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Stone Game III

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] <= 1000

Solution Explanation: Stone Game III

This problem is a variation of the classic Stone Game, incorporating negative stone values and requiring optimal play from both Alice and Bob. The optimal solution utilizes dynamic programming with memoization to efficiently determine the winner.

Approach: Memoized Dynamic Programming

The core idea is to recursively calculate the maximum score difference Alice can achieve, starting from each possible position in the stoneValue array. We use memoization (a form of dynamic programming) to avoid redundant calculations, significantly improving performance.

  1. dfs(i) function: This recursive function takes an index i representing the starting position of the remaining stones. It returns the maximum score difference Alice can achieve from position i onwards, assuming both players play optimally.

  2. Base Case: If i is beyond the array bounds (all stones have been taken), the score difference is 0.

  3. Recursive Step: The function iterates through the possible number of stones (1, 2, or 3) Alice can take in her turn. For each choice:

    • It calculates the sum of the chosen stones.
    • It recursively calls dfs() to determine the maximum score difference Bob can achieve in his subsequent turn (from the new starting position).
    • The difference between Alice's current turn score and Bob's maximum achievable score is calculated.
    • The function keeps track of the maximum score difference Alice can achieve among all possible choices.

  4. Memoization: The results of dfs(i) are stored in a memoization table (f in the code examples) to avoid recalculating the same subproblems.

  5. Final Result: dfs(0) is called to start the process, determining the maximum score difference Alice can obtain from the beginning.

    • If the result is positive, Alice wins.
    • If the result is negative, Bob wins.
    • If the result is 0, it's a tie.

Time and Space Complexity Analysis

  • Time Complexity: O(n), where n is the number of stones. Each state (represented by the index i in dfs(i)) is visited at most once due to memoization. The loop iterating through possible stone choices (1, 2, or 3) is a constant-time operation.

  • Space Complexity: O(n) to store the memoization table (f). The recursive call stack could also contribute to space complexity in the worst case, but with memoization, the depth of the recursion is at most n. Thus, the overall space complexity is dominated by the memoization table.

Code Examples (with detailed comments)

The code examples provided in the original response demonstrate the solution in various programming languages. The key elements are the dfs function, memoization, and the final logic to determine the winner based on the maximum score difference. The comments within the code illustrate the steps described above.

Remember that the efficiency gains from memoization are significant, particularly for larger input sizes, making this dynamic programming approach the optimal strategy for this problem.