<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 findBest(int target_idx, int k, vector<int>& nums, vector<long>& prefixSum) {
        int target = nums[target_idx];
        
        int i = 0;
        int j = target_idx;
        int result = target_idx;
        
        while(i <= j) {
            int mid = i + (j-i)/2;
            
            long count      = (target_idx - mid + 1);
            long windowSum  = (count * target);
            long currSum    = prefixSum[target_idx] - prefixSum[mid] + nums[mid];
            
            int ops         = windowSum - currSum;
            
            if(ops > k) {
                i = mid+1;
            } else {
                result = mid;
                j = mid-1;
            }
        }
        
        return target_idx-result+1;
    }
    
    int maxFrequency(vector<int>& nums, int k) {
        int n = nums.size();
        
        sort(begin(nums), end(nums));
        vector<long> prefixSum(n);
        prefixSum[0] = nums[0];
        
        for(int i = 1; i < n; i++) {
            prefixSum[i] = prefixSum[i-1] + nums[i];
        }
        
        int result = 0;
        
        for(int i = 0; i<n; i++) {
            result = max(result, findBest(i, k, nums, prefixSum));
        }
        
        return result;
    }
};
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