1;j < nn;j++)
            {
                if(num[i] + num[j] == k)
                {
                    res += min(count[i], count[j]);
                    count[j] = 0;
                }
            }
        }
        return res;
    }
};
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