LeetCode #1872 — HARD

Stone Game VIII

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

Alice and Bob take turns playing a game, with Alice starting first.

There are n stones arranged in a row. On each player's turn, while the number of stones is more than one, they will do the following:

  1. Choose an integer x > 1, and remove the leftmost x stones from the row.
  2. Add the sum of the removed stones' values to the player's score.
  3. Place a new stone, whose value is equal to that sum, on the left side of the row.

The game stops when only one stone is left in the row.

The score difference between Alice and Bob is (Alice's score - Bob's score). Alice's goal is to maximize the score difference, and Bob's goal is the minimize the score difference.

Given an integer array stones of length n where stones[i] represents the value of the ith stone from the left, return the score difference between Alice and Bob if they both play optimally.

Example 1:

Input: stones = [-1,2,-3,4,-5]
Output: 5
Explanation:
- Alice removes the first 4 stones, adds (-1) + 2 + (-3) + 4 = 2 to her score, and places a stone of
  value 2 on the left. stones = [2,-5].
- Bob removes the first 2 stones, adds 2 + (-5) = -3 to his score, and places a stone of value -3 on
  the left. stones = [-3].
The difference between their scores is 2 - (-3) = 5.

Example 2:

Input: stones = [7,-6,5,10,5,-2,-6]
Output: 13
Explanation:
- Alice removes all stones, adds 7 + (-6) + 5 + 10 + 5 + (-2) + (-6) = 13 to her score, and places a
  stone of value 13 on the left. stones = [13].
The difference between their scores is 13 - 0 = 13.

Example 3:

Input: stones = [-10,-12]
Output: -22
Explanation:
- Alice can only make one move, which is to remove both stones. She adds (-10) + (-12) = -22 to her
  score and places a stone of value -22 on the left. stones = [-22].
The difference between their scores is (-22) - 0 = -22.

Constraints:

  • n == stones.length
  • 2 <= n <= 105
  • -104 <= stones[i] <= 104
Patterns Used

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: Alice and Bob take turns playing a game, with Alice starting first. There are n stones arranged in a row. On each player's turn, while the number of stones is more than one, they will do the following: Choose an integer x > 1, and remove the leftmost x stones from the row. Add the sum of the removed stones' values to the player's score. Place a new stone, whose value is equal to that sum, on the left side of the row. The game stops when only one stone is left in the row. The score difference between Alice and Bob is (Alice's score - Bob's score). Alice's goal is to maximize the score difference, and Bob's goal is the minimize the score difference. Given an integer array stones of length n where stones[i] represents the value of the ith stone from the left, return the score difference between Alice and Bob if they both play optimally.

Baseline thinking

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

Pattern signal: Array · Math · Dynamic Programming

Example 1

[-1,2,-3,4,-5]

Example 2

[7,-6,5,10,5,-2,-6]

Example 3

[-10,-12]

Related Problems

  • Stone Game (stone-game)
  • Stone Game II (stone-game-ii)
  • Stone Game III (stone-game-iii)
  • Stone Game IV (stone-game-iv)
  • Stone Game V (stone-game-v)
Step 02

Core Insight

What unlocks the optimal approach

  • Let's note that the only thing that matters is how many stones were removed so we can maintain dp[numberOfRemovedStones]
  • dp[x] = max(sum of all elements up to y - dp[y]) for all y > x
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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
Largest constraint values
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 #1872: Stone Game VIII
class Solution {
    private Integer[] f;
    private int[] s;
    private int n;

    public int stoneGameVIII(int[] stones) {
        n = stones.length;
        f = new Integer[n];
        for (int i = 1; i < n; ++i) {
            stones[i] += stones[i - 1];
        }
        s = stones;
        return dfs(1);
    }

    private int dfs(int i) {
        if (i >= n - 1) {
            return s[i];
        }
        if (f[i] == null) {
            f[i] = Math.max(dfs(i + 1), s[i] - dfs(i + 1));
        }
        return f[i];
    }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(n)

Approach Breakdown

RECURSIVE
O(2ⁿ) time
O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING
O(n × m) time
O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.
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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.