LeetCode #1140 — MEDIUM

Stone Game II

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

The Problem

Problem Statement

Alice and Bob continue their games with piles of stones. There are a number of piles arranged in a row, and each pile has a positive integer number of stones piles[i]. The objective of the game is to end with the most stones.

Alice and Bob take turns, with Alice starting first.

On each player's turn, that player can take all the stones in the first X remaining piles, where 1 <= X <= 2M. Then, we set M = max(M, X). Initially, M = 1.

The game continues until all the stones have been taken.

Assuming Alice and Bob play optimally, return the maximum number of stones Alice can get.

Example 1:

Input: piles = [2,7,9,4,4]

Output: 10

Explanation:

If Alice takes one pile at the beginning, Bob takes two piles, then Alice takes 2 piles again. Alice can get 2 + 4 + 4 = 10 stones in total.
If Alice takes two piles at the beginning, then Bob can take all three piles left. In this case, Alice get 2 + 7 = 9 stones in total.

So we return 10 since it's larger.

Example 2:

Input: piles = [1,2,3,4,5,100]

Output: 104

Constraints:

1 <= piles.length <= 100
1 <= piles[i] <= 10⁴

Step 01

Brute Force Baseline

Problem summary: Alice and Bob continue their games with piles of stones. There are a number of piles arranged in a row, and each pile has a positive integer number of stones piles[i]. The objective of the game is to end with the most stones. Alice and Bob take turns, with Alice starting first. On each player's turn, that player can take all the stones in the first X remaining piles, where 1 <= X <= 2M. Then, we set M = max(M, X). Initially, M = 1. The game continues until all the stones have been taken. Assuming Alice and Bob play optimally, return the maximum number of stones Alice can get.

Baseline thinking

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

Pattern signal: Array · Math · Dynamic Programming

Example 1

[2,7,9,4,4]

Example 2

[1,2,3,4,5,100]

Core Insight

What unlocks the optimal approach

Use dynamic programming: the states are (i, m) for the answer of piles[i:] and that given m.

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 #1140: Stone Game II
class Solution {
    private int[] s;
    private Integer[][] f;
    private int n;

    public int stoneGameII(int[] piles) {
        n = piles.length;
        s = new int[n + 1];
        f = new Integer[n][n + 1];
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + piles[i];
        }
        return dfs(0, 1);
    }

    private int dfs(int i, int m) {
        if (m * 2 >= n - i) {
            return s[n] - s[i];
        }
        if (f[i][m] != null) {
            return f[i][m];
        }
        int res = 0;
        for (int x = 1; x <= m * 2; ++x) {
            res = Math.max(res, s[n] - s[i] - dfs(i + x, Math.max(m, x)));
        }
        return f[i][m] = res;
    }
}

// Accepted solution for LeetCode #1140: Stone Game II
func stoneGameII(piles []int) int {
	n := len(piles)
	s := make([]int, n+1)
	f := make([][]int, n+1)
	for i, x := range piles {
		s[i+1] = s[i] + x
		f[i] = make([]int, n+1)
	}
	var dfs func(i, m int) int
	dfs = func(i, m int) int {
		if m*2 >= n-i {
			return s[n] - s[i]
		}
		if f[i][m] > 0 {
			return f[i][m]
		}
		f[i][m] = 0
		for x := 1; x <= m<<1; x++ {
			f[i][m] = max(f[i][m], s[n]-s[i]-dfs(i+x, max(m, x)))
		}
		return f[i][m]
	}
	return dfs(0, 1)
}

# Accepted solution for LeetCode #1140: Stone Game II
class Solution:
    def stoneGameII(self, piles: List[int]) -> int:
        @cache
        def dfs(i, m):
            if m * 2 >= n - i:
                return s[n] - s[i]
            return max(
                s[n] - s[i] - dfs(i + x, max(m, x)) for x in range(1, m << 1 | 1)
            )

        n = len(piles)
        s = list(accumulate(piles, initial=0))
        return dfs(0, 1)

// Accepted solution for LeetCode #1140: Stone Game II
struct Solution;
use std::collections::HashMap;

impl Solution {
    fn stone_game_ii(piles: Vec<i32>) -> i32 {
        let n = piles.len();
        let mut memo: HashMap<(usize, usize), (i32, i32)> = HashMap::new();
        Self::dp(0, 1, &mut memo, &piles, n).0
    }

    fn dp(
        start: usize,
        m: usize,
        memo: &mut HashMap<(usize, usize), (i32, i32)>,
        piles: &[i32],
        n: usize,
    ) -> (i32, i32) {
        if start == n {
            (0, 0)
        } else {
            if let Some(&res) = memo.get(&(start, m)) {
                return res;
            }
            let mut a = 0;
            let mut res = (0, 0);
            for i in start..(start + 2 * m).min(n) {
                a += piles[i];
                let x = i - start + 1;
                let (b, c) = Self::dp(i + 1, x.max(m), memo, piles, n);
                if a + c > res.0 {
                    res = (a + c, b);
                }
            }
            memo.insert((start, m), res);
            res
        }
    }
}

#[test]
fn test() {
    let piles = vec![2, 7, 9, 4, 4];
    let res = 10;
    assert_eq!(Solution::stone_game_ii(piles), res);
}

// Accepted solution for LeetCode #1140: Stone Game II
function stoneGameII(piles: number[]): number {
    const n = piles.length;
    const f = Array.from({ length: n }, _ => new Array(n + 1).fill(0));
    const s = new Array(n + 1).fill(0);
    for (let i = 0; i < n; ++i) {
        s[i + 1] = s[i] + piles[i];
    }
    const dfs = (i: number, m: number) => {
        if (m * 2 >= n - i) {
            return s[n] - s[i];
        }
        if (f[i][m]) {
            return f[i][m];
        }
        let res = 0;
        for (let x = 1; x <= m * 2; ++x) {
            res = Math.max(res, s[n] - s[i] - dfs(i + x, Math.max(m, x)));
        }
        return (f[i][m] = res);
    };
    return dfs(0, 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^3)

Space

O(n^2)

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.