In this tutorial, we dive into a fascinating array manipulation challenge: LeetCode problem “1838. Frequency of the Most Frequent Element”. This problem tests our ability to optimize the frequency of elements in an array with a limited number of operations.

Problem Statement

We are given an array nums and an integer k. The task is to maximize the frequency of any element in the array. We can perform at most k operations, and in each operation, we can increment any element in the array by 1. Our goal is to find the maximum possible frequency of any element after performing these operations.

Examples

1. Input: nums = [1,2,4], k = 5 Output: 3 Explanation: Increment 1 three times and 2 two times to get [4,4,4], making the frequency of 4 equal to 3.
2. Input: nums = [1,4,8,13], k = 5 Output: 2 Explanation: Multiple solutions exist, e.g., increment 1 three times to get two 4s or increment 4 four times to get two 8s.
3. Input: nums = [3,9,6], k = 2 Output: 1 Explanation: The operations are limited, leading to a maximum frequency of 1 for any element.

Solution Approach

Our approach is to sort the array and use a sliding window technique to find the maximum frequency. The idea is to maintain a window where the difference between the sum of elements in the window and the maximum element times the window’s length is less than or equal to k.

Step-by-Step Solution

Sort the Array:

• We sort nums to efficiently manage the frequency count.

Initialize Variables:

• Two pointers left and right for the sliding window, result to store the maximum frequency, and total to keep track of the sum within the window.

Iterate with Sliding Window:

• Extend the window by moving right and update total.
• If the current window violates the condition (nums[right] * window length > total + k), shrink the window from the left.
• Update result with the maximum window size encountered.

Return the Result:

• The result variable holds the maximum frequency achievable.

Implementation

Here’s the Python code for the solution:

class Solution:
    def maxFrequency(self, nums: List[int], k: int) -> int:
        # Sort the array to make it easier to handle frequency optimization
        nums.sort()

        # Initialize two pointers for the sliding window: left and right
        # Initialize result to store the maximum frequency found
        # Initialize total to keep track of the sum of elements in the window
        left, right = 0, 0
        result, total = 0, 0

        # Iterate through the array with the right pointer
        while right < len(nums):
            # Add the current element to the total sum of the window
            total += nums[right]

            # Shrink the window from the left if the condition is violated
            # The condition checks if the total sum plus allowed increments (k)
            # is less than the sum needed to make all elements in the window equal to nums[right]
            while nums[right] * (right - left + 1) > total + k:
                # If the condition is violated, reduce the window size from the left
                # and adjust the total sum accordingly
                total -= nums[left]
                left += 1
            
            # Update the result with the maximum size of the valid window found so far
            # The window size is the frequency of the most frequent element in the current window
            result = max(result, right - left + 1)

            # Move the right pointer to the right to expand the window
            right += 1

        # Return the maximum frequency found
        return result

Complexity Analysis

• Time Complexity: O(n log n) due to sorting.
• Space Complexity: O(1) as we use constant extra space.

Conclusion

This problem illustrates the power of the sliding window technique in array manipulation, especially when dealing with frequency optimization under constraints. By carefully managing the window size, we can achieve the maximum frequency of an element in an efficient manner.

Reference

https://youtu.be/vgBrQ0NM5vE?si=CpGZWbXA_5-EFBCO