<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:
        nums.sort()
        left = deque()
        right = deque()
        cost = 0
        def add(num):
            nonlocal cost
            right.append(num)
            if len(right) > len(left):
                old_median = left[-1] if left else 0
                new_median = right[0]
                cost += len(left) * (new_median - old_median)
                cost -= (len(right) - 1) * (new_median - old_median)
                cost += num - new_median
                left.append(right.popleft())
            else:
                cost += num - left[-1]
        def shift():
            nonlocal cost
            if len(left) - 1 < len(right):
                old_median = left[-1]
                new_median = right[0]
                cost += len(left) * (new_median - old_median)
                cost -= len(right) * (new_median - old_median)
                cost -= new_median - left[0]
                left.popleft()
                left.append(right.popleft())
            else:
                num = left.popleft()
                cost -= left[-1] - num
        res = 0
        for num in nums:
            add(num)
            while cost > k:
                shift()
            res = max(res, len(left) + len(right))
        return res
31500