<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> and an integer <code>k</code>.</p>

<p>You can perform the following operation on the array <strong>at most</strong> <code>k</code> times:</p>

<ul>
	<li>Choose any index <code>i</code> from the array and <strong>increase</strong> or <strong>decrease</strong> <code>nums[i]</code> by <code>1</code>.</li>
</ul>

<p>The score of the final array is the <strong>frequency</strong> of the most frequent element in the array.</p>

<p>Return <em>the <strong>maximum</strong> score you can achieve</em>.</p>

<p>The frequency of an element is the number of occurences of that element in the array.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,6,4], k = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> We can do the following operations on the array:
- Choose i = 0, and increase the value of nums[0] by 1. The resulting array is [2,2,6,4].
- Choose i = 3, and decrease the value of nums[3] by 1. The resulting array is [2,2,6,3].
- Choose i = 3, and decrease the value of nums[3] by 1. The resulting array is [2,2,6,2].
The element 2 is the most frequent in the final array so our score is 3.
It can be shown that we cannot achieve a better score.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,4,4,2,4], k = 0
<strong>Output:</strong> 3
<strong>Explanation:</strong> We cannot apply any operations so our score will be the frequency of the most frequent element in the original array, which is 3.
</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>0 &lt;= k &lt;= 10<sup>14</sup></code></li>
</ul>

class Solution:
    def maxFrequencyScore(self, nums: List[int], k: int) -> int:
        if not k:
            return max(Counter(nums).values())
        
        nums.sort()
        pref = list(accumulate(nums, initial=0))
        l = r = ans = 0
        for i in range(len(nums)):
            c = nums[i] * (i - l) - (pref[i] - pref[l])
            c += (pref[r] - pref[i]) - nums[i] * (r - i) 
            
            while l < i and c > k:
                c -= nums[i] - nums[l]
                l += 1

            while r < len(nums) and c + (nums[r] - nums[i]) <= k:
                c += nums[r] - nums[i]
                r += 1

            while l < i < r < len(nums) and nums[i] - nums[l] > nums[r] - nums[i]:
                c -= (nums[i] - nums[l])
                c += (nums[r] - nums[i])
                l += 1
                r += 1
            
            while l and c + (nums[i] - nums[l - 1]) <= k:
                l -= 1
                c += nums[i] - nums[l]

            ans = max(ans, r - l)

        return ans







        



            
           

        return ans



        


                


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