Introduction

In this blog post, we will explore how to solve the LeetCode problem "1588" 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

Given an array of positive integers arr, return the sum of all possible odd-length subarrays of arr. A subarray is a contiguous subsequence of the array. Example 1: Input: arr = [1,4,2,5,3] Output: 58 Explanation: The odd-length subarrays of arr and their sums are: [1] = 1 [4] = 4 [2] = 2 [5] = 5 [3] = 3 [1,4,2] = 7 [4,2,5] = 11 [2,5,3] = 10 [1,4,2,5,3] = 15 If we add all these together we get 1 + 4 + 2 + 5 + 3 + 7 + 11 + 10 + 15 = 58 Example 2: Input: arr = [1,2] Output: 3 Explanation: There are only 2 subarrays of odd length, [1] and [2]. Their sum is 3. Example 3: Input: arr = [10,11,12] Output: 66 Constraints: 1 <= arr.length <= 100 1 <= arr[i] <= 1000 Follow up: Could you solve this problem in O(n) time complexity?

Explanation

Here's a breakdown of the approach, followed by the Python code:

• Contribution of each element: Instead of generating all subarrays, focus on how many times each element arr[i] contributes to the sum of odd-length subarrays.
• Formula for contribution: The number of odd-length subarrays that include arr[i] can be calculated directly using its index i and the array length n. The formula involves considering combinations of elements to the left and right of arr[i] to form odd-length subarrays.

• Efficient Calculation: Utilize the formula ((i + 1) * (n - i) + 1) // 2 to determine the number of odd-length subarrays each element belongs to, and multiply this count by the element's value, accumulating the result.

• Runtime & Storage Complexity: O(n) runtime complexity, O(1) storage complexity

Code

    def sum_odd_length_subarrays(arr: list[int]) -> int:
    """
    Given an array of positive integers arr, return the sum of all possible odd-length subarrays of arr.
    """
    n = len(arr)
    total_sum = 0
    for i in range(n):
        total_sum += ((i + 1) * (n - i) + 1) // 2 * arr[i]
    return total_sum