2270. Number of Ways to Split Array

Problem

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 <= 10^5

• -10^5 <= nums[i] <= 10^5

Solution (Java)

class Solution {
    public int waysToSplitArray(int[] nums) {
        long leftSum = 0;
        long rightSum = 0;
        for (int i : nums) {
            rightSum += i;
        }
        int count = 0;
        for (int i = 0; i < nums.length - 1; i++) {
            rightSum -= nums[i];
            leftSum += nums[i];
            if (leftSum >= rightSum) {
                count++;
            }
        }
        return count;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).