Introduction

In this blog post, we will explore how to solve the LeetCode problem "3318" 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 an array nums of n integers and two integers k and x. The x-sum of an array is calculated by the following procedure: Count the occurrences of all elements in the array. Keep only the occurrences of the top x most frequent elements. If two elements have the same number of occurrences, the element with the bigger value is considered more frequent. Calculate the sum of the resulting array. Note that if an array has less than x distinct elements, its x-sum is the sum of the array. Return an integer array answer of length n - k + 1 where answer[i] is the x-sum of the subarray nums[i..i + k - 1]. Example 1: Input: nums = [1,1,2,2,3,4,2,3], k = 6, x = 2 Output: [6,10,12] Explanation: For subarray [1, 1, 2, 2, 3, 4], only elements 1 and 2 will be kept in the resulting array. Hence, answer[0] = 1 + 1 + 2 + 2. For subarray [1, 2, 2, 3, 4, 2], only elements 2 and 4 will be kept in the resulting array. Hence, answer[1] = 2 + 2 + 2 + 4. Note that 4 is kept in the array since it is bigger than 3 and 1 which occur the same number of times. For subarray [2, 2, 3, 4, 2, 3], only elements 2 and 3 are kept in the resulting array. Hence, answer[2] = 2 + 2 + 2 + 3 + 3. Example 2: Input: nums = [3,8,7,8,7,5], k = 2, x = 2 Output: [11,15,15,15,12] Explanation: Since k == x, answer[i] is equal to the sum of the subarray nums[i..i + k - 1]. Constraints: 1 <= n == nums.length <= 50 1 <= nums[i] <= 50 1 <= x <= k <= nums.length

Explanation

Here's an efficient approach to solve this problem:

• Sliding Window: Iterate through the array using a sliding window of size k.
• Frequency Counting and Selection: For each window, count the frequency of each element using a dictionary or array. Then, select the top x most frequent elements, considering the larger value as having higher priority in case of ties.

• Sum Calculation: Calculate the sum of the elements in the window, considering only the occurrences of the selected top x frequent elements.

• Runtime Complexity: O(n*k), where n is the length of nums and k is the window size. Storage Complexity: O(k), for storing the frequency counts of elements within a window.

Code

    def x_sum(nums, k, x):
    n = len(nums)
    result = []

    for i in range(n - k + 1):
        subarray = nums[i:i + k]
        counts = {}
        for num in subarray:
            counts[num] = counts.get(num, 0) + 1

        sorted_counts = sorted(counts.items(), key=lambda item: (item[1], item[0]), reverse=True)

        top_x_elements = [item[0] for item in sorted_counts[:x]]

        current_sum = 0
        for num in subarray:
            if num in top_x_elements:
                current_sum += num

        result.append(current_sum)

    return result