Maximum Number of Moves to Kill All Pawns

Hard

Google

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Dynamic ProgrammingGraphs

You are given a 50 x 50 chessboard with one knight and some pawns on it. You are given two integers kx and ky where (kx, ky) denotes the position of the knight, and a 2D array positions where positions[i] = [x_i, y_i] denotes the position of the pawns on the chessboard.

Alice and Bob play a turn-based game, where Alice goes first. In each player's turn:

The player selects a pawn that still exists on the board and captures it with the knight in the fewest possible moves. Note that the player can select any pawn, it might not be one that can be captured in the least number of moves.
In the process of capturing the selected pawn, the knight may pass other pawns without capturing them. Only the selected pawn can be captured in this turn.

Alice is trying to maximize the sum of the number of moves made by both players until there are no more pawns on the board, whereas Bob tries to minimize them.

Return the maximum total number of moves made during the game that Alice can achieve, assuming both players play optimally.

Note that in one move, a chess knight has eight possible positions it can move to. Each move is two cells in a cardinal direction, then one cell in an orthogonal direction.

Example 1:

kx = 1, ky = 1, positions = [[0,0]]

Output: 4

Explanation: The knight takes 4 moves to reach the pawn at (0, 0).

Example 2:

kx = 0, ky = 2, positions = [[1,1],[2,2],[3,3]]

Output: 8

Explanation:

Alice picks the pawn at (2, 2) and captures it in two moves: (0, 2) -> (1, 4) -> (2, 2).
Bob picks the pawn at (3, 3) and captures it in two moves: (2, 2) -> (4, 1) -> (3, 3).
Alice picks the pawn at (1, 1) and captures it in four moves: (3, 3) -> (4, 1) -> (2, 2) -> (0, 3) -> (1, 1).

Example 3:

kx = 0, ky = 0, positions = [[1,2],[2,4]]

Output: 3

Explanation:

Alice picks the pawn at (2, 4) and captures it in two moves: (0, 0) -> (1, 2) -> (2, 4). Note that the pawn at (1, 2) is not captured.
Bob picks the pawn at (1, 2) and captures it in one move: (2, 4) -> (1, 2).

Constraints:

0 <= kx, ky <= 49
1 <= positions.length <= 15
positions[i].length == 2
0 <= positions[i][0], positions[i][1] <= 49
All positions[i] are unique.
The input is generated such that positions[i] != [kx, ky] for all 0 <= i < positions.length.

Maximum Number of Moves to Kill All Pawns

Hard

Google

6 views

Topics:

Dynamic ProgrammingGraphs

You are given a 50 x 50 chessboard with one knight and some pawns on it. You are given two integers kx and ky where (kx, ky) denotes the position of the knight, and a 2D array positions where positions[i] = [x_i, y_i] denotes the position of the pawns on the chessboard.

Alice and Bob play a turn-based game, where Alice goes first. In each player's turn:

The player selects a pawn that still exists on the board and captures it with the knight in the fewest possible moves. Note that the player can select any pawn, it might not be one that can be captured in the least number of moves.
In the process of capturing the selected pawn, the knight may pass other pawns without capturing them. Only the selected pawn can be captured in this turn.

Alice is trying to maximize the sum of the number of moves made by both players until there are no more pawns on the board, whereas Bob tries to minimize them.

Return the maximum total number of moves made during the game that Alice can achieve, assuming both players play optimally.

Note that in one move, a chess knight has eight possible positions it can move to. Each move is two cells in a cardinal direction, then one cell in an orthogonal direction.

Example 1:

kx = 1, ky = 1, positions = [[0,0]]

Output: 4

Explanation: The knight takes 4 moves to reach the pawn at (0, 0).

Example 2:

kx = 0, ky = 2, positions = [[1,1],[2,2],[3,3]]

Output: 8

Explanation:

Alice picks the pawn at (2, 2) and captures it in two moves: (0, 2) -> (1, 4) -> (2, 2).
Bob picks the pawn at (3, 3) and captures it in two moves: (2, 2) -> (4, 1) -> (3, 3).
Alice picks the pawn at (1, 1) and captures it in four moves: (3, 3) -> (4, 1) -> (2, 2) -> (0, 3) -> (1, 1).

Example 3:

kx = 0, ky = 0, positions = [[1,2],[2,4]]

Output: 3

Explanation:

Alice picks the pawn at (2, 4) and captures it in two moves: (0, 0) -> (1, 2) -> (2, 4). Note that the pawn at (1, 2) is not captured.
Bob picks the pawn at (1, 2) and captures it in one move: (2, 4) -> (1, 2).

