Introduction

In this blog post, we will explore how to solve the LeetCode problem "3321" 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: nums.length == n 1 <= n <= 105 1 <= nums[i] <= 109 1 <= x <= k <= nums.length

Explanation

• Sliding Window with Frequency Counting: Use a sliding window of size k to iterate through the input array. Maintain a frequency map for each window to count element occurrences.
  • Priority Queue for Top X: Utilize a priority queue (min-heap) to efficiently identify and track the top x most frequent elements based on occurrence count and value (for tie-breaking).
  • X-Sum Calculation: For each window, calculate the x-sum by summing the elements whose frequencies are within the top x.

• Runtime Complexity: O(n k log(x)), where n is the length of nums, k is the window size, and x is the number of top elements.
• Storage Complexity: O(k), primarily due to the frequency map and the priority queue.

Code

    import heapq
from collections import defaultdict

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

    for i in range(n - k + 1):
        subarray = nums[i:i + k]
        counts = defaultdict(int)
        for num in subarray:
            counts[num] += 1

        pq = []  # Priority queue (min-heap) of (frequency, value) tuples
        for num, count in counts.items():
            heapq.heappush(pq, (count, num))
            if len(pq) > x:
                heapq.heappop(pq)

        top_x_elements = set()
        while pq:
            count, num = heapq.heappop(pq)
            top_x_elements.add(num)

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

        answer.append(current_x_sum)

    return answer