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1406. Stone Game III
Description
Alice and Bob continue their games with piles of stones. There are several stones arranged in a row, and each stone has an associated value which is an integer given in the array stoneValue
.
Alice and Bob take turns, with Alice starting first. On each player's turn, that player can take 1
, 2
, or 3
stones from the first remaining stones in the row.
The score of each player is the sum of the values of the stones taken. The score of each player is 0
initially.
The objective of the game is to end with the highest score, and the winner is the player with the highest score and there could be a tie. The game continues until all the stones have been taken.
Assume Alice and Bob play optimally.
Return "Alice"
if Alice will win, "Bob"
if Bob will win, or "Tie"
if they will end the game with the same score.
Example 1:
Input: stoneValue = [1,2,3,7] Output: "Bob" Explanation: Alice will always lose. Her best move will be to take three piles and the score become 6. Now the score of Bob is 7 and Bob wins.
Example 2:
Input: stoneValue = [1,2,3,9] Output: "Alice" Explanation: Alice must choose all the three piles at the first move to win and leave Bob with negative score. If Alice chooses one pile her score will be 1 and the next move Bob's score becomes 5. In the next move, Alice will take the pile with value = 9 and lose. If Alice chooses two piles her score will be 3 and the next move Bob's score becomes 3. In the next move, Alice will take the pile with value = 9 and also lose. Remember that both play optimally so here Alice will choose the scenario that makes her win.
Example 3:
Input: stoneValue = [1,2,3,6] Output: "Tie" Explanation: Alice cannot win this game. She can end the game in a draw if she decided to choose all the first three piles, otherwise she will lose.
Constraints:
1 <= stoneValue.length <= 5 * 10^{4}
1000 <= stoneValue[i] <= 1000
Solutions
Solution 1: Memoization Search
We design a function $dfs(i)$, which represents the maximum score difference that the current player can obtain when playing the game in the range $[i, n)$. If $dfs(0) > 0$, it means that the first player Alice can win; if $dfs(0) < 0$, it means that the second player Bob can win; otherwise, it means that the two players tie.
The execution logic of the function $dfs(i)$ is as follows:
 If $i \geq n$, it means that there are no stones to take now, so we can directly return $0$;
 Otherwise, we enumerate that the current player takes the first $j+1$ piles of stones, where $j \in {0, 1, 2}$. Then the score difference that the other player can get in the next round is $dfs(i + j + 1)$, so the score difference that the current player can get is $\sum_{k=i}^{i+j} stoneValue[k]  dfs(i + j + 1)$. We want to maximize the score difference of the current player, so we can use the $\max$ function to get the maximum score difference, that is:
To prevent repeated calculations, we can use memoization search.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the number of piles of stones.

class Solution { private int[] stoneValue; private Integer[] f; private int n; public String stoneGameIII(int[] stoneValue) { n = stoneValue.length; f = new Integer[n]; this.stoneValue = stoneValue; int ans = dfs(0); if (ans == 0) { return "Tie"; } return ans > 0 ? "Alice" : "Bob"; } private int dfs(int i) { if (i >= n) { return 0; } if (f[i] != null) { return f[i]; } int ans = (1 << 30); int s = 0; for (int j = 0; j < 3 && i + j < n; ++j) { s += stoneValue[i + j]; ans = Math.max(ans, s  dfs(i + j + 1)); } return f[i] = ans; } }

class Solution { public: string stoneGameIII(vector<int>& stoneValue) { int n = stoneValue.size(); int f[n]; memset(f, 0x3f, sizeof(f)); function<int(int)> dfs = [&](int i) > int { if (i >= n) { return 0; } if (f[i] != 0x3f3f3f3f) { return f[i]; } int ans = (1 << 30), s = 0; for (int j = 0; j < 3 && i + j < n; ++j) { s += stoneValue[i + j]; ans = max(ans, s  dfs(i + j + 1)); } return f[i] = ans; }; int ans = dfs(0); if (ans == 0) { return "Tie"; } return ans > 0 ? "Alice" : "Bob"; } };

class Solution: def stoneGameIII(self, stoneValue: List[int]) > str: @cache def dfs(i: int) > int: if i >= n: return 0 ans, s = inf, 0 for j in range(3): if i + j >= n: break s += stoneValue[i + j] ans = max(ans, s  dfs(i + j + 1)) return ans n = len(stoneValue) ans = dfs(0) if ans == 0: return 'Tie' return 'Alice' if ans > 0 else 'Bob'

func stoneGameIII(stoneValue []int) string { n := len(stoneValue) f := make([]int, n) const inf = 1 << 30 for i := range f { f[i] = inf } var dfs func(int) int dfs = func(i int) int { if i >= n { return 0 } if f[i] != inf { return f[i] } ans, s := (1 << 30), 0 for j := 0; j < 3 && i+j < n; j++ { s += stoneValue[i+j] ans = max(ans, sdfs(i+j+1)) } f[i] = ans return ans } ans := dfs(0) if ans == 0 { return "Tie" } if ans > 0 { return "Alice" } return "Bob" }

function stoneGameIII(stoneValue: number[]): string { const n = stoneValue.length; const inf = 1 << 30; const f: number[] = new Array(n).fill(inf); const dfs = (i: number): number => { if (i >= n) { return 0; } if (f[i] !== inf) { return f[i]; } let ans = inf; let s = 0; for (let j = 0; j < 3 && i + j < n; ++j) { s += stoneValue[i + j]; ans = Math.max(ans, s  dfs(i + j + 1)); } return (f[i] = ans); }; const ans = dfs(0); if (ans === 0) { return 'Tie'; } return ans > 0 ? 'Alice' : 'Bob'; }