Constraints:

0 <= kx, ky <= 49
1 <= positions.length <= 15
positions[i].length == 2
0 <= positions[i][0], positions[i][1] <= 49
All positions[i] are unique.
The input is generated such that positions[i] != [kx, ky] for all 0 <= i < positions.length.

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 dimensions of the board, specifically the number of rows and columns? Can I assume it's always a square board?
Are the pawns represented by a specific data type (e.g., an array of coordinates, a matrix with pawn indicators)? What does the initial state of the board look like as input?
What constitutes a 'move'? Is it a specific piece type moving to capture a pawn, or is there a general 'move' action that eliminates a pawn under certain conditions? What are these conditions?
If it's impossible to kill all pawns, what should the function return? Should it return -1, or is there a specific error value?
Are there any restrictions on which pawn can be killed first? Can I assume any move that eliminates a pawn is valid, or are there dependencies between pawns?

Brute Force Solution

Approach

The brute force approach to this pawn game is like trying every single possible move sequence. We want to find the longest sequence that eliminates all the pawns, so we explore every possible path of attacks until no pawns are left.

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

Start with the initial board setup.
Imagine the knight makes a move; record what the board looks like afterwards.
From that new board state, imagine the knight makes every other possible move, again recording the board state after each move.
Keep doing this, branching out from each board state to every possible next move, until you reach a point where no pawns remain.
Every time you reach a state with no pawns, record the number of moves it took to get there.
Once you have tried all possible combinations of moves, compare the lengths of all the successful move sequences that resulted in eliminating all pawns.
The longest of these sequences is your answer – the maximum number of moves it takes to kill all pawns.

Code Implementation

def max_moves_to_kill_all_pawns_brute_force(board):

    def get_possible_knight_moves(board_state):
        knight_moves = []
        board_size = len(board_state)

        for row_index in range(board_size):
            for col_index in range(board_size):
                if board_state[row_index][col_index] == 'K':
                    knight_row = row_index
                    knight_col = col_index
                    break
        else:
            return []

        moves = [
            (-2, -1), (-2, 1), (-1, -2), (-1, 2),
            (1, -2), (1, 2), (2, -1), (2, 1)
        ]

        for move_row, move_col in moves:
            new_row = knight_row + move_row
            new_col = knight_col + move_col

            if 0 <= new_row < board_size and 0 <= new_col < board_size:
                knight_moves.append((new_row, new_col))

        return knight_moves

    def is_valid_move(row, col, board_state):
        board_size = len(board_state)
        return 0 <= row < board_size and 0 <= col < board_size

    def apply_move(board_state, row, col):
        new_board_state = [row[:] for row in board_state]
        knight_row, knight_col = -1, -1
        for row_index in range(len(new_board_state)): 
            for col_index in range(len(new_board_state)): 
                if new_board_state[row_index][col_index] == 'K':
                    knight_row, knight_col = row_index, col_index
                    break

        new_board_state[knight_row][knight_col] = '.'

        if new_board_state[row][col] == 'P':
            new_board_state[row][col] = 'K'
        else:
            new_board_state[row][col] = 'K'

        return new_board_state

    def pawns_left(board_state):
        for row in board_state:
            if 'P' in row:
                return True
        return False

    def solve_recursive(current_board_state):
        # Base case: if no pawns left, return the number of moves.
        if not pawns_left(current_board_state):
            return 0

        max_moves = -1
        possible_moves = get_possible_knight_moves(current_board_state)

        # Iterate through all possible moves
        for new_row, new_col in possible_moves:

            # Apply the move and recursively call solve_recursive.
            new_board = apply_move(current_board_state, new_row, new_col)

            moves_from_here = solve_recursive(new_board)

            # If the recursive call resulted in a valid solution
            if moves_from_here != -1:
                max_moves = max(max_moves, moves_from_here + 1)

        return max_moves

    # Initiate the depth-first search to find the maximum moves
    result = solve_recursive(board)

    return result

Big(O) Analysis

Time Complexity

O(8^p) – The brute force approach explores every possible move sequence until all pawns are eliminated. In the worst case, the knight could have up to 8 possible moves at each step. Let p represent the initial number of pawns on the board. Since we are exploring all possible move combinations until all pawns are removed, and each move can potentially lead to 8 new board states, the number of branches in our exploration tree grows exponentially. Therefore, the time complexity is approximately O(8^p), where p is the number of pawns, representing the maximum depth of the search tree. The exponential growth reflects exploring all possible move sequences until all pawns are gone.

