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Stone Game IV

Hard

Asked by:

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Topics:

Dynamic Programming

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

Try coding it in LeetCode

Stone Game IV

Hard

Asked by:

5 views

Topics:

Dynamic Programming

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum possible value of 'n'? Understanding the range of 'n' will help me choose an appropriate algorithm and data structure.
Is 'n' always a non-negative integer? Should I handle cases where 'n' is zero or negative?
If 'n' is zero, should Alice be considered to have lost immediately, or should I consider it as a valid state where no move can be made?
Could you provide a few example values of 'n' and the corresponding win/loss outcome for Alice, to ensure I fully grasp the problem's winning conditions?
Are we concerned about memory usage, given that the value of 'n' can be large?

Brute Force Solution

Approach

The brute force approach to this stone game is all about exploring every single possible move, step by step. We imagine each player taking turns, and we try out every possible number of stones they could remove on their turn. We continue until someone wins, and repeat for every possible combination of moves from the start.

Here's how the algorithm would work step-by-step:

Start with the initial pile of stones.
Figure out all the perfect squares that are less than or equal to the number of stones currently remaining.
For each of those perfect squares, imagine the first player removes that many stones.
Now, pretend it's the second player's turn, and repeat the process: find the perfect squares less than or equal to the remaining stones, and imagine the second player removes each of those.
Keep alternating turns between the players, trying every possible number of stones they could take at each step.
For each sequence of moves, check if the first player won (meaning the pile is now empty and it was the second player's turn).
If a winning sequence of moves for the first player exists, then the first player can win. Otherwise, the second player is guaranteed to win.

Code Implementation

def stone_game_iv_brute_force(number_of_stones):
    def can_win(remaining_stones):
        # If no stones left, current player loses
        if remaining_stones == 0:
            return False

        # Try all possible moves
        max_square_root = int(remaining_stones**0.5)
        for stone_to_remove in range(1, max_square_root + 1):
            # Simulate the move and check if it leads to a loss for opponent
            if not can_win(remaining_stones - stone_to_remove**2):

                return True

        # If all moves lead to opponent winning, then we lose
        return False

    # Start the game and determine if first player can win
    return can_win(number_of_stones)

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach explores every possible combination of moves by using recursion. For a pile of n stones, the first player can remove 1, 4, 9, ..., k^2 stones where k^2 <= n. In the worst case, the number of choices at each level of recursion is proportional to the square root of n, but for simplicity, we can conservatively bound it by 2. Since each player alternates turns until the pile is empty, there could be up to n levels of recursion. Thus, the time complexity is approximately O(2^n) because we explore an exponential number of game states.

Space Complexity

O(N) – The brute force approach explores every possible move using recursion. The maximum depth of the recursion can be N, where N is the initial number of stones. Each recursive call requires space on the call stack to store function arguments and local variables. Thus, the auxiliary space required is proportional to the maximum recursion depth, which is O(N).

Optimal Solution

Approach

This game involves two players taking stones. The key to winning is to realize that we can determine if a player can win from a given number of stones by knowing if the other player *loses* in the subsequent turn. We use this knowledge to build a table of who wins and loses from the beginning.

Here's how the algorithm would work step-by-step:

Create a table to keep track of whether the first player can win starting from a particular number of stones.
We know the first player wins if they start with zero stones because there are no stones to take, so the second player loses.
Now, loop through each possible number of stones from one up to the total number of stones.
For each number of stones, check all possible perfect squares that are less than or equal to the current number of stones. These represent valid moves the first player can make.
If the first player can make a move that forces the *second* player to lose (meaning the remaining number of stones after the move corresponds to a losing position for the second player based on our table), then the first player can win from the starting position.
If the first player cannot make any move that forces the second player to lose, the first player will lose starting from that particular number of stones.
Continue filling the table until you reach the total number of stones. The value in the table at the total number of stones will tell you whether the first player can win the game.

