Problem Description

Alice and Bob are playing a game with an arrangement of stones in a row. Each stone has a value, and these values are listed in an array called stoneValue. The players take turns, with Alice going first, and on each player's turn, they may take 1, 2, or 3 stones from the beginning of the remaining row of stones.

The score for a player increases by the value of the stones they take, and everyone starts with a score of 0. The goal is to finish the game with the highest score, and it is possible that the game ends in a tie if both players have the same score when all the stones have been taken.

The key part of the problem is that both players will play optimally, meaning they will make the best possible moves to maximize their own score.

The objective is to determine the winner of the game or to find out if it will end in a tie, based on the values of the stones.

Intuition

To solve this problem, we need to use dynamic programming, as we're looking for the best strategy over several turns, considering the impact of each possible move.

The intuition behind the solution is to look at each position in the array and decide the best move for the player whose turn it is. Since each player is trying to maximize their own score, they should be looking to maximize their current score minus whatever score the opponent would get after that move. This is because while a player aims to increase their score, they should also try to minimize the future opportunities for the opponent.

The dynamic programming function, here defined as dfs(i), returns the maximum score difference (player score - opponent score) from the ith stone to the end of the array. When it's the current player's turn, they look ahead at up to 3 moves and calculate the score difference they would get if the next player played optimally from that point. We recursively calculate dfs(i) for the possible moves and choose the one with the maximum score difference.

Since Python's @cache decorator is being used, computations for each position are remembered and not recalculated multiple times, which greatly improves the efficiency.

After performing the DFS, the final answer is compared against 0. If the final score difference resulting from dfs(0) (starting at the first stone) is zero, it is a 'Tie.' If greater than zero, the current player (Alice, since she starts) wins, and if it is less than zero, it means the second player (Bob) wins.

Learn more about Math and Dynamic Programming patterns.

Solution Approach

The solution is implemented using dynamic programming, one of the most powerful algorithms in solving optimization problems, especially those involving the best strategy to play a game. In this case, memoization via Python's @cache decorator is used to store the results of subproblems.

The dfs(i) function defined within the stoneGameIII method is the core of the solution, which represents the best score difference that can be obtained starting with the ith stone.

Here's a breakdown of how the dfs(i) function operates:

• It takes a parameter i which represents the index of the starting stone in the stoneValue list.
• The function is initially called with i = 0 as we start evaluating from the first stone.
• A check is made to see if i is beyond the last stone in the array, in which case the function returns 0, indicating no more scores can be obtained.
• The function maintains two variables: ans, which will store the maximum score difference possible from this position, and s, which is the cumulative sum of stone values that have been picked.
• A loop explores taking 1, 2, or 3 stones from the current position by incrementing j. The constraint is to break out of the loop if the end of the stones array is reached.
• For each possible move (taking j stones), we calculate the sum s of values of stones taken, and we calculate a potential answer as s - dfs(i + j + 1). This calculation ensures we consider the opponent's optimal play after our move.
• We update ans to be the maximum of itself or the newly calculated potential answer.
• Once we loop through all possible moves, we return the maximum answer.

The memoization ensures that we don't compute the same state multiple times, reducing the problem to linear complexity.

Finally, when the dfs(0) is called, we check if the result is 0, indicating a tie. If it's positive, Alice wins because it means she can have a greater score difference by playing optimally. If it's negative, Bob wins for the opposite reason.

The data structure used here is essentially the list called stoneValue. The @cache decorator creates an implicit hashtable-like structure to remember previously calculated results of the dfs function.

This approach uses the concept of minimax, which is widely used in decision-making and turn-based games, to always make the best move considering the opponent will also play optimally.

Example Walkthrough

Let's illustrate the solution approach with an example. Suppose we have the stoneValue array as follows: [1,2,3,7].

Alice and Bob will play optimally, taking turns starting with Alice. Let's see how the dfs function would be used to determine the game's outcome.

