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Solving Leetcode Interviews in Seconds with AI: Maximum and Minimum Sums of at Most Size K Subarrays

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2 min read

Introduction

In this blog post, we will explore how to solve the LeetCode problem "3430" 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 a positive integer k. Return the sum of the maximum and minimum elements of all subarrays with at most k elements. Example 1: Input: nums = [1,2,3], k = 2 Output: 20 Explanation: The subarrays of nums with at most 2 elements are: Subarray Minimum Maximum Sum [1] 1 1 2 [2] 2 2 4 [3] 3 3 6 [1, 2] 1 2 3 [2, 3] 2 3 5 Final Total 20 The output would be 20. Example 2: Input: nums = [1,-3,1], k = 2 Output: -6 Explanation: The subarrays of nums with at most 2 elements are: Subarray Minimum Maximum Sum [1] 1 1 2 [-3] -3 -3 -6 [1] 1 1 2 [1, -3] -3 1 -2 [-3, 1] -3 1 -2 Final Total -6 The output would be -6. Constraints: 1 <= nums.length <= 80000 1 <= k <= nums.length -106 <= nums[i] <= 106

Explanation

Here's the solution:

  • Iterate through all possible subarrays of nums with length at most k.
  • For each subarray, find the minimum and maximum elements.
  • Add the sum of the minimum and maximum to the running total.

  • Time Complexity: O(n*k), Space Complexity: O(1)

Code

    def sub_array_max_min(nums, k):
    """
    Calculates the sum of the maximum and minimum elements of all subarrays with at most k elements.

    Args:
        nums (list[int]): An integer array.
        k (int): A positive integer representing the maximum length of subarrays.

    Returns:
        int: The sum of the maximum and minimum elements of all subarrays with at most k elements.
    """

    total_sum = 0
    n = len(nums)

    for i in range(n):
        for j in range(i, min(i + k, n)):
            sub_array = nums[i : j+1]
            min_val = min(sub_array)
            max_val = max(sub_array)
            total_sum += min_val + max_val

    return total_sum

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