#
# Create by Hua on 5/28/22.
#

"""
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

"""


class Solution(object):
    def waysToSplitArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int

        thought: prefix sum.
        medium - easy
        5 min.
        """
        total = sum(nums)
        left = ret = 0
        for i in range(len(nums)-1):
            left += nums[i]
            if left >= total - left:
                ret += 1
        return ret

