<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, vector<int>& nums, int k, vector<long> &prefixSum) {
        int target= nums[target_idx];
        int l = 0;
        int r = target_idx;
        int best_idx= target_idx;
        while(l<=r) {
            int mid = l+(r-l)/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) {
                l=mid+1;
            } else{
                best_idx = mid;
                r = 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> prefixSum(n);
        prefixSum[0] = nums[0];
        for(int i = 1;i<n; i++) {
            prefixSum[i] = prefixSum[i-1]+nums[i];
        }
        int res = 0;
        for(int target=0;target<n;target++) {
            int freq = bSearch(target, nums, k, prefixSum);
            res = max(res, freq);
        }
        return res;
    }
};
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