# Solving Leetcode Interviews in Seconds with AI: Number of Ways to Split Array


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2270" 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 a 0-indexed integer array nums of length n. nums contains a valid split at index i if the following are true:  The sum of the first i + 1 elements is greater than or equal to the sum of the last n - i - 1 elements. There is at least one element to the right of i. That is, 0 <= i < n - 1.  Return the number of valid splits in nums.   Example 1:  Input: nums = [10,4,-8,7] Output: 2 Explanation:  There are three ways of splitting nums into two non-empty parts: - Split nums at index 0. Then, the first part is [10], and its sum is 10. The second part is [4,-8,7], and its sum is 3. Since 10 >= 3, i = 0 is a valid split. - Split nums at index 1. Then, the first part is [10,4], and its sum is 14. The second part is [-8,7], and its sum is -1. Since 14 >= -1, i = 1 is a valid split. - Split nums at index 2. Then, the first part is [10,4,-8], and its sum is 6. The second part is [7], and its sum is 7. Since 6 < 7, i = 2 is not a valid split. Thus, the number of valid splits in nums is 2.  Example 2:  Input: nums = [2,3,1,0] Output: 2 Explanation:  There are two valid splits in nums: - Split nums at index 1. Then, the first part is [2,3], and its sum is 5. The second part is [1,0], and its sum is 1. Since 5 >= 1, i = 1 is a valid split.  - Split nums at index 2. Then, the first part is [2,3,1], and its sum is 6. The second part is [0], and its sum is 0. Since 6 >= 0, i = 2 is a valid split.    Constraints:  2 <= nums.length <= 105 -105 <= nums[i] <= 105  

	# Explanation
	Here's the breakdown of the solution:

*   **Prefix Sum:** Calculate prefix sums to efficiently determine the sum of the first `i + 1` elements.
*   **Total Sum:** Calculate the total sum of the array to determine the sum of the last `n - i - 1` elements efficiently using `total_sum - prefix_sum[i]`.
*   **Iterate and Check:** Iterate through possible split indices and check if the condition for a valid split is met.

*   **Runtime Complexity:** O(n)
*   **Storage Complexity:** O(n)

	
	# Code
	```python
	def valid_splits(nums):
    n = len(nums)
    prefix_sum = [0] * n
    prefix_sum[0] = nums[0]
    for i in range(1, n):
        prefix_sum[i] = prefix_sum[i - 1] + nums[i]

    total_sum = prefix_sum[n - 1]
    count = 0
    for i in range(n - 1):
        left_sum = prefix_sum[i]
        right_sum = total_sum - left_sum
        if left_sum >= right_sum:
            count += 1

    return count
	```