Apply Operations to Maximize Frequency Score
<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>
Array; Binary Search; Prefix Sum; Sorting; Sliding Window
func maxFrequencyScore(nums []int, k int64) int {
	sort.Ints(nums)
	n := len(nums)
	s := make([]int64, n+1)
	for i := 1; i <= n; i++ {
		s[i] = s[i-1] + int64(nums[i-1])
	}

	l, r := 0, n
	for l < r {
		mid := (l + r + 1) >> 1
		ok := false

		for i := 0; i <= n-mid; i++ {
			j := i + mid
			x := int64(nums[(i+j)/2])
			left := (int64((i+j)/2-i) * x) - (s[(i+j)/2] - s[i])
			right := (s[j] - s[(i+j)/2]) - (int64(j-(i+j)/2) * x)

			if left+right <= k {
				ok = true
				break
			}
		}

		if ok {
			l = mid
		} else {
			r = mid - 1
		}
	}

	return l
}