Code Implementation

def stone_game_iv(number_of_stones):    winning_positions = [False] * (number_of_stones + 1)
    # Base case: 0 stones means the first player loses.
    winning_positions[0] = False
    for current_stone_count in range(1, number_of_stones + 1):
        can_win = False
        perfect_square = 1

        # Iterate through possible moves (perfect squares).
        while perfect_square * perfect_square <= current_stone_count:
            # If after the move, the other player loses, we win.
            if not winning_positions[current_stone_count - perfect_square * perfect_square]:
                can_win = True
                break
            perfect_square += 1

        winning_positions[current_stone_count] = can_win

    #The value at the number of stones index shows who wins.
    return winning_positions[number_of_stones]

Big(O) Analysis

Time Complexity

O(n * sqrt(n)) – The algorithm iterates from 1 to n, where n is the total number of stones. For each number of stones 'i', it iterates through all possible perfect squares less than or equal to 'i'. The number of perfect squares less than or equal to 'i' is approximately sqrt(i), and in the worst case, it can be sqrt(n). Therefore, the nested loop structure results in a time complexity of approximately n * sqrt(n). Thus, the overall time complexity is O(n * sqrt(n)).

Space Complexity

O(N) – The algorithm creates a boolean table (or array) to store whether the first player can win starting from a particular number of stones. This table's size is directly proportional to the total number of stones, denoted as N. Therefore, the auxiliary space required to store this table is N. Hence, the space complexity is O(N).

Edge Cases

Case	How to Handle
n = 0	Alice loses immediately since there are no moves possible, so return false.
n = 1	Alice removes 1, wins, return true.
n is a perfect square	Alice can remove n, winning immediately, return true.
Large n (e.g., close to integer limit)	Dynamic programming approach should scale efficiently, avoiding stack overflow or excessive recursion depth.
n = 2, 3	Alice cannot remove a perfect square from 2 or 3, so Bob wins and Alice loses, return false.
Memory constraints with very large n	Consider using bitset or optimized boolean array implementation to reduce memory footprint if necessary.
Integer overflow when calculating perfect squares	Prevent potential integer overflow by using long data type or check before calculating larger perfect squares.
Checking if a number is a perfect square can be slow, especially when checking a lot of squares	Precompute the perfect squares less than or equal to n to speed up the check during moves.

n = 0

How to Handle:

Alice loses immediately since there are no moves possible, so return false.

n = 1

How to Handle:

Alice removes 1, wins, return true.

n is a perfect square

How to Handle:

Alice can remove n, winning immediately, return true.

Large n (e.g., close to integer limit)

How to Handle:

Dynamic programming approach should scale efficiently, avoiding stack overflow or excessive recursion depth.

n = 2, 3

How to Handle:

Alice cannot remove a perfect square from 2 or 3, so Bob wins and Alice loses, return false.

Memory constraints with very large n

How to Handle:

Consider using bitset or optimized boolean array implementation to reduce memory footprint if necessary.

Integer overflow when calculating perfect squares

How to Handle:

Prevent potential integer overflow by using long data type or check before calculating larger perfect squares.

Checking if a number is a perfect square can be slow, especially when checking a lot of squares

How to Handle:

Precompute the perfect squares less than or equal to n to speed up the check during moves.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum possible value of 'n'? Understanding the range of 'n' will help me choose an appropriate algorithm and data structure.
Is 'n' always a non-negative integer? Should I handle cases where 'n' is zero or negative?
If 'n' is zero, should Alice be considered to have lost immediately, or should I consider it as a valid state where no move can be made?
Could you provide a few example values of 'n' and the corresponding win/loss outcome for Alice, to ensure I fully grasp the problem's winning conditions?
Are we concerned about memory usage, given that the value of 'n' can be large?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start with the initial pile of stones.
Figure out all the perfect squares that are less than or equal to the number of stones currently remaining.
For each of those perfect squares, imagine the first player removes that many stones.
Now, pretend it's the second player's turn, and repeat the process: find the perfect squares less than or equal to the remaining stones, and imagine the second player removes each of those.
Keep alternating turns between the players, trying every possible number of stones they could take at each step.
For each sequence of moves, check if the first player won (meaning the pile is now empty and it was the second player's turn).
If a winning sequence of moves for the first player exists, then the first player can win. Otherwise, the second player is guaranteed to win.

