<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:
    // Helper function to check if it's possible to make 'mid' elements equal to nums[i] with k operations
    bool isPossible(vector<int>& nums, int k, int i, int mid, vector<long long>& prefixSum) {
        int startIdx = i - (mid - 1);
        long long sumNeeded = static_cast<long long>(nums[i]) * mid;
        long long currentSum = prefixSum[i] - (startIdx > 0 ? prefixSum[startIdx - 1] : 0);

        return sumNeeded - currentSum <= k;
    }

    // Binary search function to find the maximum length for making elements equal to nums[i]
    int findMaxFrequency(vector<int>& nums, int k, int i, vector<long long>& prefixSum) {
        int low = 1;
        int high = i + 1;

        while (low <= high) {
            int mid = low + (high - low) / 2;

            if (isPossible(nums, k, i, mid, prefixSum)) {
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }

        return high;
    }

    // Main function to find the maximum frequency of any element in the array
    int maxFrequency(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> maxFreq(n, 0);
        sort(nums.begin(), nums.end());

        vector<long 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) {
            int currentMaxFreq;

            if (i > 0 && nums[i] != nums[i - 1]) {
                currentMaxFreq = findMaxFrequency(nums, k, i, prefixSum);
            } else {
                currentMaxFreq = (i == 0) ? 1 : maxFreq[i - 1] + 1;
            }

            maxFreq[i] = currentMaxFreq;
            result = max(result, currentMaxFreq);
        }

        return result;
    }
};

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