LeetCode #2270 — MEDIUM

Number of Ways to Split Array

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[10,4,-8,7]

Example 2

[2,3,1,0]

Core Insight

What unlocks the optimal approach

For any index i, how can we find the sum of the first (i+1) elements from the sum of the first i elements?
If the total sum of the array is known, how can we check if the sum of the first (i+1) elements greater than or equal to the remaining elements?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2270: Number of Ways to Split Array
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;
    }
}

// Accepted solution for LeetCode #2270: Number of Ways to Split Array
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
}

# Accepted solution for LeetCode #2270: Number of Ways to Split Array
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

// Accepted solution for LeetCode #2270: Number of Ways to Split Array
/**
 * [2270] Number of Ways to Split Array
 *
 * 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
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/number-of-ways-to-split-array/
// discuss: https://leetcode.com/problems/number-of-ways-to-split-array/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn ways_to_split_array(nums: Vec<i32>) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2270_example_1() {
        let nums = vec![10, 4, -8, 7];

        let result = 2;

        assert_eq!(Solution::ways_to_split_array(nums), result);
    }

    #[test]
    #[ignore]
    fn test_2270_example_2() {
        let nums = vec![2, 3, 1, 0];

        let result = 2;

        assert_eq!(Solution::ways_to_split_array(nums), result);
    }
}

// Accepted solution for LeetCode #2270: Number of Ways to Split Array
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.