N-Queen.py

N = 4

""" ld is an array where its indices indicate row-col+N-1
(N-1) is for shifting the difference to store negative
indices """
ld = [0] * 30

""" rd is an array where its indices indicate row+col
and used to check whether a queen can be placed on
right diagonal or not"""
rd = [0] * 30

"""column array where its indices indicates column and
used to check whether a queen can be placed in that
	row or not"""
cl = [0] * 30

""" A utility function to print solution """
def printSolution(board):
	for i in range(N):
		for j in range(N):
			print(board[i][j], end = " ")
		print()

""" A recursive utility function to solve N
Queen problem """
def solveNQUtil(board, col):
	
	""" base case: If all queens are placed
		then return True """
	if (col >= N):
		return True
		
	""" Consider this column and try placing
		this queen in all rows one by one """
	for i in range(N):
		
		""" Check if the queen can be placed on board[i][col] """
		""" A check if a queen can be placed on board[row][col].
		We just need to check ld[row-col+n-1] and rd[row+coln]
		where ld and rd are for left and right diagonal respectively"""
		if ((ld[i - col + N - 1] != 1 and
			rd[i + col] != 1) and cl[i] != 1):
				
			""" Place this queen in board[i][col] """
			board[i][col] = 1
			ld[i - col + N - 1] = rd[i + col] = cl[i] = 1
			
			""" recur to place rest of the queens """
			if (solveNQUtil(board, col + 1)):
				return True
				
			""" If placing queen in board[i][col]
			doesn't lead to a solution,
			then remove queen from board[i][col] """
			board[i][col] = 0 # BACKTRACK
			ld[i - col + N - 1] = rd[i + col] = cl[i] = 0
			
			""" If the queen cannot be placed in
			any row in this column col then return False """
	return False
	
""" This function solves the N Queen problem using
Backtracking. It mainly uses solveNQUtil() to
solve the problem. It returns False if queens
cannot be placed, otherwise, return True and
prints placement of queens in the form of 1s.
Please note that there may be more than one
solutions, this function prints one of the
feasible solutions."""
def solveNQ():
	board = [[0, 0, 0, 0],
			[0, 0, 0, 0],
			[0, 0, 0, 0],
			[0, 0, 0, 0]]
	if (solveNQUtil(board, 0) == False):
		printf("Solution does not exist")
		return False
	printSolution(board)
	return True
	
# Driver Code
solveNQ()

# This code is contributed by SHUBHAMSINGH10