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 II

Difficulty: Hard

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 (with ties broken by the larger value), and calculating the sum of this resulting array. Your task is to 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

  • Use a sliding window approach to efficiently calculate the x-sum for each subarray of length k.
  • Utilize a frequency map (hash table) to count occurrences of elements in the current window.
  • Use a max-heap (priority queue) to efficiently retrieve the top x most frequent elements.
  • Handle edge cases where the number of distinct elements is less than x.

Space and Time Complexity

Time Complexity: O(n log k)
Space Complexity: O(n)

Solution

To solve this problem, we can use a sliding window approach, which allows us to efficiently calculate the x-sum for each subarray of length k. We maintain a frequency map to count the occurrences of elements in the current window. As we slide the window, we update this frequency map by adding the new element entering the window and removing the element that is leaving.

To find the top x most frequent elements, we can use a max-heap (priority queue). The heap will store pairs of (frequency, value), allowing us to pop the top x elements efficiently. The final x-sum is computed by adding the contributions of these top elements.

Code Solutions

def find_x_sum(nums, k, x):
    from collections import defaultdict
    import heapq

    n = len(nums)
    answer = []
    freq = defaultdict(int)

    # Initialize frequency map for the first window
    for i in range(k):
        freq[nums[i]] += 1

    # Function to calculate x-sum for the current frequency map
    def calculate_x_sum(freq, x):
        # Use a max-heap to get the top x frequent elements
        max_heap = []
        for value, count in freq.items():
            # Push (-count, value) to create a max-heap based on count and value
            heapq.heappush(max_heap, (-count, value))

        x_sum = 0
        for _ in range(min(x, len(max_heap))):
            count, value = heapq.heappop(max_heap)
            x_sum += -count * value  # Negate count to get the actual frequency
        return x_sum

    # Calculate x-sum for the first window
    answer.append(calculate_x_sum(freq, x))

    # Slide the window across the array
    for i in range(1, n - k + 1):
        # Remove the element going out of the window
        outgoing = nums[i - 1]
        freq[outgoing] -= 1
        if freq[outgoing] == 0:
            del freq[outgoing]

        # Add the new element coming into the window
        incoming = nums[i + k - 1]
        freq[incoming] += 1

        # Calculate x-sum for the current window
        answer.append(calculate_x_sum(freq, x))

    return answer
← Back to All Questions