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
function maxFrequencyScore(nums: number[], k: number): number {
    nums.sort((a, b) => a - b);
    const n = nums.length;
    const s: number[] = Array(n + 1).fill(0);
    for (let i = 1; i <= n; i++) {
        s[i] = s[i - 1] + nums[i - 1];
    }

    let l: number = 0;
    let r: number = n;
    while (l < r) {
        const mid: number = (l + r + 1) >> 1;
        let ok: boolean = false;
        for (let i = 0; i <= n - mid; i++) {
            const j = i + mid;
            const x = nums[Math.floor((i + j) / 2)];
            const left = (Math.floor((i + j) / 2) - i) * x - (s[Math.floor((i + j) / 2)] - s[i]);
            const right = s[j] - s[Math.floor((i + j) / 2)] - (j - Math.floor((i + j) / 2)) * x;
            if (left + right <= k) {
                ok = true;
                break;
            }
        }
        if (ok) {
            l = mid;
        } else {
            r = mid - 1;
        }
    }

    return l;
}
