<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 bSearch(int target_idx, int k, vector<int>&nums, vector<long long>&prefix ){

        int target = nums[target_idx];

        int left = 0;
        int right = target_idx;

        int best_idx = target_idx;

        while(left<=right){
            int mid = (left+right)/2;

            long count = (target_idx - mid +1);
            long windowSum = count*target;

            long currSum = prefix[target_idx] - prefix[mid] +nums[mid];

            int ops = windowSum - currSum;


            if(ops > k){
                left = mid+1;
            }
            else{
                best_idx = mid;
                right = mid-1;
            }
        }

        return target_idx - best_idx +1;

    }





    int maxFrequency(vector<int>& nums, int k) {

        int n = nums.size();

        sort(nums.begin(),nums.end());

        vector<long long>prefix(n);
        prefix[0] = nums[0];

        for(int i =1;i<n;i++){
            prefix[i] = prefix[i-1] + nums[i];
        }

        int result = 0;

        for(int target_idx =0;target_idx<n;target_idx++){
            result = max(result, bSearch(target_idx,k,nums,prefix));
        }

        return result;
        
    }
};
114900