<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 {
    #define ll long long
public:
    int maxFrequency(vector<int>& nums, int k) {
        ios::sync_with_stdio(false);
        cin.tie(0);

        sort(nums.rbegin() , nums.rend());

        ll cnt = 1;
        ll val = k;
        int n = nums.size();
        ll ans = 1, start = 0, end = 1;

        // for(auto i : nums) cout<<i;
        // cout<<'\n';


        while(end < n) {
            if( val >= (nums[start] - nums[end]) ) {
                cnt++;
                ans = max(ans , cnt);
                val -= (nums[start] - nums[end]);
                end++;
            } else {
                start++;
                end = start+1;
                ans = max(ans , cnt);
                cnt = 1;
                val = k;
            }
            if(end >= n) break;
            if(nums[start] == nums[end]) start++;
            // cout<<nums[start]<<" "<<nums[end]<<" "<<cnt<<" "<<ans<<'\n';
        }

        ans = max(ans , cnt);

        return ans;
    }
};
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