<p>The <strong>frequency</strong> of an element is the number of times it occurs in an array.</p>

<p>You are given an integer array <code>nums</code> and an integer <code>k</code>. In one operation, you can choose an index of <code>nums</code> and increment the element at that index by <code>1</code>.</p>

<p>Return <em>the <strong>maximum possible frequency</strong> of an element after performing <strong>at most</strong> </em><code>k</code><em> operations</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,4], k = 5
<strong>Output:</strong> 3<strong>
Explanation:</strong> Increment the first element three times and the second element two times to make nums = [4,4,4].
4 has a frequency of 3.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,4,8,13], k = 5
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are multiple optimal solutions:
- Increment the first element three times to make nums = [4,4,8,13]. 4 has a frequency of 2.
- Increment the second element four times to make nums = [1,8,8,13]. 8 has a frequency of 2.
- Increment the third element five times to make nums = [1,4,13,13]. 13 has a frequency of 2.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,9,6], k = 2
<strong>Output:</strong> 1
</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>5</sup></code></li>
	<li><code>1 &lt;= k &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    vector<int> f(vector<int>& a, vector<long long>& p, long long k) {
        int n = a.size();
        vector<int> res(n, -1);

        auto check = [&](int i, int j) {
            long long sum = p[i] - (j > 0 ? p[j - 1] : 0);
            return (long long)a[i] * (i - j + 1) <= k + sum;
        };

        for (int i = 0; i < n; ++i) {
            int l = 0, r = i;
            while (r - l > 1) {
                int m = l + (r - l) / 2;
                check(i, m) ? r = m : l = m + 1;
            }
            res[i] = check(i, l) ? l : r;
        }
        return res;
    }

    int maxFrequency(vector<int>& nums, int k) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        vector<long long> pref(n);
        pref[0] = nums[0];
        for (int i = 1; i < n; ++i)
            pref[i] = pref[i - 1] + nums[i];

        vector<int> v = f(nums, pref, k);

        int ans = 0;
        for (int i = 0; i < n; ++i) {
            ans = max(ans, i - v[i] + 1);
        }

        return ans;
    }
};
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