<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()
        prefix = []
        curr = 0
        for n in nums:
            curr+=n
            prefix.append(curr)

        def getMedianAndMedianIdx(x, start):
            end = start+x-1
            mi = (end-start)/2
            return ((nums[(x//2)+start]+nums[((x//2)-1)+start])/2, mi+start)  if x%2==0 else (nums[(x//2)+start], int(mi)+start)

        def canMakeXSame(x):
            for j in range(x-1, len(nums)):
                start = j - x + 1 
                med, medIdx = getMedianAndMedianIdx(x, start)
                if x%2:
                    leftHalfSum = prefix[medIdx-1] if start==0 else prefix[medIdx-1] - prefix[start-1]
                    leftHalfCount = medIdx - start
                    rightHalfSum = prefix[j]-prefix[medIdx]
                    rightHalfCount = j - medIdx
                else:
                    leftHalfSum = prefix[int(medIdx)] if start ==0 else prefix[int(medIdx)] - prefix[start-1]
                    leftHalfCount = int(medIdx) - start + 1
                    rightHalfSum = prefix[j] - prefix[int(medIdx)]
                    rightHalfCount = j - int(medIdx)

                if (leftHalfCount * med - leftHalfSum) + (rightHalfSum - rightHalfCount * med) <=k:
                    return True
            return False
        
        low = 1
        high = len(nums)
        ans = 0
        while low<=high:
            mid = (low+high)//2
            if canMakeXSame(mid):
                ans = mid
                low = mid + 1
            else:
                high = mid - 1
        return ans
30900