<p>The <strong>frequency</strong> of an element is the number of times it occurs in an array.</p>

<p>You are given an integer array <code>nums</code> and an integer <code>k</code>. In one operation, you can choose an index of <code>nums</code> and increment the element at that index by <code>1</code>.</p>

<p>Return <em>the <strong>maximum possible frequency</strong> of an element after performing <strong>at most</strong> </em><code>k</code><em> operations</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,4], k = 5
<strong>Output:</strong> 3<strong>
Explanation:</strong> Increment the first element three times and the second element two times to make nums = [4,4,4].
4 has a frequency of 3.</pre>

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

<pre>
<strong>Input:</strong> nums = [1,4,8,13], k = 5
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are multiple optimal solutions:
- Increment the first element three times to make nums = [4,4,8,13]. 4 has a frequency of 2.
- Increment the second element four times to make nums = [1,8,8,13]. 8 has a frequency of 2.
- Increment the third element five times to make nums = [1,4,13,13]. 13 has a frequency of 2.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,9,6], k = 2
<strong>Output:</strong> 1
</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>5</sup></code></li>
	<li><code>1 &lt;= k &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    int maxFrequency(vector<int>& nums, int k) {
        vector<int> a;
        sort(nums.begin(),nums.end(), greater<int>());
        int l=0;
        for(int i=0;i<nums.size()-1;i++){
            a.push_back(nums[i]-nums[i+1]);
        }
        for(int i=1;i<a.size();i++){
            a[i]+=a[i-1];
        }
        int max=0,p=0,s=0;
        int curr=0;
        for(int i=0;i<a.size();i++){
            if(l>0){
                s=s+a[i]-a[l-1];
            }
            else{
                s=s+a[i];
            }
            if(s>k){
                if(curr>max){
                    max=curr;
                }
                if(l==0){
                    s-=(i-l+1)*a[l];
                }
                else{
                    s-=(i-l+1)*(a[l]-a[l-1]);
                }
                l++;
                curr=i-l;
            }
            curr++;
        }
        if(curr>max){
            max=curr;
        }

        return max+1;
    }
};
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