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 * 104-1000 <= stoneValue[i] <= 1000This approach involves using dynamic programming to evaluate possible outcomes from the end of the stone array to the beginning. This allows us to make optimal decisions at each step because we know the best scores for the rest of the game. The idea is to calculate the maximum score difference Alice can enforce using a suffix-based strategy.
The code initializes a DP array to store maximum scores achievable from each position. It iterates backward through the stone array, calculating for each position the maximum score difference possible by taking 1, 2, or 3 stones and subtracting the opponent's best response.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n), where n is the number of stones. Each stone position is computed once.
Space Complexity: O(n), where n is the number of stones. It uses an array of size n+1 to store computed values.
This approach leverages the recursive computation of outcomes with memoization. We recursively compute the score differences at each step, caching the results to avoid redundant calculations. At each stone index, the player's optimal score is determined by considering 1 to 3 stone choices.
The Python solution employs a memoization decorator to cache previously computed results, optimizing recursive calls. This avoids recalculating results for subproblems, ensuring efficiency. Each call makes decisions based on potential stone selections and returns the best achievable score differential for Alice.
Time Complexity: O(n), due to computation of score differences for unique indices once per index.
Space Complexity: O(n), from the recursive call stack and memoization cache storing outcomes for each state.
| Approach | Complexity |
|---|---|
| Dynamic Programming with Suffix Evaluation | Time Complexity: O(n), where n is the number of stones. Each stone position is computed once. |
| Recursive Memoization | Time Complexity: O(n), due to computation of score differences for unique indices once per index. |
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