<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:
    // Function to find the appropriate index using binary search
    int bs(int i, int j, int k, vector<int>& nums, vector<long long>& prefixSum) {
        int left = 0, right = j, idx = j;
        while (left <= right) {
            int mid = (left + right) / 2;
            // Use long long to avoid overflow
            long long total = (long long)nums[j] * (j - mid + 1) - (prefixSum[j] - (mid > 0 ? prefixSum[mid - 1] : 0));
            if (total <= k) {
                idx = mid;
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return idx;
    }

    int maxFrequency(vector<int>& nums, int k) {
        int n = nums.size();
        sort(nums.begin(), nums.end());

        // Create the prefix sum array
        vector<long long> prefixSum(n, 0); // Use long long to store large sums
        prefixSum[0] = nums[0];
        for (int i = 1; i < n; i++) {
            prefixSum[i] = prefixSum[i - 1] + nums[i];
        }

        vector<int> ans(n, 0);
        int maxFreq = 1;

        for (int i = 1; i < n; i++) {
            // Binary search to find the valid left index
            int idx = bs(0, i, k, nums, prefixSum);

            // Calculate the maximum frequency
            ans[i] = i - idx + 1;
            maxFreq = max(maxFreq, ans[i]);
        }

        return maxFreq;
    }
};

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