LeetCode #1712 — MEDIUM

Ways to Split Array Into Three Subarrays

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

The Problem

Problem Statement

A split of an integer array is good if:

The array is split into three non-empty contiguous subarrays - named left, mid, right respectively from left to right.
The sum of the elements in left is less than or equal to the sum of the elements in mid, and the sum of the elements in mid is less than or equal to the sum of the elements in right.

Given nums, an array of non-negative integers, return the number of good ways to split nums. As the number may be too large, return it modulo 10⁹+ 7.

Example 1:

Input: nums = [1,1,1]
Output: 1
Explanation: The only good way to split nums is [1] [1] [1].

Example 2:

Input: nums = [1,2,2,2,5,0]
Output: 3
Explanation: There are three good ways of splitting nums:
[1] [2] [2,2,5,0]
[1] [2,2] [2,5,0]
[1,2] [2,2] [5,0]

Example 3:

Input: nums = [3,2,1]
Output: 0
Explanation: There is no good way to split nums.

Constraints:

3 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁴

Step 01

Brute Force Baseline

Problem summary: A split of an integer array is good if: The array is split into three non-empty contiguous subarrays - named left, mid, right respectively from left to right. The sum of the elements in left is less than or equal to the sum of the elements in mid, and the sum of the elements in mid is less than or equal to the sum of the elements in right. Given nums, an array of non-negative integers, return the number of good ways to split nums. As the number may be too large, return it modulo 109 + 7.

Baseline thinking

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

Pattern signal: Array · Two Pointers · Binary Search

Example 1

[1,1,1]

Example 2

[1,2,2,2,5,0]

Example 3

[3,2,1]

Core Insight

What unlocks the optimal approach

Create a prefix array to efficiently find the sum of subarrays.
As we are dividing the array into three subarrays, there are two "walls". Iterate over the right wall positions and find where the left wall could be for each right wall position.
Use binary search to find the left-most position and right-most position the left wall could be.

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 #1712: Ways to Split Array Into Three Subarrays
class Solution {
    private static final int MOD = (int) 1e9 + 7;

    public int waysToSplit(int[] nums) {
        int n = nums.length;
        int[] s = new int[n];
        s[0] = nums[0];
        for (int i = 1; i < n; ++i) {
            s[i] = s[i - 1] + nums[i];
        }
        int ans = 0;
        for (int i = 0; i < n - 2; ++i) {
            int j = search(s, s[i] << 1, i + 1, n - 1);
            int k = search(s, ((s[n - 1] + s[i]) >> 1) + 1, j, n - 1);
            ans = (ans + k - j) % MOD;
        }
        return ans;
    }

    private int search(int[] s, int x, int left, int right) {
        while (left < right) {
            int mid = (left + right) >> 1;
            if (s[mid] >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

// Accepted solution for LeetCode #1712: Ways to Split Array Into Three Subarrays
func waysToSplit(nums []int) (ans int) {
	const mod int = 1e9 + 7
	n := len(nums)
	s := make([]int, n)
	s[0] = nums[0]
	for i := 1; i < n; i++ {
		s[i] = s[i-1] + nums[i]
	}
	for i := 0; i < n-2; i++ {
		j := sort.Search(n-1, func(h int) bool { return h > i && s[h] >= (s[i]<<1) })
		k := sort.Search(n-1, func(h int) bool { return h >= j && s[h] > (s[n-1]+s[i])>>1 })
		ans = (ans + k - j) % mod
	}
	return
}

# Accepted solution for LeetCode #1712: Ways to Split Array Into Three Subarrays
class Solution:
    def waysToSplit(self, nums: List[int]) -> int:
        mod = 10**9 + 7
        s = list(accumulate(nums))
        ans, n = 0, len(nums)
        for i in range(n - 2):
            j = bisect_left(s, s[i] << 1, i + 1, n - 1)
            k = bisect_right(s, (s[-1] + s[i]) >> 1, j, n - 1)
            ans += k - j
        return ans % mod

// Accepted solution for LeetCode #1712: Ways to Split Array Into Three Subarrays
struct Solution;

const MOD: i64 = 1_000_000_007;

impl Solution {
    fn ways_to_split(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut prefix: Vec<i32> = vec![];
        let mut prev = 0;
        let mut res = 0;
        for i in 0..n {
            prev += nums[i];
            prefix.push(prev);
        }
        let total = prev;
        let mut mid_start = 0;
        let mut mid_end = 0;
        for i in 0..n {
            let left = prefix[i];
            mid_start = mid_start.max(i + 1);
            while mid_start < n && prefix[mid_start] - left < left {
                mid_start += 1;
            }
            while mid_end + 1 < n && total - prefix[mid_end] >= prefix[mid_end] - left {
                mid_end += 1;
            }
            if mid_end >= mid_start {
                res += (mid_end - mid_start) as i64;
                res %= MOD;
            }
        }
        res as i32
    }
}

#[test]
fn test() {
    let nums = vec![1, 1, 1];
    let res = 1;
    assert_eq!(Solution::ways_to_split(nums), res);
    let nums = vec![1, 2, 2, 2, 5, 0];
    let res = 3;
    assert_eq!(Solution::ways_to_split(nums), res);
    let nums = vec![3, 2, 1];
    let res = 0;
    assert_eq!(Solution::ways_to_split(nums), res);
    let nums = vec![0, 3, 3];
    let res = 1;
    assert_eq!(Solution::ways_to_split(nums), res);
}

// Accepted solution for LeetCode #1712: Ways to Split Array Into Three Subarrays
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1712: Ways to Split Array Into Three Subarrays
// class Solution {
//     private static final int MOD = (int) 1e9 + 7;
// 
//     public int waysToSplit(int[] nums) {
//         int n = nums.length;
//         int[] s = new int[n];
//         s[0] = nums[0];
//         for (int i = 1; i < n; ++i) {
//             s[i] = s[i - 1] + nums[i];
//         }
//         int ans = 0;
//         for (int i = 0; i < n - 2; ++i) {
//             int j = search(s, s[i] << 1, i + 1, n - 1);
//             int k = search(s, ((s[n - 1] + s[i]) >> 1) + 1, j, n - 1);
//             ans = (ans + k - j) % MOD;
//         }
//         return ans;
//     }
// 
//     private int search(int[] s, int x, int left, int right) {
//         while (left < right) {
//             int mid = (left + right) >> 1;
//             if (s[mid] >= x) {
//                 right = mid;
//             } else {
//                 left = mid + 1;
//             }
//         }
//         return left;
//     }
// }

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 of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

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.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.