1. Initially, dfs(0) is called since Alice starts with the first stone.
2. At i = 0, there are three choices for Alice:
   • Take stoneValue[0] which is 1, and then dfs(1) will be called for Bob.
   • Take stoneValue[0] and stoneValue[1] which sums to 3, and then dfs(2) will be called for Bob.
   • Take stoneValue[0], stoneValue[1], and stoneValue[2] which sums to 6, and then dfs(3) will be called for Bob.
3. dfs(1) will be Bob's turn with remaining stones [2,3,7]. Bob has the same choice to take 1, 2, or 3 stones, and after each choice, a new dfs with the appropriate index would be called for Alice.
4. This process continues until i reaches the length of the stoneValue array, at which point the function returns 0 since no stones are left to take.
5. During each call, dfs calculates the maximum score difference Alice or Bob can have at that point. For instance, if Alice takes one stone, then Bob can optimally choose the best outcome from the remaining stones, and the score difference is updated accordingly.
6. Eventually, all possible options and their score differences are computed, and dfs(i) will return the optimal score difference the current player can achieve from i to the end of the array.

The game would unfold as follows, showing the choices and corresponding dfs calls:

• dfs(0):
  • Choice 1: Take 1. dfs(1) (Bob's turn) is now evaluated.
  • Choice 2: Take 1 + 2 = 3. dfs(2) (Bob's turn) is now evaluated.
  • Choice 3: Take 1 + 2 + 3 = 6. dfs(3) (Bob's turn) is now evaluated.
• dfs(1) (Bob's turn with [2,3,7]):
  • Bob will look ahead and follow similar steps, aiming to minimize Alice's score following his moves.

After evaluating all the possibilities, dfs(0) will yield the best score difference Alice can achieve by taking stones optimally from the beginning of the array. If the score difference is:

• Greater than 0: Alice wins.
• Less than 0: Bob wins.
• Exactly 0: The game results in a tie.

For this example, Alice can secure a win by taking the first stone (with a value of 1), because now the score difference (dfs(0)) can be maximized as follows:

• Alice takes 1, and dfs(1) is called.
• If Bob takes 2, Alice can take 3 and 7 and wins. If Bob takes 2 and 3, Alice takes 7 and still wins. Hence Bob will choose the option that minimizes loss, which might be taking 2, making the sequence of plays [1], [2], [3,7], and Alice wins by 7 (Alice's total: 11 - Bob's total: 4 = 7).

Using this strategy, we understand that dfs(0) would return a positive score indicating Alice as the winner. Each call to dfs was efficiently processed due to memoization, which saved the state to prevent re-evaluation.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the function stoneGameIII is determined by the recursion process performed by the helper function dfs. In this function, the algorithm attempts to maximize the score for the current player by choosing up to three consecutive stones (which is implemented in the for loop that iterates at most three times).

The memoization (through the @cache decorator) stores the results of the subproblems, which are the optimal scores starting from each index i. There are n possible indices to start from (where n is the length of stoneValue), and since each function call of dfs examines at most 3 different scenarios before recursion, the total number of operations needed is proportional to n.

Hence, the time complexity is O(n) because each state is computed only once due to memoization, and the recursive calls made from each state are at most three.

Space Complexity

The space complexity of stoneGameIII is determined by two factors: the space used for recursion (the recursion depth) and the space used to store the results of subproblems (due to memoization).

1. Recursion Depth: Since the function dfs is called recursively and can potentially make up to three recursive calls for each call made, the depth of the recursion tree could theoretically go up to n in a worst-case scenario. However, due to memoization, many recursive calls are pruned, and thus, the true recursion depth is limited. Generally, the recursion does not go beyond n.

2. Memoization Storage: The memoization of dfs subproblem results is implemented through the @cache decorator, which could potentially store results for each of the n starting indices.

Combining these two factors, the space complexity is also O(n), primarily due to the space needed to store the computed values for memoization for each possible starting index and the call stack for the recursive calls.

Learn more about how to find time and space complexity quickly using problem constraints.