1510. Stone Game IV

Problem Description

Alice and Bob are playing a turn-based game where they remove stones from a pile. The game begins with n stones in the pile and in each turn, a player removes a square number of stones (like 1, 4, 9, 16, ...) from the pile. The players take alternate turns, with Alice always starting the game. The objective of the game is not to be the person who is unable to make a move (meaning that player can't remove a square number of stones because it doesn't exist in the pile). If a player can't make a move, they lose. The problem asks us to determine if Alice can win the game, assuming both Alice and Bob play optimally (making the best moves possible at every turn).

Intuition

To solve this game algorithmically, we can analyze it as a recursive problem, where each state of the game (the number of stones left in the pile) can lead to several possible next states (depending on which square number of stones is taken). The key insight is to recognize that the game has optimal substructure, meaning the optimal decision at a given state depends only on the states reachable from it and not on the path taken to reach that state. We can use dynamic programming to avoid recomputing the outcomes of these substates.

We want to determine if Alice can guarantee a win given n stones. We can do this by simulating each player's moves recursively and memoizing (caching) the results to make the solution efficient. The base case of our recursive function is when there are 0 stones left, meaning the current player loses (since they can't make a move).

During Alice's turn (or Bob's), we check all possible moves (removing a square number of stones) and recursively call the function to simulate the opponent's turn with the remaining stones. If there's at least one move that leads Bob to a losing state (he can't force a win from that state), Alice wins by taking that move.

The dfs function in the provided solution is a typical depth-first search with memoization (caching) that implements the above logic. The cache decorator is used to memoize the results of the recursive calls, storing whether a particular number of stones leads to a win (True) or loss (False). During the recursion, if we find a state (number of stones) that makes the next player lose, we return True (indicating the current player can win).

In summary, we use depth-first search with memoization to check every possible move, remembering the outcomes of sub-problems to determine if Alice can win with n starting stones.

Learn more about Math and Dynamic Programming patterns.

Solution Approach

The solution uses a recursive function that employs depth-first search (DFS) to explore the state space of the game and memoization to save the results of subproblems. Let's walk through the implementation as found above in the Solution class:

1. The winnerSquareGame function is the starting point that takes an integer n, denoting the number of stones in the pile initially.

2. It defines a nested function dfs which is a recursive function designed to return True if the current player can force a win with i stones remaining, otherwise it returns False.

3. The dfs function is decorated with @cache from Python's standard library. This decorator automatically memoizes the results of the dfs function to avoid recalculating the same subproblems. Memoization is a key feature that improves the efficiency of the solution by storing the outcome of each state after it is computed for the first time.

4. Inside dfs, the base case is checked: if i == 0, the function returns False as the current player cannot make a move and thus loses.

5. If the base case is not met, the function enters a loop to try all possible square number moves (j * j) that are less than or equal to the current stone count (i).

6. For each possible move, it performs a recursive call: dfs(i - j * j). This call represents the next player's turn with the remaining stones after the current player removes j * j stones.

7. The key part of the logic is checking if not dfs(i - j * j). If the result is True, this means that the opponent (next player) will lose in the state after the current move. Since the opponent loses, the current player wins, so the function returns True.

8. If none of the potential moves lead to a winning state (the opponent can force a win no matter what the current player does), the loop finishes and the function returns False, indicating that the current player cannot force a win from this state.

9. Finally, the initial call return dfs(n) kicks off the recursive process for the initial state where Alice starts with n stones.

The recursive depth-first search, coupled with memoization, is a classic dynamic programming approach, and it is very effective in this case for solving combinatorial game problems. The solution systematically explores all possible outcomes and is able to avoid redundant work thanks to the caching mechanism, making it feasible to compute the answer for larger values of n.

Example Walkthrough

Let's take n = 7 stones as an example to illustrate how the solution algorithm works. According to the rules, Alice can only take square numbers of stones - like 1, 4, or 9, and so on. Since we are concerned with the number 7, the only square numbers less than or equal to 7 are 1 (1x1) and 4 (2x2).

1. Alice starts the game and has the option to take 1 or 4 stones.

2. If Alice takes 1 stone, we're left with 6 stones. Now, it's Bob's turn. The recursive function (dfs) will try to determine if Bob can win given that 6 stones remain.

3. If Alice takes 4 stones instead, 3 stones are left. Bob has only one choice here, which is to take 1 stone (as 4 is too many for the remaining 3), leaving 2 stones.

4. If at any point the dfs function is called with 0, it would return False, indicating the current player lost the game (as they can't make a move).

Let's trace the recursive calls from Alice's perspective when she takes 1 stone:

• dfs(7) - Alice's turn.
  • Alice takes 1: dfs(6) - Bob's turn.
    • Bob can take 1: dfs(5) - Alice's turn.
      • If Alice takes 1: dfs(4) - Bob's turn.
        • If Bob takes 1: dfs(3) - Alice's turn.
          • Alice will have no choice but to leave a losing state for Bob eventually, as none of the paths from this state lead to a win for her.
        • If Bob takes 4: dfs(0) - Alice wins (Bob loses since no stones are left).
      • If Alice takes 4: dfs(1) - Bob's turn.
        • No matter what Bob does (taking 1), he leaves a win for Alice.

From this example, with 7 stones, we can see that if Alice starts by taking 1 stone and at each step makes sure to leave a number of stones such that no square number can subtract to zero, she will win. The recursive calls will discover these winning strategies through its depth-first exploration, and the memoization will ensure we don't recompute the results for subproblems we've already seen.

Alice's winning strategy will look like this:

• Alice takes 1 (6 stones left).
• Bob could take 1 (5 stones left) or 4 (2 stones left).
• If 5 stones are left, Alice would take 4 (1 left), forcing Bob to take the last stone and lose.
• If 2 stones are left, Alice just needs to take 1, and Bob will be forced to take the last and lose.

By running through all possible scenarios like this, the algorithm concludes that Alice can win when the game starts with 7 stones.

Solution Implementation

Time and Space Complexity

The given Python code aims to determine if a player can win a game where they can remove any square number of stones from a total of n stones. The code uses a depth-first search (DFS) algorithm with memoization to prevent recomputing states that have already been solved. The memoization is implemented using the cache decorator from Python's functools.

Time Complexity

The time complexity of the solution depends on the number of distinct states i that will be passed to the dfs function and the number of iterations within each function call.

• Since each state from n to 1 is computed only once due to memoization, there will be n distinct states.
• For each state i, the inner while loop runs for sqrt(i) times since it enumerates through all possible square numbers that are less than or equal to i.

Combining these factors, the overall time complexity of the recursive calls can be represented by the sum of square roots for all numbers from 1 to n, which can be approximated by an integral from 1 to n, resulting in O(n^(3/2)).

Space Complexity

The space complexity of the solution includes the space used by the recursion stack and the memoization cache.

• The recursion stack will use O(sqrt(n)) space since the maximum depth of recursion is limited by the maximum number of perfect squares you can subtract from n consecutively before reaching zero.
• The memoization cache will hold a result for each state from n to 1, therefore requiring O(n) space.

As a result, the overall space complexity is O(n) because the cache dominates the space complexity due to storing the result for each state compared to the stack space in depth.

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