<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) {
        sort(nums.begin(), nums.end());
        int n = nums.size();

        vector<pair<int,int>>vec;

        int cnt = 1;
        int ele = nums[0];

        for(int i=1; i<n; i++){
            if(nums[i]==ele) cnt++;
            else{
                vec.push_back({ele, cnt});
                ele = nums[i];
                cnt = 1;
            }
        }
        vec.push_back({ele, cnt});

        int ans = 0;

        int l = 0, r = vec.size()-1;

        while(l<=r){
            if(vec[l].first + vec[r].first > k){
                r--;
            }
            else if(vec[l].first + vec[r].first < k){
                l++;
            }
            else{
                if(l!=r){
                    ans += min(vec[l].second, vec[r].second);
                }
                else{
                    ans += vec[l].second/2;
                }
                l++;
                r--;
            }

        }

        return ans;



    }
};
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