<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>& nums, int k) {
        sort(nums.begin(), nums.end());
        queue<pair<int, int>> queue;
        int queuecount = 0;
        long kused = 0;
        int ans = 0;
        int n = nums.size();
        for (int x = 0;;) {
            int y = x + 1;
            int num = nums[x];
            for (;y < n && nums[y] == num; y++);
            int count = y - x;
            ans = max(ans, count + queuecount);
            if (y == n) {
                return ans;
            }
            x = y;
            queue.push({num, count});
            queuecount += count;
            int num2 = nums[x];
            kused += queuecount * (long)(num2 - num);
            if (kused <= k) {
                continue;
            }
            for (;!queue.empty();queue.pop()) {
                auto [num, count] = queue.front();
                int numdif = num2 - num;
                kused -= count * (long)numdif;
                queuecount -= count;
                if (kused > k) {
                    continue;
                }
                int cnt = (k - kused) / numdif; 
                queuecount += cnt;
                queue.front() = {num, cnt};
                kused += cnt * numdif;
                break;
            }
        }
    }
};
108200