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.
Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
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:
x > 1, and remove the leftmost x stones from 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.length2 <= n <= 105-104 <= stones[i] <= 104Problem 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Math · Dynamic Programming
[-1,2,-3,4,-5]
[7,-6,5,10,5,-2,-6]
[-10,-12]
stone-game)stone-game-ii)stone-game-iii)stone-game-iv)stone-game-v)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];
}
}
// Accepted solution for LeetCode #1872: Stone Game VIII
func stoneGameVIII(stones []int) int {
n := len(stones)
f := make([]int, n)
for i := range f {
f[i] = -1
}
for i := 1; i < n; i++ {
stones[i] += stones[i-1]
}
var dfs func(int) int
dfs = func(i int) int {
if i >= n-1 {
return stones[i]
}
if f[i] == -1 {
f[i] = max(dfs(i+1), stones[i]-dfs(i+1))
}
return f[i]
}
return dfs(1)
}
# Accepted solution for LeetCode #1872: Stone Game VIII
class Solution:
def stoneGameVIII(self, stones: List[int]) -> int:
@cache
def dfs(i: int) -> int:
if i >= len(stones) - 1:
return s[-1]
return max(dfs(i + 1), s[i] - dfs(i + 1))
s = list(accumulate(stones))
return dfs(1)
// Accepted solution for LeetCode #1872: Stone Game VIII
/**
* [1872] Stone Game VIII
*
* 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:
*
* <ol>
* 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.
* </ol>
*
* 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 i^th 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 <= 10^5
* -10^4 <= stones[i] <= 10^4
*
*/
pub struct Solution {}
// problem: https://leetcode.com/problems/stone-game-viii/
// discuss: https://leetcode.com/problems/stone-game-viii/discuss/?currentPage=1&orderBy=most_votes&query=
// submission codes start here
impl Solution {
pub fn stone_game_viii(stones: Vec<i32>) -> i32 {
0
}
}
// submission codes end
#[cfg(test)]
mod tests {
use super::*;
#[test]
#[ignore]
fn test_1872_example_1() {
let stones = vec![-1, 2, -3, 4, -5];
let result = 5;
assert_eq!(Solution::stone_game_viii(stones), result);
}
#[test]
#[ignore]
fn test_1872_example_2() {
let stones = vec![7, -6, 5, 10, 5, -2, -6];
let result = 13;
assert_eq!(Solution::stone_game_viii(stones), result);
}
#[test]
#[ignore]
fn test_1872_example_3() {
let stones = vec![-10, -12];
let result = -22;
assert_eq!(Solution::stone_game_viii(stones), result);
}
}
// Accepted solution for LeetCode #1872: Stone Game VIII
function stoneGameVIII(stones: number[]): number {
const n = stones.length;
const f: number[] = Array(n).fill(-1);
for (let i = 1; i < n; ++i) {
stones[i] += stones[i - 1];
}
const dfs = (i: number): number => {
if (i >= n - 1) {
return stones[i];
}
if (f[i] === -1) {
f[i] = Math.max(dfs(i + 1), stones[i] - dfs(i + 1));
}
return f[i];
};
return dfs(1);
}
Use this to step through a reusable interview workflow for this problem.
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.
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.
Review these before coding to avoid predictable interview regressions.
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.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
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.