LeetCode #1690 — MEDIUM

Stone Game VII

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

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, they can remove either the leftmost stone or the rightmost stone from the row and receive points equal to the sum of the remaining stones' values in the row. The winner is the one with the higher score when there are no stones left to remove.

Bob found that he will always lose this game (poor Bob, he always loses), so he decided to minimize the score's difference. Alice's goal is to maximize the difference in the score.

Given an array of integers stones where stones[i] represents the value of the i^th stone from the left, return the difference in Alice and Bob's score if they both play optimally.

Example 1:

Input: stones = [5,3,1,4,2]
Output: 6
Explanation: 
- Alice removes 2 and gets 5 + 3 + 1 + 4 = 13 points. Alice = 13, Bob = 0, stones = [5,3,1,4].
- Bob removes 5 and gets 3 + 1 + 4 = 8 points. Alice = 13, Bob = 8, stones = [3,1,4].
- Alice removes 3 and gets 1 + 4 = 5 points. Alice = 18, Bob = 8, stones = [1,4].
- Bob removes 1 and gets 4 points. Alice = 18, Bob = 12, stones = [4].
- Alice removes 4 and gets 0 points. Alice = 18, Bob = 12, stones = [].
The score difference is 18 - 12 = 6.

Example 2:

Input: stones = [7,90,5,1,100,10,10,2]
Output: 122

Constraints:

n == stones.length
2 <= n <= 1000
1 <= stones[i] <= 1000

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, they can remove either the leftmost stone or the rightmost stone from the row and receive points equal to the sum of the remaining stones' values in the row. The winner is the one with the higher score when there are no stones left to remove. Bob found that he will always lose this game (poor Bob, he always loses), so he decided to minimize the score's difference. Alice's goal is to maximize the difference in the score. Given an array of integers stones where stones[i] represents the value of the ith stone from the left, return the difference in Alice and Bob's score 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

[5,3,1,4,2]

Example 2

[7,90,5,1,100,10,10,2]

Core Insight

What unlocks the optimal approach

The constraints are small enough for an N^2 solution.
Try using dynamic programming.

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 #1690: Stone Game VII
class Solution {
    private int[] s;
    private Integer[][] f;

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

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

// Accepted solution for LeetCode #1690: Stone Game VII
func stoneGameVII(stones []int) int {
	n := len(stones)
	s := make([]int, n+1)
	f := make([][]int, n)
	for i, x := range stones {
		s[i+1] = s[i] + x
		f[i] = make([]int, n)
	}
	var dfs func(int, int) int
	dfs = func(i, j int) int {
		if i > j {
			return 0
		}
		if f[i][j] != 0 {
			return f[i][j]
		}
		a := s[j+1] - s[i+1] - dfs(i+1, j)
		b := s[j] - s[i] - dfs(i, j-1)
		f[i][j] = max(a, b)
		return f[i][j]
	}
	return dfs(0, n-1)
}

# Accepted solution for LeetCode #1690: Stone Game VII
class Solution:
    def stoneGameVII(self, stones: List[int]) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            if i > j:
                return 0
            a = s[j + 1] - s[i + 1] - dfs(i + 1, j)
            b = s[j] - s[i] - dfs(i, j - 1)
            return max(a, b)

        s = list(accumulate(stones, initial=0))
        ans = dfs(0, len(stones) - 1)
        dfs.cache_clear()
        return ans

// Accepted solution for LeetCode #1690: Stone Game VII
struct Solution;

impl Solution {
    fn stone_game_vii(stones: Vec<i32>) -> i32 {
        let n = stones.len();
        let mut memo: Vec<Vec<i32>> = vec![vec![0; n]; n];
        let sum = stones.iter().sum();
        Self::dp(0, n - 1, &mut memo, &stones, sum)
    }

    fn dp(l: usize, r: usize, memo: &mut Vec<Vec<i32>>, stones: &[i32], sum: i32) -> i32 {
        if l == r {
            0
        } else {
            if memo[l][r] != 0 {
                memo[l][r]
            } else {
                let mut res = 0;
                res = res.max(sum - stones[l] - Self::dp(l + 1, r, memo, stones, sum - stones[l]));
                res = res.max(sum - stones[r] - Self::dp(l, r - 1, memo, stones, sum - stones[r]));
                memo[l][r] = res;
                res
            }
        }
    }
}

#[test]
fn test() {
    let stones = vec![5, 3, 1, 4, 2];
    let res = 6;
    assert_eq!(Solution::stone_game_vii(stones), res);
    let stones = vec![7, 90, 5, 1, 100, 10, 10, 2];
    let res = 122;
    assert_eq!(Solution::stone_game_vii(stones), res);
}

// Accepted solution for LeetCode #1690: Stone Game VII
function stoneGameVII(stones: number[]): number {
    const n = stones.length;
    const s: number[] = Array(n + 1).fill(0);
    for (let i = 0; i < n; ++i) {
        s[i + 1] = s[i] + stones[i];
    }
    const f: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
    const dfs = (i: number, j: number): number => {
        if (i > j) {
            return 0;
        }
        if (f[i][j]) {
            return f[i][j];
        }
        const a = s[j + 1] - s[i + 1] - dfs(i + 1, j);
        const b = s[j] - s[i] - dfs(i, j - 1);
        return (f[i][j] = Math.max(a, b));
    };
    return dfs(0, n - 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^2)

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.