Frequency of the Most Frequent Element Explanation.md

Apologies for the confusion! Here's the explanation in proper Markdown format with correct code blocks and proper formatting.

---

Problem: 1838. Frequency of the Most Frequent Element

Problem Statement

The frequency of an element is the number of times it occurs in an array.

You are given an integer array nums and an integer k. In one operation, you can choose an index of nums and increment the element at that index by 1.

Return the maximum possible frequency of an element after performing at most k operations.

---

Example 1:

Input: nums = [1, 2, 4], k = 5
Output: 3

Explanation:
Increment the first element three times and the second element two times to make nums = [4, 4, 4].
The frequency of 4 is 3.

Example 2:

Input: nums = [1, 4, 8, 13], k = 5
Output: 2

Explanation:
There are multiple optimal solutions:

• Increment the first element three times to make nums = [4, 4, 8, 13], where 4 has a frequency of 2.
• Increment the second element four times to make nums = [1, 8, 8, 13], where 8 has a frequency of 2.
• Increment the third element five times to make nums = [1, 4, 13, 13], where 13 has a frequency of 2.

Example 3:

Input: nums = [3, 9, 6], k = 2
Output: 1

Explanation:
With k = 2, it's impossible to increase any number enough times to make two elements the same. So the maximum frequency is 1.

---

Solution Overview

Approach:

1. Sorting the array helps by grouping similar numbers together. Once sorted, we can efficiently increase the smaller numbers to match the larger ones.
2. We use a sliding window technique to find the largest window of numbers that can be made identical using at most k operations.
3. If the number of operations exceeds k, we contract the window by moving the left pointer.

Formula Insight:

For any subarray [nums[l], nums[l+1], ..., nums[r]], we check if we can make all elements equal to nums[r] by ensuring:

nums[r] * (r - l + 1) ≤ total + k

Where total is the sum of the elements in the current window.

Algorithm:

1. Sort the array nums.
2. Use a sliding window with two pointers left and right, both starting at 0, and a total to store the sum of the elements in the current window.
3. Expand the window by moving the right pointer.
4. If the current window violates the condition, shrink the window by moving the left pointer.
5. Track the maximum window size, which gives the maximum frequency.

---

Solution Code (C++)

class Solution {
public:
    int maxFrequency(vector<int>& nums, int k) {
        long long l = 0, r = 0, n = nums.size(), res = 0, total = 0;
        sort(nums.begin(), nums.end());
        while (r < n) {
            total += nums[r];
            while (nums[r] * (r - l + 1) > total + k) {
                total -= nums[l];
                l++;
            }
            res = max(res, (r - l + 1));
            r++;
        }
        return res;
    }
};

---

Key Steps

1. Sort the Input
   Sorting helps group numbers together, making it easier to handle in the sliding window approach.

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

2. Sliding Window
   Use two pointers l and r to define the window. Calculate the total sum of elements in the window and use the formula nums[r] * (r - l + 1) > total + k to check if we need to adjust the window.

   while (r < n) {
       total += nums[r];
       while (nums[r] * (r - l + 1) > total + k) {
           total -= nums[l];
           l++;
       }
       res = max(res, (r - l + 1));
       r++;
   }

3. Track Maximum Frequency
   Update res whenever the current window size (i.e., r - l + 1) is larger than the previous maximum.

---

Time and Space Complexity

• Time Complexity:
  Sorting the array takes O(n log n).
  The sliding window technique takes O(n), so the overall time complexity is O(n log n).

• Space Complexity:
  The space complexity is O(1) (not counting the input space).

---

Conclusion

This sliding window approach efficiently solves the problem by dynamically adjusting the window size and ensuring that the total number of operations performed does not exceed k.

---

If you found this helpful, don't forget to ⭐ star the repository and share it with others!