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
class Solution {
public:
    int maxFrequencyScore(vector<int>& nums, long long k) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        vector<long long> s(n + 1, 0);
        for (int i = 1; i <= n; i++) {
            s[i] = s[i - 1] + nums[i - 1];
        }

        int l = 0, r = n;
        while (l < r) {
            int mid = (l + r + 1) >> 1;
            bool ok = false;

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

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

            if (ok) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }

        return l;
    }
};