<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:
        nlen = len(nums)

        def find_nops(start, end):
            if start == end:
                return 0
            mid = (start + end) // 2
            sum_upper = sums[end + 1] - sums[mid + 1]
            sum_lower = (sums[mid] if (end - start + 1) % 2 == 1 else sums[mid + 1]) - sums[start]          
            return sum_upper - sum_lower

        def can_make_same(nsame):
            for i in range(0, nlen - nsame + 1):
                if find_nops(i, i + nsame - 1) <= k:
                    return True
            return False

        nums.sort()
        sums = [0] * (nlen + 1)
        for i, num in enumerate(nums):
            sums[i + 1] = sums[i] + nums[i]

        hi = nlen
        lo = 1
        while lo <= hi:
            mid = (lo + hi) // 2
            if can_make_same(mid):
                lo = mid + 1
            else:
                hi = mid - 1

        return hi


        
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