<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:
    int maxFrequency(vector<int>& v, int k) {
        sort(v.begin(), v.end());
        int n = v.size();
        long long int prefix[n + 1];
        memset(prefix, 0, sizeof(prefix));
        prefix[0] = v[0];
        for (int i = 1; i < n; i++) {
            prefix[i] = prefix[i - 1] + (long long)v[i];
        }

        int ans = 1;
        for (int i = n - 1; i >= 0; i--) {
            // lets assume that we want to make all elements equal to v[i]
            
            int l = 1, r = i;
            while (r >= l) {
                int mid = l + (r - l) / 2;
                int start = i - 1;
                int end = (i - 1 - mid);
                long long int req = prefix[start] - (end >= 0 ? prefix[end] : 0);
                long long int temp = (long long)mid * (long long)v[i] - req;
                if (temp > k) {
                    r = mid - 1;
                } else {
                    l = mid + 1;
                    ans = max(ans, (mid + 1));
                }
            }
            // cout<<"---->"<<i<<"---->"<<ans<<endl;
        }
        return ans;
    }
};
106500