# Solving Leetcode Interviews in Seconds with AI: Sum of Variable Length Subarrays


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3427" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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 of size n. For each index i where 0 <= i < n, define a subarray nums[start ... i] where start = max(0, i - nums[i]). Return the total sum of all elements from the subarray defined for each index in the array.   Example 1:  Input: nums = [2,3,1] Output: 11 Explanation:    i Subarray Sum   0 nums[0] = [2] 2   1 nums[0 ... 1] = [2, 3] 5   2 nums[1 ... 2] = [3, 1] 4   Total Sum   11    The total sum is 11. Hence, 11 is the output.  Example 2:  Input: nums = [3,1,1,2] Output: 13 Explanation:    i Subarray Sum   0 nums[0] = [3] 3   1 nums[0 ... 1] = [3, 1] 4   2 nums[1 ... 2] = [1, 1] 2   3 nums[1 ... 3] = [1, 1, 2] 4   Total Sum   13    The total sum is 13. Hence, 13 is the output.    Constraints:  1 <= n == nums.length <= 100 1 <= nums[i] <= 1000  

	# Explanation
	Here's an efficient solution to the problem:

*   **Iterate and Calculate Subarray Start:** For each index `i`, calculate the start index of the subarray as `max(0, i - nums[i])`.
*   **Sum Subarray Elements:** Sum the elements of the subarray from the calculated start index up to the current index `i`.
*   **Accumulate Total Sum:** Add the subarray sum to a running total, which will be the final result.

*   **Runtime Complexity:** O(n), where n is the length of the input array. **Storage Complexity:** O(1).

	
	# Code
	```python
	def subarray_sum(nums):
    """
    Calculates the total sum of elements from subarrays defined for each index.

    Args:
        nums: An integer array.

    Returns:
        The total sum of elements from the subarrays.
    """

    total_sum = 0
    n = len(nums)

    for i in range(n):
        start = max(0, i - nums[i])
        subarray_sum = 0
        for j in range(start, i + 1):
            subarray_sum += nums[j]
        total_sum += subarray_sum

    return total_sum
	```