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Stone Game IV

Alice and Bob take turns playing a game, with Alice starting first.

Initially, there are n stones in a pile. On each player's turn, that player makes a move consisting of removing any non-zero square number of stones in the pile.

Also, if a player cannot make a move, he/she loses the game.

Given a positive integer n, return true if and only if Alice wins the game otherwise return false, assuming both players play optimally.

 

Example 1:

Input: n = 1
Output: true
Explanation: Alice can remove 1 stone winning the game because Bob doesn't have any moves.

Example 2:

Input: n = 2
Output: false
Explanation: Alice can only remove 1 stone, after that Bob removes the last one winning the game (2 -> 1 -> 0).

Example 3:

Input: n = 4
Output: true
Explanation: n is already a perfect square, Alice can win with one move, removing 4 stones (4 -> 0).

 

Constraints:

  • 1 <= n <= 105

Solution Explanation for Stone Game IV

This problem asks whether Alice wins the Stone Game, assuming optimal play from both players. The game involves removing square numbers of stones from a pile. We can solve this using dynamic programming (DP) or memoization.

Understanding the Game

The core idea is that a player loses if they can't make a move (the pile is empty). Winning depends on whether a player can force a losing position for the other player. If Alice can make a move that leaves Bob in a losing position, Alice wins.

Approach 1: Memoization (Top-Down DP)

This approach uses recursion with memoization to avoid redundant calculations.

  • dfs(i) function: This recursive function determines if the current player (Alice or Bob) can win with i stones remaining.
  • Base Case: If i is 0, the current player loses (False).
  • Recursive Step: The function iterates through all possible square numbers less than or equal to i. For each square number j*j, it recursively calls dfs(i - j*j). If the recursive call returns False (meaning the next player loses), the current player wins (True).
  • Memoization: The @cache decorator in Python (or equivalent mechanisms in other languages) stores the results of dfs(i) to avoid recalculating them.

Time Complexity: O(n√n) - The outer loop iterates up to n, and the inner loop iterates up to √n. The memoization ensures that each state is visited at most once.

Space Complexity: O(n) - The space is dominated by the memoization table (or recursion stack in the absence of memoization).

Approach 2: Bottom-Up DP (Iteration)

This approach iteratively builds a DP table to determine the winning states.

  • f array: A boolean array where f[i] indicates whether the current player can win with i stones.
  • Iteration: The loop iterates from 1 to n. For each i, it checks if there's a square number j*j such that f[i - j*j] is False (the next player loses). If such a square number exists, f[i] is set to True.

Time Complexity: O(n√n) - Similar to Approach 1, the nested loops contribute to this complexity.

Space Complexity: O(n) - The space is used by the DP table f.

Code Examples (Python, Java, C++, Go, TypeScript):

The code examples provided illustrate both approaches (memoization and iterative DP) in multiple languages. The iterative approach is generally slightly more efficient in practice due to the overhead of recursive function calls. However, the memoization approach often leads to more concise and readable code. Choose the approach best suited to your preferences and the context of the problem.

Note: The provided code snippets demonstrate both methods, allowing you to compare their implementations and performance characteristics. Remember to choose the solution that best fits your needs in terms of readability and efficiency.