Introduction

In this blog post, we will explore how to solve the LeetCode problem "1712" 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

A split of an integer array is good if: The array is split into three non-empty contiguous subarrays - named left, mid, right respectively from left to right. The sum of the elements in left is less than or equal to the sum of the elements in mid, and the sum of the elements in mid is less than or equal to the sum of the elements in right. Given nums, an array of non-negative integers, return the number of good ways to split nums. As the number may be too large, return it modulo 109 + 7. Example 1: Input: nums = [1,1,1] Output: 1 Explanation: The only good way to split nums is [1] [1] [1]. Example 2: Input: nums = [1,2,2,2,5,0] Output: 3 Explanation: There are three good ways of splitting nums: [1] [2] [2,2,5,0] [1] [2,2] [2,5,0] [1,2] [2,2] [5,0] Example 3: Input: nums = [3,2,1] Output: 0 Explanation: There is no good way to split nums. Constraints: 3 <= nums.length <= 105 0 <= nums[i] <= 104

Explanation

Here's the breakdown of the solution:

• Prefix Sums: Calculate prefix sums to efficiently compute subarray sums. This avoids repeatedly summing elements within subarrays.
• Two Pointers: Iterate through possible split points for the left subarray and use two pointers (mid_start and mid_end) to find the valid range for the middle subarray's starting position that satisfies the good split conditions.

• Counting Valid Splits: For each valid left subarray split, the number of valid splits is the range between the valid mid_start and mid_end positions. Accumulate the count of good splits modulo 10^9 + 7.

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

• Storage Complexity: O(n)

Code

    def waysToSplit(nums):
    n = len(nums)
    prefix_sums = [0] * (n + 1)
    for i in range(n):
        prefix_sums[i + 1] = prefix_sums[i] + nums[i]

    count = 0
    mod = 10**9 + 7

    for i in range(1, n - 1):
        left_sum = prefix_sums[i]

        # Find the valid range for mid_start
        mid_start = i + 1
        while mid_start < n and prefix_sums[mid_start] - left_sum < left_sum:
            mid_start += 1

        # Find the valid range for mid_end
        mid_end = i + 1
        while mid_end < n and \
                prefix_sums[n] - prefix_sums[mid_end] >= prefix_sums[mid_end] - left_sum:
            mid_end += 1

        if mid_start < n and mid_start <= mid_end:
            count = (count + (mid_end - mid_start)) % mod

    return count