Introduction

In this blog post, we will explore how to solve the LeetCode problem "1838" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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]. 4 has a frequency of 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]. 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. Example 3: Input: nums = [3,9,6], k = 2 Output: 1 Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 105 1 <= k <= 105

Explanation

Here's the breakdown of the solution:

• Sorting: Sort the input array nums in ascending order. This allows us to efficiently use a sliding window approach.
• Sliding Window: Use a sliding window to find the maximum frequency. Expand the window to the right, and if the operations needed to make all elements in the window equal to the rightmost element exceed k, shrink the window from the left.

• Operations Calculation: Efficiently calculate the number of operations needed using the current window size and the sum of elements in the window.

• Runtime Complexity: O(n log n) due to sorting. Storage Complexity: O(1) excluding the input array (in-place sorting possible).

Code

    def maxFrequency(nums, k):
    nums.sort()
    left = 0
    right = 0
    window_sum = 0
    max_freq = 0

    while right < len(nums):
        window_sum += nums[right]

        while (right - left + 1) * nums[right] - window_sum > k:
            window_sum -= nums[left]
            left += 1

        max_freq = max(max_freq, right - left + 1)
        right += 1

    return max_freq