Create RatInAMaze.py

Author: ShadabKhan7253

backtracking/RatInAMaze.py

@@ -0,0 +1,74 @@
+class Solution:
+	def __init__(self):
+		# Initialize a string direction which represents all the directions.
+		self.direction = "DLRU"
+		# Arrays to represent changes in rows and columns
+		self.dr = [1, 0, 0, -1]
+		self.dc = [0, -1, 1, 0]
+
+	# Function to check if cell (r, c) is inside the maze and unblocked
+	def is_valid(self, r, c, n, maze):
+		return 0 <= r < n and 0 <= c < n and maze[r] == 1
+
+	# Function to get all valid paths
+	def solve(self, r, c, maze, n, ans, current_path):
+		# If we reach the bottom right cell of the matrix, add the current path to ans and return
+		if r == n - 1 and c == n - 1:
+			ans.append(current_path)
+			return
+
+		# Mark the current cell as blocked
+		maze[r] = 0
+
+		for i in range(4):
+			# Find the next row based on the current row (r) and the dr[] array
+			nextr = r + self.dr[i]
+			# Find the next column based on the current column (c) and the dc[] array
+			nextc = c + self.dc[i]
+
+			# Check if the next cell is valid or not
+			if self.is_valid(nextr, nextc, n, maze):
+				current_path += self.direction[i]
+				# Recursively call the solve function for the next cell
+				self.solve(nextr, nextc, maze, n, ans, current_path)
+				current_path = current_path[:-1]
+
+		# Mark the current cell as unblocked
+		maze[r] = 1
+
+	def find_path(self, maze, n):
+		# List to store all the valid paths
+		ans = []
+
+		# Check if the top left cell is unblocked
+		if maze[0][0] == 1:
+			current_path = ""
+			self.solve(0, 0, maze, n, ans, current_path)
+		return ans
+
+# Main function
+
+
+def main():
+	n = 4
+
+	m = [
+		[1, 0, 0, 0],
+		[1, 1, 0, 1],
+		[1, 1, 0, 0],
+		[0, 1, 1, 1]
+	]
+
+	obj = Solution()
+	result = obj.find_path(m, n)
+
+	if not result:
+		print(-1)
+	else:
+		for path in result:
+			print(path, end=" ")
+		print()
+
+
+if __name__ == "__main__":
+	main()