LeetCode #3698 — MEDIUM

Split Array With Minimum Difference

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

The Problem

Problem Statement

You are given an integer array nums.

Split the array into exactly two subarrays, left and right, such that left is strictly increasing and right is strictly decreasing.

Return the minimum possible absolute difference between the sums of left and right. If no valid split exists, return -1.

Example 1:

Input: nums = [1,3,2]

Output: 2

Explanation:

`i`	`left`	`right`	Validity	`left` sum	`right` sum	Absolute difference
0	[1]	[3, 2]	Yes	1	5	`\|1 - 5\| = 4`
1	[1, 3]	[2]	Yes	4	2	`\|4 - 2\| = 2`

Thus, the minimum absolute difference is 2.

Example 2:

Input: nums = [1,2,4,3]

Output: 4

Explanation:

`i`	`left`	`right`	Validity	`left` sum	`right` sum	Absolute difference
0	[1]	[2, 4, 3]	No	1	9	-
1	[1, 2]	[4, 3]	Yes	3	7	`\|3 - 7\| = 4`
2	[1, 2, 4]	[3]	Yes	7	3	`\|7 - 3\| = 4`

Thus, the minimum absolute difference is 4.

Example 3:

Input: nums = [3,1,2]

Output: -1

Explanation:

No valid split exists, so the answer is -1.

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums. Split the array into exactly two subarrays, left and right, such that left is strictly increasing and right is strictly decreasing. Return the minimum possible absolute difference between the sums of left and right. If no valid split exists, return -1.

Baseline thinking

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

Pattern signal: Array

Example 1

[1,3,2]

Example 2

[1,2,4,3]

Example 3

[3,1,2]

Step 02

Core Insight

What unlocks the optimal approach

Build a prefix boolean <code>inc[i]</code> that is true iff the subarray <code>nums[0..i]</code> is strictly increasing.
Build a suffix boolean <code>dec[i]</code> that is true iff the subarray <code>nums[i..n - 1]</code> is strictly decreasing.
A split after index <code>i</code> (where <code>0 <= i < n - 1</code>) is valid iff <code>inc[i] && dec[i + 1]</code>.
Build a prefix-sum array <code>pref</code>, and use it to check if a valid split exists. If no valid split exists return <code>-1</code>.

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 #3698: Split Array With Minimum Difference
class Solution {
    public long splitArray(int[] nums) {
        int n = nums.length;
        long[] s = new long[n];
        s[0] = nums[0];
        boolean[] f = new boolean[n];
        Arrays.fill(f, true);
        boolean[] g = new boolean[n];
        Arrays.fill(g, true);
        for (int i = 1; i < n; ++i) {
            s[i] = s[i - 1] + nums[i];
            f[i] = f[i - 1];
            if (nums[i] <= nums[i - 1]) {
                f[i] = false;
            }
        }
        for (int i = n - 2; i >= 0; --i) {
            g[i] = g[i + 1];
            if (nums[i] <= nums[i + 1]) {
                g[i] = false;
            }
        }
        final long inf = Long.MAX_VALUE;
        long ans = inf;
        for (int i = 0; i < n - 1; ++i) {
            if (f[i] && g[i + 1]) {
                long s1 = s[i];
                long s2 = s[n - 1] - s[i];
                ans = Math.min(ans, Math.abs(s1 - s2));
            }
        }
        return ans < inf ? ans : -1;
    }
}

// Accepted solution for LeetCode #3698: Split Array With Minimum Difference
func splitArray(nums []int) int64 {
	n := len(nums)
	s := make([]int64, n)
	f := make([]bool, n)
	g := make([]bool, n)
	for i := range f {
		f[i] = true
		g[i] = true
	}

	s[0] = int64(nums[0])
	for i := 1; i < n; i++ {
		s[i] = s[i-1] + int64(nums[i])
		f[i] = f[i-1]
		if nums[i] <= nums[i-1] {
			f[i] = false
		}
	}
	for i := n - 2; i >= 0; i-- {
		g[i] = g[i+1]
		if nums[i] <= nums[i+1] {
			g[i] = false
		}
	}

	const inf = int64(^uint64(0) >> 1)
	ans := inf
	for i := 0; i < n-1; i++ {
		if f[i] && g[i+1] {
			s1 := s[i]
			s2 := s[n-1] - s[i]
			ans = min(ans, abs(s1-s2))
		}
	}
	if ans < inf {
		return ans
	}
	return -1
}

func abs(x int64) int64 {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #3698: Split Array With Minimum Difference
class Solution:
    def splitArray(self, nums: List[int]) -> int:
        s = list(accumulate(nums))
        n = len(nums)
        f = [True] * n
        for i in range(1, n):
            f[i] = f[i - 1]
            if nums[i] <= nums[i - 1]:
                f[i] = False
        g = [True] * n
        for i in range(n - 2, -1, -1):
            g[i] = g[i + 1]
            if nums[i] <= nums[i + 1]:
                g[i] = False
        ans = inf
        for i in range(n - 1):
            if f[i] and g[i + 1]:
                s1 = s[i]
                s2 = s[n - 1] - s[i]
                ans = min(ans, abs(s1 - s2))
        return ans if ans < inf else -1

// Accepted solution for LeetCode #3698: Split Array With Minimum Difference
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3698: Split Array With Minimum Difference
// class Solution {
//     public long splitArray(int[] nums) {
//         int n = nums.length;
//         long[] s = new long[n];
//         s[0] = nums[0];
//         boolean[] f = new boolean[n];
//         Arrays.fill(f, true);
//         boolean[] g = new boolean[n];
//         Arrays.fill(g, true);
//         for (int i = 1; i < n; ++i) {
//             s[i] = s[i - 1] + nums[i];
//             f[i] = f[i - 1];
//             if (nums[i] <= nums[i - 1]) {
//                 f[i] = false;
//             }
//         }
//         for (int i = n - 2; i >= 0; --i) {
//             g[i] = g[i + 1];
//             if (nums[i] <= nums[i + 1]) {
//                 g[i] = false;
//             }
//         }
//         final long inf = Long.MAX_VALUE;
//         long ans = inf;
//         for (int i = 0; i < n - 1; ++i) {
//             if (f[i] && g[i + 1]) {
//                 long s1 = s[i];
//                 long s2 = s[n - 1] - s[i];
//                 ans = Math.min(ans, Math.abs(s1 - s2));
//             }
//         }
//         return ans < inf ? ans : -1;
//     }
// }

// Accepted solution for LeetCode #3698: Split Array With Minimum Difference
function splitArray(nums: number[]): number {
    const n = nums.length;
    const s: number[] = Array(n);
    const f: boolean[] = Array(n).fill(true);
    const g: boolean[] = Array(n).fill(true);

    s[0] = nums[0];
    for (let i = 1; i < n; ++i) {
        s[i] = s[i - 1] + nums[i];
        f[i] = f[i - 1];
        if (nums[i] <= nums[i - 1]) {
            f[i] = false;
        }
    }

    for (let i = n - 2; i >= 0; --i) {
        g[i] = g[i + 1];
        if (nums[i] <= nums[i + 1]) {
            g[i] = false;
        }
    }

    const INF = Number.MAX_SAFE_INTEGER;
    let ans = INF;

    for (let i = 0; i < n - 1; ++i) {
        if (f[i] && g[i + 1]) {
            const s1 = s[i];
            const s2 = s[n - 1] - s[i];
            ans = Math.min(ans, Math.abs(s1 - s2));
        }
    }

    return ans < INF ? ans : -1;
}

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(n)

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.