LeetCode #877 — MEDIUM

Stone Game

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

The Problem

Problem Statement

Alice and Bob play a game with piles of stones. There are an even 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. The total number of stones across all the piles is odd, so there are no ties.

Alice and Bob take turns, with Alice starting first. Each turn, a player takes the entire pile of stones either from the beginning or from the end of the row. This continues until there are no more piles left, at which point the person with the most stones wins.

Assuming Alice and Bob play optimally, return true if Alice wins the game, or false if Bob wins.

Example 1:

Input: piles = [5,3,4,5]
Output: true
Explanation: 
Alice starts first, and can only take the first 5 or the last 5.
Say she takes the first 5, so that the row becomes [3, 4, 5].
If Bob takes 3, then the board is [4, 5], and Alice takes 5 to win with 10 points.
If Bob takes the last 5, then the board is [3, 4], and Alice takes 4 to win with 9 points.
This demonstrated that taking the first 5 was a winning move for Alice, so we return true.

Example 2:

Input: piles = [3,7,2,3]
Output: true

Constraints:

2 <= piles.length <= 500
piles.length is even.
1 <= piles[i] <= 500
sum(piles[i]) is odd.

Step 01

Brute Force Baseline

Problem summary: Alice and Bob play a game with piles of stones. There are an even 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. The total number of stones across all the piles is odd, so there are no ties. Alice and Bob take turns, with Alice starting first. Each turn, a player takes the entire pile of stones either from the beginning or from the end of the row. This continues until there are no more piles left, at which point the person with the most stones wins. Assuming Alice and Bob play optimally, return true if Alice wins the game, or false if Bob wins.

Baseline thinking

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

Pattern signal: Array · Math · Dynamic Programming

Example 1

[5,3,4,5]

Example 2

[3,7,2,3]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #877: Stone Game
class Solution {
    private int[] piles;
    private int[][] f;

    public boolean stoneGame(int[] piles) {
        this.piles = piles;
        int n = piles.length;
        f = new int[n][n];
        return dfs(0, n - 1) > 0;
    }

    private int dfs(int i, int j) {
        if (i > j) {
            return 0;
        }
        if (f[i][j] != 0) {
            return f[i][j];
        }
        return f[i][j] = Math.max(piles[i] - dfs(i + 1, j), piles[j] - dfs(i, j - 1));
    }
}

// Accepted solution for LeetCode #877: Stone Game
func stoneGame(piles []int) bool {
	n := len(piles)
	f := make([][]int, n)
	for i := range f {
		f[i] = make([]int, n)
	}
	var dfs func(i, j int) int
	dfs = func(i, j int) int {
		if i > j {
			return 0
		}
		if f[i][j] == 0 {
			f[i][j] = max(piles[i]-dfs(i+1, j), piles[j]-dfs(i, j-1))
		}
		return f[i][j]
	}
	return dfs(0, n-1) > 0
}

# Accepted solution for LeetCode #877: Stone Game
class Solution:
    def stoneGame(self, piles: List[int]) -> bool:
        @cache
        def dfs(i: int, j: int) -> int:
            if i > j:
                return 0
            return max(piles[i] - dfs(i + 1, j), piles[j] - dfs(i, j - 1))

        return dfs(0, len(piles) - 1) > 0

// Accepted solution for LeetCode #877: Stone Game
struct Solution;

impl Solution {
    fn stone_game(_: Vec<i32>) -> bool {
        true
    }
}

#[test]
fn test() {
    let piles = vec![5, 3, 4, 5];
    let res = true;
    assert_eq!(Solution::stone_game(piles), res);
}

// Accepted solution for LeetCode #877: Stone Game
function stoneGame(piles: number[]): boolean {
    const n = piles.length;
    const f: number[][] = new Array(n).fill(0).map(() => new Array(n).fill(0));
    const dfs = (i: number, j: number): number => {
        if (i > j) {
            return 0;
        }
        if (f[i][j] === 0) {
            f[i][j] = Math.max(piles[i] - dfs(i + 1, j), piles[j] - dfs(i, j - 1));
        }
        return f[i][j];
    };
    return dfs(0, n - 1) > 0;
}

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 × m)

Space

O(n × m)

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.