2270. Number of Ways to Split Array 

Description

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

Solutions

Solution 1: Prefix Sum

First, we calculate the total sum $s$ of the array $\textit{nums}$. Then, we traverse the first $n-1$ elements of the array $\textit{nums}$, using the variable $t$ to record the prefix sum. If $t \geq s - t$, we increment the answer by one.

After the traversal, we return the answer.

The time complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.

Python3

class Solution:
    def waysToSplitArray(self, nums: List[int]) -> int:
        s = sum(nums)
        ans = t = 0
        for x in nums[:-1]:
            t += x
            ans += t >= s - t
        return ans

Java

class Solution {
    public int waysToSplitArray(int[] nums) {
        long s = 0;
        for (int x : nums) {
            s += x;
        }
        long t = 0;
        int ans = 0;
        for (int i = 0; i + 1 < nums.length; ++i) {
            t += nums[i];
            ans += t >= s - t ? 1 : 0;
        }
        return ans;
    }
}

C++

class Solution {
public:
    int waysToSplitArray(vector<int>& nums) {
        long long s = accumulate(nums.begin(), nums.end(), 0LL);
        long long t = 0;
        int ans = 0;
        for (int i = 0; i + 1 < nums.size(); ++i) {
            t += nums[i];
            ans += t >= s - t;
        }
        return ans;
    }
};

Go

func waysToSplitArray(nums []int) (ans int) {
	var s, t int
	for _, x := range nums {
		s += x
	}
	for _, x := range nums[:len(nums)-1] {
		t += x
		if t >= s-t {
			ans++
		}
	}
	return
}

TypeScript

function waysToSplitArray(nums: number[]): number {
    const s = nums.reduce((acc, cur) => acc + cur, 0);
    let [ans, t] = [0, 0];
    for (const x of nums.slice(0, -1)) {
        t += x;
        if (t >= s - t) {
            ++ans;
        }
    }
    return ans;
}