Number of Ways to Split Array

Problem Statement

In this task, we are given a 0-indexed integer array nums with n elements. Our goal is to determine the number of valid splits in this array. A split at index i is considered valid if it satisfies two conditions:

• The summation of the elements from the beginning of the array up to index i (inclusive) must be greater than or equal to the sum of the elements from index i + 1 to the end.
• The split index i must leave at least one element to the right, which means it must satisfy the condition 0 <= i < n - 1.

We need to count and return the number of such valid splits in the array nums.

Examples

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

Approach and Intuition

To efficiently determine the number of valid splits in the array nums, consider the following approach:

1. Calculate the total sum of the array. This gives us a reference to quickly calculate any segment of the array without iterating through it multiple times.

2. Track the cumulative sum of elements as you iterate from the start. This allows you to know the sum of the subarray from the beginning up to any index i instantaneously.

3. Determine the sum of the elements to the right of an index i using the formula right_sum = total_sum - cumulative_sum[i], where cumulative_sum[i] is the sum of elements from the start up to the index i.

4. For each index starting from 0 up to n - 2, check if cumulative_sum[i] is greater than or equal to right_sum. If true, it constitutes a valid split.

Using cumulative sums reduces time complexity, as each index check becomes a constant-time operation after the initial sum calculations.

Illustration with Examples:

• For nums = [10, 4, -8, 7], after calculating the total sum and progressively calculating the cumulative sums, we can directly compare these sums split-wise:
  • Index 0 split gives the sums for [10] and [4, -8, 7]. Check if the sum of [10] ≥ sum of [4, -8, 7].
  • Continue this for indices 1 and 2 as described, ensuring the cumulative sum is always greater or equal to the right sum for a valid split.

This model allows quick evaluations of potential splits and is well-suited to handle the worst case scenarios defined by the constraints, considering the lengths up to 10^5 and element bounds from -10^5 to 10^5.

Solutions

class Solution {
public:
    int countValidSplits(vector<int>& numbers) {
        long long sumLeft = 0, sumRight = 0;
    
        // Initially calculate the total of all elements
        for (int value : numbers) {
            sumRight += value;
        }
    
        int validSplits = 0;
        // Evaluate each split point
        for (int i = 0; i < numbers.size() - 1; i++) {
            sumLeft += numbers[i];
            sumRight -= numbers[i];
    
            // Check the conditions for a valid split
            if (sumLeft >= sumRight) {
                validSplits++;
            }
        }
    
        return validSplits;
    }
};