<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) {
        int count[100001]={0};
        
        for(int i=0;i<nums.size();i++)
        {
            count[nums[i]]++;
        }
        
        sort(nums.begin(),nums.end());
        
        int i=count[nums[0]];
        int max=i;
        if(i==nums.size())
        {
            return nums.size();
        }
        
        int freq;
        for(;i<nums.size();)
        {
            freq=count[nums[i]];
            int m=k;
            for(int j=i-1;j>=0;j--)
            {
                if(nums[i]-nums[j]<=m)
                {
                    m-=(nums[i]-nums[j]);
                    freq++;
                }
                else
                {
                    break;
                }
            }
            if(max<freq)
            {
                max=freq;
            }
            i+=count[nums[i]];
            

        }
        cout<<count[1];
        return max;


    }
};
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