<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>

from typing import List
class Solution:
    def maxFrequencyScore(self, nums: List[int], k: int) -> int:
        nums.sort()
        res = 0
        n = len(nums)

        sum_before = 0
        sum_after = 0
        left = 0
        right = 0

        for i in range(n):            
            
            if i > 0:
                sum_before += (i - left) * (nums[i] - nums[i-1])

                if i < right:
                    sum_after -= (right + 1 - i) * (nums[i] - nums[i-1])
                else:
                    right = max(right,i)
                    sum_after = 0
            
            while left < i and (right < n-1 and nums[i] - nums[left] > nums[right+1] - nums[i]) or (sum_before > k):
                sum_before -= nums[i] - nums[left]
                left += 1

            while k >= sum_after + sum_before:
                
                best_right = None
                if left == 0 and right == n-1:
                    best_right == None
                elif left == 0 or (right < n-1 and nums[i] -  nums[left-1] > nums[right+1] - nums[i]):
                    best_right = True
                else: 
                    best_right = False

                if best_right == None:
                    break
                if best_right:
                    if k - (sum_after + sum_before) >= nums[right+1] - nums[i]:
                        sum_after += nums[right+1] - nums[i]
                        right += 1
                    else:
                        break
                else:
                    if k - (sum_after + sum_before) >= nums[i] - nums[left-1]:
                        sum_before += nums[i] - nums[left-1]
                        left -= 1
                    else:
                        break
            res = max(res, right - left + 1)
        
        return res
1039