The nqueens problem is of placing N queens on a N  N chess board such that no queen can attack any other queens placed on that chess board. This means that one queen cannot have any other queen on its horizontal, vertical and diagonal lines. This function returns a boolean value True if it is safe to place a queen there considering the current state of the board. Parameters: board 2D matrix: The chessboard row, column: Coordinates of the cell on the board Returns: Boolean Value  issafe0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1 True  issafe1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1 False Check if there is any queen in the same row, column, left upper diagonal, and right upper diagonal This function creates a state space tree and calls the safe function until it receives a False Boolean and terminates that branch and backtracks to the next possible solution branch. If the row number exceeds N, we have a board with a successful combination and that combination is appended to the solution list and the board is printed. For every row, it iterates through each column to check if it is feasible to place a queen there. If all the combinations for that particular branch are successful, the board is reinitialized for the next possible combination. Prints the boards that have a successful combination. Number of queens e.g., n8 for an 8x8 board
from __future__ import annotations

solution = []


def is_safe(board: list[list[int]], row: int, column: int) -> bool:
    """
    This function returns a boolean value True if it is safe to place a queen there
    considering the current state of the board.

    Parameters:
    board (2D matrix): The chessboard
    row, column: Coordinates of the cell on the board

    Returns:
    Boolean Value

    >>> is_safe([[0, 0, 0], [0, 0, 0], [0, 0, 0]], 1, 1)
    True
    >>> is_safe([[1, 0, 0], [0, 0, 0], [0, 0, 0]], 1, 1)
    False
    """

    n = len(board)  # Size of the board

    # Check if there is any queen in the same row, column,
    # left upper diagonal, and right upper diagonal
    return (
        all(board[i][j] != 1 for i, j in zip(range(row, -1, -1), range(column, n)))
        and all(
            board[i][j] != 1 for i, j in zip(range(row, -1, -1), range(column, -1, -1))
        )
        and all(board[i][j] != 1 for i, j in zip(range(row, n), range(column, n)))
        and all(board[i][j] != 1 for i, j in zip(range(row, n), range(column, -1, -1)))
    )


def solve(board: list[list[int]], row: int) -> bool:
    """
    This function creates a state space tree and calls the safe function until it
    receives a False Boolean and terminates that branch and backtracks to the next
    possible solution branch.
    """
    if row >= len(board):
        """
        If the row number exceeds N, we have a board with a successful combination
        and that combination is appended to the solution list and the board is printed.
        """
        solution.append(board)
        printboard(board)
        print()
        return True
    for i in range(len(board)):
        """
        For every row, it iterates through each column to check if it is feasible to
        place a queen there.
        If all the combinations for that particular branch are successful, the board is
        reinitialized for the next possible combination.
        """
        if is_safe(board, row, i):
            board[row][i] = 1
            solve(board, row + 1)
            board[row][i] = 0
    return False


def printboard(board: list[list[int]]) -> None:
    """
    Prints the boards that have a successful combination.
    """
    for i in range(len(board)):
        for j in range(len(board)):
            if board[i][j] == 1:
                print("Q", end=" ")  # Queen is present
            else:
                print(".", end=" ")  # Empty cell
        print()


# Number of queens (e.g., n=8 for an 8x8 board)
n = 8
board = [[0 for i in range(n)] for j in range(n)]
solve(board, 0)
print("The total number of solutions are:", len(solution))
