Alice and Bob Playing Flower Game

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Greedy AlgorithmsArrays

Alice and Bob are playing a turn-based game on a circular field surrounded by flowers. The circle represents the field, and there are x flowers in the clockwise direction between Alice and Bob, and y flowers in the anti-clockwise direction between them.

The game proceeds as follows:

Alice takes the first turn.
In each turn, a player must choose either the clockwise or anti-clockwise direction and pick one flower from that side.
At the end of the turn, if there are no flowers left at all, the current player captures their opponent and wins the game.

Given two integers, n and m, the task is to compute the number of possible pairs (x, y) that satisfy the conditions:

Alice must win the game according to the described rules.
The number of flowers x in the clockwise direction must be in the range [1,n].
The number of flowers y in the anti-clockwise direction must be in the range [1,m].

Return the number of possible pairs (x, y) that satisfy the conditions mentioned in the statement.

Example 1:

Input: n = 3, m = 2
Output: 3
Explanation: The following pairs satisfy conditions described in the statement: (1,2), (3,2), (2,1).

Example 2:

Input: n = 1, m = 1
Output: 0
Explanation: No pairs satisfy the conditions described in the statement.

Constraints:

1 <= n, m <= 10⁵

Try coding it in LeetCode

Alice and Bob Playing Flower Game

Medium

Asked by:

18 views

Topics:

Greedy AlgorithmsArrays

Alice and Bob are playing a turn-based game on a circular field surrounded by flowers. The circle represents the field, and there are x flowers in the clockwise direction between Alice and Bob, and y flowers in the anti-clockwise direction between them.

The game proceeds as follows:

Alice takes the first turn.
In each turn, a player must choose either the clockwise or anti-clockwise direction and pick one flower from that side.
At the end of the turn, if there are no flowers left at all, the current player captures their opponent and wins the game.

Given two integers, n and m, the task is to compute the number of possible pairs (x, y) that satisfy the conditions:

Alice must win the game according to the described rules.
The number of flowers x in the clockwise direction must be in the range [1,n].
The number of flowers y in the anti-clockwise direction must be in the range [1,m].

Return the number of possible pairs (x, y) that satisfy the conditions mentioned in the statement.

Example 1:

Input: n = 3, m = 2
Output: 3
Explanation: The following pairs satisfy conditions described in the statement: (1,2), (3,2), (2,1).

Example 2:

Input: n = 1, m = 1
Output: 0
Explanation: No pairs satisfy the conditions described in the statement.

Constraints:

1 <= n, m <= 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 are the constraints on the values of `n` and `m`? Can they be zero or negative?
Could you clarify what constitutes a 'possible flower arrangement'? Does the order in which Alice and Bob place the flowers matter, or only the final arrangement of flowers in the pots?
If either `n` or `m` is zero, should I return zero, or is there another specific value expected?
Are we looking for the total number of *distinct* possible flower arrangements, or is repetition allowed (i.e., if there's some symmetry in the arrangement that can be achieved in multiple ways)?
Are the flower pots distinguishable (e.g., numbered 1 to n+m), or are we only concerned with the total number of arrangements of even and odd flowers?

Brute Force Solution

Approach

The brute force solution is like trying out every single possible way Alice and Bob can place flowers. We check each placement to see if it's a valid game and count how many valid games there are.

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

Imagine Alice picks the first spot for a flower.
Then, Bob picks a spot.
Now, Alice picks another spot, and Bob picks another.
Keep going like this until they can't place any more flowers following the game's rules.
Record whether the game was valid or not, meaning if they filled the entire field.
Repeat this process for every single possible combination of flower placements Alice and Bob could choose.
Count how many of these possibilities resulted in a valid game (they completely filled the grid). That number is your answer.

Code Implementation

def flower_game_brute_force(number_rows, number_columns):
    total_cells = number_rows * number_columns
    valid_games_count = 0

    def is_valid_placement(placement, current_grid):
        row, col = placement
        if row < 0 or row >= number_rows or col < 0 or col >= number_columns:
            return False
        return current_grid[row][col] == 0

    def play_game(current_grid, moves):
        alice_moves = []
        bob_moves = []

        for i in range(len(moves)):
            if i % 2 == 0:
                alice_moves.append(moves[i])
            else:
                bob_moves.append(moves[i])

        for row in range(number_rows):
            for col in range(number_columns):
                current_grid[row][col] = 0

        alice_cells_filled = 0
        for alice_move in alice_moves:
            row, col = alice_move
            if is_valid_placement(alice_move, current_grid):
                current_grid[row][col] = 1
                alice_cells_filled += 1
            else:
                return False

