2270. Number of Ways to Split Array

Featured Summary: To find the number of ways to split an array such that the left sum is greater than or equal to the right sum, you should use a prefix sum algorithm. By calculating the total sum of the array first, you can iterate through the array, dynamically maintaining the left sum and the right sum in O(1) time per step. This yields an optimal O(n) time complexity and O(1) space complexity if modifying the sums in-place. A split is valid if the left sum is at least the right sum and there is at least one element on the right.

Understanding the Problem

You are given a 0-indexed integer array nums of length n. The goal is to determine the number of valid splits. nums contains a valid split at index i if the following conditions 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.

Returning the number of valid splits is essential for many algorithmic applications involving prefix sums, subarray partitioning, and dynamic programming optimizations.

Examples Walkthrough

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 Analysis

2 <= nums.length <= 10⁵
-10⁵ <= nums[i] <= 10⁵

These constraints strongly indicate that an O(n) approach is required, as an O(n^2) algorithm would exceed the time limit on large inputs.

Optimal C++ Solution Explained

Below is the highly optimized C++ solution that solves the problem in linear time. We use an intelligent mathematical technique called Prefix Sum. By tracking the sum of the array sequentially, we avoid redundant calculations.

class Solution {
public:
    int waysToSplitArray(vector<int>& nums) {
        // Keep track of sum of elements on left and right sides
        long long leftSum = 0, rightSum = 0;

        // Initially all elements are on right side
        for (int num : nums) {
            rightSum += num;
        }

        int count = 0;
        // Try each possible split position
        for (int i = 0; i < nums.size() - 1; i++) {
            // Move current element from right to left side
            leftSum += nums[i];
            rightSum -= nums[i];

            // Check if this creates a valid split
            if (leftSum >= rightSum) {
                count++;
            }
        }

        return count;
    }
};

Why This Algorithm Works Intuitively

First, we calculate the total sum of the array and store it in rightSum. Initially, the leftSum is 0. As we iterate through the array, we simulate moving an element from the right part to the left part by subtracting the element's value from rightSum and adding it to leftSum. Then, we simply compare if leftSum >= rightSum. Because we only iterate through the array twice—once for the total sum, and once to evaluate split points—our time complexity is strictly O(n). The space complexity is O(1) since we only use a few long long variables to store sums, bypassing the need for an explicit prefix sum array allocation.

Frequently Asked Questions

What is the optimal time complexity for Number of Ways to Split Array?

The optimal time complexity is O(n). We can achieve this by using a prefix sum array to calculate the sum of elements on the left and right sides in constant time O(1) after an initial pass.

How do you handle negative numbers in the array?

The prefix sum approach naturally handles negative numbers. The sum of the left part and the right part is calculated algebraically, so negative numbers simply decrease the total sum as expected.

What are common pitfalls when solving this algorithmic challenge?

A major pitfall is integer overflow. Notice how the code uses long long for leftSum and rightSum. With constraints allowing values up to 10^5 and an array length of 10^5, the maximum possible sum could be 10^10, which exceeds the maximum value of a standard 32-bit signed integer (~2 * 10^9). Thus, utilizing a 64-bit integer type is critical for a correct implementation.

Conclusion

Mastering prefix sums is crucial for competitive programming and technical interviews. By reducing redundant array sweeps, we achieve an elegant, linear-time solution. Practice this pattern to identify similar optimization opportunities in other array partitioning problems. This algorithmic strategy not only applies to arrays but also to a wide array of sliding window problems.