<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:
    def maxFrequency(self, nums: List[int], k: int) -> int:
        n = len(nums)
        nums.sort()
        l = r = 0
        mx = 0
        have = nums[0]+k
        while(r<n):
            length = (r-l+1)
            want = nums[r]*length
            if(want <= have):
                mx = max(mx,length)
                r+=1
                if(r>=n):
                    break
                have += nums[r]
            else:
                have -= nums[l]
                l+=1
        return mx


        
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