We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsDisclaimersPrivacyTermsContactAbout Us

Find X-Sum of All K-Long Subarrays I

Difficulty: Easy

Problem Description

You are given an array nums of n integers and two integers k and x. The x-sum of an array is calculated by counting the occurrences of all elements, keeping only the occurrences of the top x most frequent elements, and calculating the sum of the resulting array. 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].

Key Insights

  • The problem involves sliding window technique to evaluate each subarray of length k.
  • Counting occurrences of elements can be efficiently handled using a hash map (dictionary).
  • To find the top x frequent elements, a max heap or sorting can be utilized.
  • Care must be taken with elements that have the same frequency; larger values should be prioritized.

Space and Time Complexity

Time Complexity: O(n * k log k) - For each subarray, counting and sorting the top x elements takes O(k log k) time. Space Complexity: O(n) - For storing the counts of elements in the hash map.

Solution

To solve this problem, we'll use a sliding window approach to iterate through each subarray of length k. For each subarray, we will:

  1. Count the occurrences of each element using a hash map.
  2. Use a priority queue (max heap) to extract the top x most frequent elements, prioritizing larger values in case of ties.
  3. Calculate the sum of the selected elements and store it in the result list.

The main data structures used here are a hash map for counting occurrences and a max heap for efficiently retrieving the top x elements.

Code Solutions

from collections import Counter
import heapq

def x_sum_of_subarrays(nums, k, x):
    n = len(nums)
    answer = []
    
    for i in range(n - k + 1):
        subarray = nums[i:i + k]
        count = Counter(subarray)
        
        # Using a max heap to get the top x frequent elements
        max_heap = [(-freq, num) for num, freq in count.items()]
        heapq.heapify(max_heap)
        
        total_sum = 0
        for _ in range(min(x, len(max_heap))):
            freq, num = heapq.heappop(max_heap)
            total_sum += -freq * num
        
        answer.append(total_sum)
    
    return answer
← Back to All Questions