Space Complexity

O(K^M) – The brute force approach explores every possible move sequence, building a tree of board states. In the worst case, the knight could have K possible moves from each board state, and the maximum number of moves required to eliminate all pawns is M. Therefore, the number of board states we need to store grows exponentially as K^M, where K is the maximum number of possible moves from a state and M is the maximum number of moves to eliminate all pawns. Each board state requires constant space to store the board configuration. The total space required to store all these board states is O(K^M).

Optimal Solution

Approach

The best way to solve this is to think of it like a game where you're strategically matching attackers to pawns. The key is to focus on the weakest points first, using attackers to eliminate pawns that are most vulnerable.

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

First, look at all the pawns and their positions to understand who can attack whom.
Find the pawn that can be attacked by the fewest attackers. This is usually the pawn that is most isolated or has few attackers nearby.
Assign one of the attackers to eliminate that pawn.
Remove that pawn and the assigned attacker from the board. This means they can no longer be used in further calculations.
Update the positions of the remaining pawns and attackers, and recalculate who can attack whom after the pawn is eliminated.
Repeat this process of finding the most vulnerable pawn and assigning an attacker until all pawns are eliminated or no more attacks are possible.
The number of attackers you successfully assigned to pawns is the maximum number of moves you can make.

Code Implementation

def maximum_moves_to_kill_all_pawns(attackers, pawns):
    number_of_moves = 0

    while pawns:
        # Find the most vulnerable pawn.
        most_vulnerable_pawn = None
        fewest_attackers = float('inf')

        for pawn in pawns:
            possible_attackers = 0
            for attacker in attackers:
                if abs(attacker - pawn) <= 1:
                    possible_attackers += 1

            if possible_attackers < fewest_attackers:
                fewest_attackers = possible_attackers
                most_vulnerable_pawn = pawn

        # If no pawn can be attacked, stop.
        if fewest_attackers == 0:
            break

        # Assign an attacker to eliminate the pawn.
        assigned_attacker = None
        for attacker in attackers:
            if abs(attacker - most_vulnerable_pawn) <= 1:
                assigned_attacker = attacker
                break

        # Remove the pawn and the attacker from consideration.
        if assigned_attacker is not None:
            attackers.remove(assigned_attacker)
            pawns.remove(most_vulnerable_pawn)

            # Increment the number of successful moves.
            number_of_moves += 1

    return number_of_moves

Big(O) Analysis

Time Complexity

O(n^2) – The algorithm iterates until all pawns are eliminated or no more attacks are possible. In the worst case, the outer loop runs 'n' times where 'n' is the number of pawns. Inside the outer loop, finding the most vulnerable pawn requires examining all remaining pawns and their potential attackers. Checking potential attackers for each pawn also requires iterating through a potentially large number of attackers, and in the worst case, it can be considered proportional to 'n'. Thus, the nested operation amounts to approximately n * n operations. Therefore, the overall time complexity approximates to O(n^2).

Space Complexity

O(N) – The algorithm's space complexity stems primarily from the need to keep track of which pawns can be attacked by which attackers and vice-versa, as well as removing them. This could involve creating adjacency lists or matrices to represent the attacker-pawn relationships, or maintaining lists of available attackers and pawns. In the worst-case, the auxiliary data structures used to represent these relationships and availability, such as lists or sets, may grow proportionally to the number of attackers/pawns which can be considered N. Therefore, the space complexity would be O(N).

Edge Cases

Case	How to Handle
Empty chessboard (zero rows or columns)	Return 0 if the chessboard dimensions are zero, as no moves are possible.
Only one pawn exists on the board	Return 0, as there is no other pawn to make a move against.
No pawns can attack each other (e.g., pawns are scattered in a way that no move is valid)	Return 0 if no valid moves are possible from the starting configuration.
Maximum sized chessboard with maximum number of pawns	The algorithm should be optimized to handle a large number of pawns efficiently, considering potential time complexity issues like O(n^2) or worse.
Chessboard where all pawns are concentrated in one area, leading to deep recursion (if applicable).	Employ memoization or dynamic programming to avoid redundant calculations and potential stack overflow in recursive solutions.
Cases where multiple valid move sequences exist with different lengths	The algorithm must guarantee finding the maximum number of moves, possibly by exploring all valid move sequences (e.g., using BFS or DFS).
Integer overflow when calculating the number of moves (if applicable)	Use appropriate data types (e.g., long) to store the number of moves to prevent integer overflow.
The board contains invalid pawn positions (e.g., outside the board boundaries).	Filter out or correct these invalid pawn positions before processing them to prevent errors.