 # Leetcode 1838. Frequency of the Most Frequent Element Solution

## Problem Explanation

You are given an integer array `nums` and an integer `k`. The goal is to perform at most 'k' operations on the array, where each operation consists of incrementing an element by 1, and find the maximum possible frequency of an element after performing these operations.

## Solution Approach

To solve this problem, we will use the sliding window approach:

1. Sort the input array `nums`.
2. Initialize two pointers, `l` and `r`, and a variable `sum` to keep track of the sum of elements in the window.
3. Iterate through the array using the `r` pointer, and for each element, add its value to `sum`.
4. If the sum + k is less than the element at position `r` times the window size (r - l + 1), we remove the element at position `l` from the sum and increment the `l` pointer. This condition checks if the window's sum + k is large enough to perform 'k' operations to make all elements in the window equal to the current element at position `r`.
5. Calculate the maximum frequency by taking the maximum between the current maximum and the window size (r - l + 1).

Let's walk through an example:

``1nums = [1, 4, 8, 13], k = 5``
1. Sort the array: `nums = [1, 4, 8, 13]`. The array is already sorted in this case.
2. Initialize `l = 0`, `r = 0`, and `sum = 0`.
3. Start iterating through the array.
• For `r = 0`, sum += nums = 1.
• For `r = 1`, sum += nums = 5. With `sum + k = 10`, the condition in step 4 does not apply, so the maximum frequency is updated to `r - l + 1 = 2`.
• For `r = 2`, sum += nums = 13. `sum + k = 18`, which is less than `nums * (r - l + 1)` or `24`. So, we remove `nums[l]` from the sum and increment `l`. Now, `sum = 12`, and the maximum frequency remains 2.
• For `r = 3`, sum += nums = 25. `sum + k = 30`, which is less than `nums * (r - l + 1)` or `52`. So, we remove `nums[l]` from the sum and increment `l`. Now, `sum = 21`, `l = 2`, and the maximum frequency remains 2.
4. The iteration is complete, and the maximum frequency is 2.

## Python Solution

``````1class Solution:
2    def maxFrequency(self, nums: List[int], k: int) -> int:
3        ans = 0
4        sum = 0
5
6        nums.sort()
7
8        l = 0
9        for r in range(len(nums)):
10            sum += nums[r]
11            while sum + k < nums[r] * (r - l + 1):
12                sum -= nums[l]
13                l += 1
14            ans = max(ans, r - l + 1)
15
16        return ans``````

## Java Solution

``````1class Solution {
2    public int maxFrequency(int[] nums, int k) {
3        int ans = 0;
4        long sum = 0;
5
6        Arrays.sort(nums);
7
8        int l = 0;
9        for (int r = 0; r < nums.length; r++) {
10            sum += nums[r];
11            while (sum + k < (long)nums[r] * (r - l + 1)) {
12                sum -= nums[l++];
13            }
14            ans = Math.max(ans, r - l + 1);
15        }
16
17        return ans;
18    }
19}``````

## JavaScript Solution

``````1class Solution {
2  maxFrequency(nums, k) {
3    let ans = 0;
4    let sum = 0;
5
6    nums.sort((a, b) => a - b);
7
8    let l = 0;
9    for (let r = 0; r < nums.length; r++) {
10      sum += nums[r];
11      while (sum + k < nums[r] * (r - l + 1)) {
12        sum -= nums[l++];
13      }
14      ans = Math.max(ans, r - l + 1);
15    }
16
17    return ans;
18  }
19}``````

## C++ Solution

``````1class Solution {
2 public:
3  int maxFrequency(vector<int>& nums, int k) {
4    int ans = 0;
5    long sum = 0;
6
7    sort(nums.begin(), nums.end());
8
9    int l = 0;
10    for (int r = 0; r < nums.size(); r++) {
11      sum += nums[r];
12      while (sum + k < static_cast<long>(nums[r]) * (r - l + 1)) {
13        sum -= nums[l++];
14      }
15      ans = max(ans, r - l + 1);
16    }
17
18    return ans;
19  }
20};``````

## C# Solution

``````1public class Solution {
2    public int MaxFrequency(int[] nums, int k) {
3        int ans = 0;
4        long sum = 0;
5
6        Array.Sort(nums);
7
8        int l = 0;
9        for (int r = 0; r < nums.Length; r++) {
10            sum += nums[r];
11            while (sum + k < (long)nums[r] * (r - l + 1)) {
12                sum -= nums[l++];
13            }
14            ans = Math.Max(ans, r - l + 1);
15        }
16
17        return ans;
18    }
19}``````

In conclusion, we have provided the optimal solution for the problem using the sliding window approach. The solution has been implemented in Python, Java, JavaScript, C++, and C#. The time complexity of the solution is O(N * log(N)), where N is the number of elements in the input array, due to the sorting of the array. The space complexity is O(1), as no additional data structures are required.