        bob_cells_filled = 0
        for bob_move in bob_moves:
            row, col = bob_move
            if is_valid_placement(bob_move, current_grid):
                current_grid[row][col] = 1
                bob_cells_filled += 1
            else:
                return False

        return (alice_cells_filled + bob_cells_filled) == total_cells

    def generate_moves(index, moves, number_rows, number_columns):
        nonlocal valid_games_count

        if index == total_cells:
            current_grid = [[0] * number_columns for _ in range(number_rows)]
            if play_game(current_grid, moves):
                valid_games_count += 1
            return

        for row in range(number_rows):
            for col in range(number_columns):
                generate_moves(index + 1, moves + [(row, col)], number_rows, number_columns)

    # Initialize the recursive function for all valid moves
    generate_moves(0, [], number_rows, number_columns)

    # Returns the count of valid game outcomes
    return valid_games_count

Big(O) Analysis

Time Complexity

O((m*n)!) – The brute force approach explores all possible placements of flowers in the m x n grid. Each placement involves considering all permutations of occupied cells by Alice and Bob. There are (m*n) cells to consider, and in the worst-case where all cells are filled, we are essentially generating permutations of all possible placements. Therefore, the time complexity is proportional to the factorial of the total number of cells, (m*n)!. This simplifies to O((m*n)!).

Space Complexity

O(m*n) – The brute force solution implicitly explores every possible combination of flower placements. To track the flower placements in a game, a grid of size m*n (where 'm' and 'n' are likely dimensions representing rows and columns if this were a rectangular grid, implied but not clearly defined in the problem description) may be needed to represent the board state at each step. The maximum memory required would be to store the arrangement of flowers placed for a single game, representing a board. This requires space proportional to the grid size. Therefore, the auxiliary space complexity is O(m*n), assuming that the dimensions of the board are 'm' and 'n'.

Optimal Solution

Approach

The most efficient way to solve this game is to realize that the number of ways Alice and Bob can pick flower pairs only depends on the number of even and odd choices they have. We can calculate the number of even and odd numbers within the given ranges to determine the answer quickly. This avoids simulating the game.

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

Determine the number of even numbers Alice can choose from.
Determine the number of odd numbers Alice can choose from.
Determine the number of even numbers Bob can choose from.
Determine the number of odd numbers Bob can choose from.
The total number of ways they can pick a valid flower pair is the product of Alice's even choices and Bob's odd choices, plus the product of Alice's odd choices and Bob's even choices.
That final number is your answer.

Code Implementation

def solve_flower_game(alice_choices, bob_choices):
    alice_even_choices_count = alice_choices // 2

    alice_odd_choices_count = (alice_choices + 1) // 2

    bob_even_choices_count = bob_choices // 2

    bob_odd_choices_count = (bob_choices + 1) // 2

    # Calculate total valid pairs
    total_valid_pairs = (
        alice_even_choices_count * bob_odd_choices_count
        + alice_odd_choices_count * bob_even_choices_count
    )

    # Return result after doing products
    return total_valid_pairs

Big(O) Analysis

Time Complexity

O(1) – The solution involves calculating the number of even and odd numbers within given ranges (n, m). This calculation uses simple arithmetic operations such as division and subtraction. These operations take constant time regardless of the input ranges. Therefore, the time complexity of the solution is O(1).

Space Complexity

O(1) – The algorithm calculates the number of even and odd numbers within given ranges. It only requires storing a few integer variables to hold these counts. The space used for these variables (number of even Alice choices, number of odd Alice choices, number of even Bob choices, and number of odd Bob choices) remains constant regardless of the input size, which we can consider as the ranges for Alice and Bob. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
n or m is zero	If either n or m is zero, the result is zero because no flowers can be placed.
n and m are both very large numbers (integer overflow)	The result of n * m might exceed the maximum integer value, so use long data type to store the result to prevent overflow.
n and m are both one	The solution should correctly calculate the number of arrangements as 1 (1 * 1).
n is one and m is a large number	The number of arrangements should scale correctly to m (1 * m).
m is one and n is a large number	The number of arrangements should scale correctly to n (n * 1).
Negative input for n or m	Invalid input, should either throw an exception or return 0, clarifying the expected behavior beforehand.
Extremely large inputs for both n and m near the limits of the long data type.	Ensure calculations using long don't result in further overflows; consider BigInteger if long is insufficient.
m and n are equal	The solution needs to correctly handle cases where the number of even and odd pots are the same.