Number of Ways to Split Array

#leetcode #dsa #theabbie

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

SOLUTION:

class Solution:
    def waysToSplitArray(self, nums: List[int]) -> int:
        n = len(nums)
        p = [0]
        for num in nums:
            p.append(p[-1] + num)
        ctr = 0
        for i in range(n - 1):
            l = p[i + 1]
            r = p[-1] - l
            if l >= r:
                ctr += 1
        return ctr