Code Implementation

def stone_game_iv_brute_force(number_of_stones):
    def can_win(remaining_stones):
        # If no stones left, current player loses
        if remaining_stones == 0:
            return False

        # Try all possible moves
        max_square_root = int(remaining_stones**0.5)
        for stone_to_remove in range(1, max_square_root + 1):
            # Simulate the move and check if it leads to a loss for opponent
            if not can_win(remaining_stones - stone_to_remove**2):

                return True

        # If all moves lead to opponent winning, then we lose
        return False

    # Start the game and determine if first player can win
    return can_win(number_of_stones)

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Create a table to keep track of whether the first player can win starting from a particular number of stones.
We know the first player wins if they start with zero stones because there are no stones to take, so the second player loses.
Now, loop through each possible number of stones from one up to the total number of stones.
For each number of stones, check all possible perfect squares that are less than or equal to the current number of stones. These represent valid moves the first player can make.
If the first player can make a move that forces the *second* player to lose (meaning the remaining number of stones after the move corresponds to a losing position for the second player based on our table), then the first player can win from the starting position.
If the first player cannot make any move that forces the second player to lose, the first player will lose starting from that particular number of stones.
Continue filling the table until you reach the total number of stones. The value in the table at the total number of stones will tell you whether the first player can win the game.

Code Implementation

def stone_game_iv(number_of_stones):    winning_positions = [False] * (number_of_stones + 1)
    # Base case: 0 stones means the first player loses.
    winning_positions[0] = False
    for current_stone_count in range(1, number_of_stones + 1):
        can_win = False
        perfect_square = 1

        # Iterate through possible moves (perfect squares).
        while perfect_square * perfect_square <= current_stone_count:
            # If after the move, the other player loses, we win.
            if not winning_positions[current_stone_count - perfect_square * perfect_square]:
                can_win = True
                break
            perfect_square += 1

        winning_positions[current_stone_count] = can_win

    #The value at the number of stones index shows who wins.
    return winning_positions[number_of_stones]

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
n = 0	Alice loses immediately since there are no moves possible, so return false.
n = 1	Alice removes 1, wins, return true.
n is a perfect square	Alice can remove n, winning immediately, return true.
Large n (e.g., close to integer limit)	Dynamic programming approach should scale efficiently, avoiding stack overflow or excessive recursion depth.
n = 2, 3	Alice cannot remove a perfect square from 2 or 3, so Bob wins and Alice loses, return false.
Memory constraints with very large n	Consider using bitset or optimized boolean array implementation to reduce memory footprint if necessary.
Integer overflow when calculating perfect squares	Prevent potential integer overflow by using long data type or check before calculating larger perfect squares.
Checking if a number is a perfect square can be slow, especially when checking a lot of squares	Precompute the perfect squares less than or equal to n to speed up the check during moves.

n = 0

How to Handle:

Alice loses immediately since there are no moves possible, so return false.

n = 1

How to Handle:

Alice removes 1, wins, return true.

n is a perfect square

How to Handle:

Alice can remove n, winning immediately, return true.

Large n (e.g., close to integer limit)

How to Handle:

Dynamic programming approach should scale efficiently, avoiding stack overflow or excessive recursion depth.

n = 2, 3

How to Handle:

Alice cannot remove a perfect square from 2 or 3, so Bob wins and Alice loses, return false.

Memory constraints with very large n

How to Handle:

Consider using bitset or optimized boolean array implementation to reduce memory footprint if necessary.

Integer overflow when calculating perfect squares

How to Handle:

Prevent potential integer overflow by using long data type or check before calculating larger perfect squares.

Checking if a number is a perfect square can be slow, especially when checking a lot of squares

How to Handle:

Precompute the perfect squares less than or equal to n to speed up the check during moves.