LeetCode #3788 — MEDIUM

Maximum Score of a Split

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

Solve on LeetCode

The Problem

Problem Statement

You are given an integer array nums of length n.

Choose an index i such that 0 <= i < n - 1.

For a chosen split index i:

Let prefixSum(i) be the sum of nums[0] + nums[1] + ... + nums[i].
Let suffixMin(i) be the minimum value among nums[i + 1], nums[i + 2], ..., nums[n - 1].

The score of a split at index i is defined as:

score(i) = prefixSum(i) - suffixMin(i)

Return an integer denoting the maximum score over all valid split indices.

Example 1:

Input: nums = [10,-1,3,-4,-5]

Output: 17

Explanation:

The optimal split is at i = 2, score(2) = prefixSum(2) - suffixMin(2) = (10 + (-1) + 3) - (-5) = 17.

Example 2:

Input: nums = [-7,-5,3]

Output: -2

Explanation:

The optimal split is at i = 0, score(0) = prefixSum(0) - suffixMin(0) = (-7) - (-5) = -2.

Example 3:

Input: nums = [1,1]

Output: 0

Explanation:

The only valid split is at i = 0, score(0) = prefixSum(0) - suffixMin(0) = 1 - 1 = 0.

Constraints:

2 <= nums.length <= 10⁵
-10⁹ <= nums[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums of length n. Choose an index i such that 0 <= i < n - 1. For a chosen split index i: Let prefixSum(i) be the sum of nums[0] + nums[1] + ... + nums[i]. Let suffixMin(i) be the minimum value among nums[i + 1], nums[i + 2], ..., nums[n - 1]. The score of a split at index i is defined as: score(i) = prefixSum(i) - suffixMin(i) Return an integer denoting the maximum score over all valid split indices.

Baseline thinking

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

Pattern signal: Array

Example 1

[10,-1,3,-4,-5]

Example 2

[-7,-5,3]

Example 3

[1,1]

Step 02

Core Insight

What unlocks the optimal approach

Use prefix sum and suffix minimum arrays.
Precompute the prefix sums of <code>nums</code> and the suffix minimums for all valid split positions.
For each split index <code>i</code>, compute <code>score(i) = prefixSum(i) - suffixMin(i)</code> and keep track of the maximum value.

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 #3788: Maximum Score of a Split
class Solution {
    public long maximumScore(int[] nums) {
        int n = nums.length;
        long[] suf = new long[n];
        suf[n - 1] = nums[n - 1];
        for (int i = n - 2; i >= 0; --i) {
            suf[i] = Math.min(nums[i], suf[i + 1]);
        }
        long ans = Long.MIN_VALUE;
        long pre = 0;
        for (int i = 0; i < n - 1; ++i) {
            pre += nums[i];
            ans = Math.max(ans, pre - suf[i + 1]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3788: Maximum Score of a Split
func maximumScore(nums []int) int64 {
	n := len(nums)
	suf := make([]int64, n)
	suf[n-1] = int64(nums[n-1])
	for i := n - 2; i >= 0; i-- {
		suf[i] = min(int64(nums[i]), suf[i+1])
	}
	var pre int64 = 0
	var ans int64 = math.MinInt64
	for i := 0; i < n-1; i++ {
		pre += int64(nums[i])
		ans = max(ans, pre-suf[i+1])
	}
	return ans
}

# Accepted solution for LeetCode #3788: Maximum Score of a Split
class Solution:
    def maximumScore(self, nums: List[int]) -> int:
        n = len(nums)
        suf = [nums[-1]] * n
        for i in range(n - 2, -1, -1):
            suf[i] = min(nums[i], suf[i + 1])
        ans = -inf
        pre = 0
        for i in range(n - 1):
            pre += nums[i]
            ans = max(ans, pre - suf[i + 1])
        return ans

// Accepted solution for LeetCode #3788: Maximum Score of a Split
// 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 #3788: Maximum Score of a Split
// class Solution {
//     public long maximumScore(int[] nums) {
//         int n = nums.length;
//         long[] suf = new long[n];
//         suf[n - 1] = nums[n - 1];
//         for (int i = n - 2; i >= 0; --i) {
//             suf[i] = Math.min(nums[i], suf[i + 1]);
//         }
//         long ans = Long.MIN_VALUE;
//         long pre = 0;
//         for (int i = 0; i < n - 1; ++i) {
//             pre += nums[i];
//             ans = Math.max(ans, pre - suf[i + 1]);
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3788: Maximum Score of a Split
function maximumScore(nums: number[]): number {
    const n = nums.length;
    const suf: number[] = new Array(n);
    suf[n - 1] = nums[n - 1];
    for (let i = n - 2; i >= 0; --i) {
        suf[i] = Math.min(nums[i], suf[i + 1]);
    }
    let ans = Number.NEGATIVE_INFINITY;
    let pre = 0;
    for (let i = 0; i < n - 1; ++i) {
        pre += nums[i];
        ans = Math.max(ans, pre - suf[i + 1]);
    }
    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(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.