Introduction

In this blog post, we will explore how to solve the LeetCode problem "2968" 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

You are given a 0-indexed integer array nums and an integer k. You can perform the following operation on the array at most k times: Choose any index i from the array and increase or decrease nums[i] by 1. The score of the final array is the frequency of the most frequent element in the array. Return the maximum score you can achieve. The frequency of an element is the number of occurences of that element in the array. Example 1: Input: nums = [1,2,6,4], k = 3 Output: 3 Explanation: We can do the following operations on the array: - Choose i = 0, and increase the value of nums[0] by 1. The resulting array is [2,2,6,4]. - Choose i = 3, and decrease the value of nums[3] by 1. The resulting array is [2,2,6,3]. - Choose i = 3, and decrease the value of nums[3] by 1. The resulting array is [2,2,6,2]. The element 2 is the most frequent in the final array so our score is 3. It can be shown that we cannot achieve a better score. Example 2: Input: nums = [1,4,4,2,4], k = 0 Output: 3 Explanation: We cannot apply any operations so our score will be the frequency of the most frequent element in the original array, which is 3. Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 109 0 <= k <= 1014

Explanation

Here's the approach, complexity analysis, and Python code for efficiently solving this problem:

• Sort and Iterate: Sort the input array nums. Iterate through the sorted array, considering each element as a potential most frequent element.
• Sliding Window: For each element, use a sliding window to find the maximum window size (frequency) that can be achieved with at most k operations. The window expands to include more elements, and the cost of making those elements equal to the current target element is calculated. If the cost exceeds k, the window shrinks from the left.

• Prefix Sum Optimization: Use a prefix sum array to efficiently calculate the cost of operations within the sliding window.

• Time Complexity: O(n log n) due to sorting. The sliding window iteration is O(n). Space Complexity: O(n) for the prefix sum array.

Code

    def maxFrequency(nums, k):
    nums.sort()
    n = len(nums)
    prefix_sum = [0] * (n + 1)
    for i in range(n):
        prefix_sum[i + 1] = prefix_sum[i] + nums[i]

    max_freq = 0
    left = 0
    for right in range(n):
        window_size = right - left + 1
        cost = nums[right] * window_size - (prefix_sum[right + 1] - prefix_sum[left])

        while cost > k:
            cost -= nums[right] - nums[left]
            left += 1
            window_size = right - left + 1
        max_freq = max(max_freq, window_size)

    return max_freq