<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 maxFrequency(vector<int>& nums, int k) {
        // sort(nums.begin(), nums.end());
        // int max_freq = 1;
        // for (int i = 0; i < nums.size(); i++) {
        //     int temp = k, freq = 1;
        //     for (int j = i + 1; j < nums.size() && temp >= 0; j++) {
        //         int diff = (nums[j] - nums[j - 1]) * (j - i);
        //         if (diff <= temp) {
        //             freq++;
        //             temp -= diff;
        //         } else {
        //             break;
        //         }
        //     }
        //     max_freq = max(freq, max_freq);
        // }
        // return max_freq;

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

        long long total = 0; // Total difference cost
        int max_freq = 1;    // At least one element has a frequency of 1
        int i = 0;           // Left pointer for the sliding window

        // Step 2: Use sliding window approach
        for (int j = 1; j < nums.size(); ++j) {
            // Add the cost of making nums[i...j-1] into nums[j]
            total += (long long)(nums[j] - nums[j - 1]) * (j - i);

            // If the total cost exceeds k, shrink the window from the left
            while (total > k) {
                total -= nums[j] - nums[i];  // Remove the cost of nums[i]
                i++;  // Move the left pointer to the right
            }

            // Update the maximum frequency
            max_freq = max(max_freq, j - i + 1);
        }

        return max_freq;
    }
};
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