Introduction

In this blog post, we will explore how to solve the LeetCode problem "2461" 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 integer array nums and an integer k. Find the maximum subarray sum of all the subarrays of nums that meet the following conditions: The length of the subarray is k, and All the elements of the subarray are distinct. Return the maximum subarray sum of all the subarrays that meet the conditions. If no subarray meets the conditions, return 0. A subarray is a contiguous non-empty sequence of elements within an array. Example 1: Input: nums = [1,5,4,2,9,9,9], k = 3 Output: 15 Explanation: The subarrays of nums with length 3 are: - [1,5,4] which meets the requirements and has a sum of 10. - [5,4,2] which meets the requirements and has a sum of 11. - [4,2,9] which meets the requirements and has a sum of 15. - [2,9,9] which does not meet the requirements because the element 9 is repeated. - [9,9,9] which does not meet the requirements because the element 9 is repeated. We return 15 because it is the maximum subarray sum of all the subarrays that meet the conditions Example 2: Input: nums = [4,4,4], k = 3 Output: 0 Explanation: The subarrays of nums with length 3 are: - [4,4,4] which does not meet the requirements because the element 4 is repeated. We return 0 because no subarrays meet the conditions. Constraints: 1 <= k <= nums.length <= 105 1 <= nums[i] <= 105

Explanation

Here's the breakdown of the solution:

• Sliding Window: Use a sliding window of size k to iterate through the array.
• Distinct Element Check: Within the window, maintain a frequency count of each element using a dictionary (or hash map). This allows for efficient checking of whether all elements in the window are distinct.

• Maximum Sum Tracking: If the window contains distinct elements, calculate its sum and update the maximum sum found so far.

• Runtime Complexity: O(n), where n is the length of the input array nums.

• Storage Complexity: O(k) in the worst case, where k is the size of the window. This is due to the dictionary potentially storing k distinct elements.

Code

    def max_subarray_sum(nums, k):
    """
    Finds the maximum subarray sum of all the subarrays of nums that meet the following conditions:
    The length of the subarray is k, and All the elements of the subarray are distinct.
    Return the maximum subarray sum of all the subarrays that meet the conditions. If no subarray meets the conditions, return 0.

    Args:
        nums (list[int]): An integer array.
        k (int): An integer representing the length of the subarray.

    Returns:
        int: The maximum subarray sum of all the subarrays that meet the conditions.
    """
    if len(nums) < k:
        return 0

    max_sum = 0
    window_sum = 0
    freq = {}

    # Initialize the window
    for i in range(k):
        if nums[i] in freq:
            freq[nums[i]] += 1
        else:
            freq[nums[i]] = 1
        window_sum += nums[i]

    # Check if initial window has distinct elements
    distinct = True
    for count in freq.values():
        if count > 1:
            distinct = False
            break

    if distinct:
        max_sum = window_sum

    # Slide the window
    for i in range(k, len(nums)):
        # Remove the leftmost element from the window
        left = nums[i - k]
        freq[left] -= 1
        if freq[left] == 0:
            del freq[left]
        window_sum -= left

        # Add the rightmost element to the window
        right = nums[i]
        if right in freq:
            freq[right] += 1
        else:
            freq[right] = 1
        window_sum += right

        # Check if the current window has distinct elements
        distinct = True
        for count in freq.values():
            if count > 1:
                distinct = False
                break

        if distinct:
            max_sum = max(max_sum, window_sum)

    return max_sum