# Solving the Rat In A Maze Leetcode Problem

At first glance, the “Rat in a Maze” problem might seem like a simple maze traversal problem. However, when you dive deeper into its complexities, you’ll find that it’s a challenging puzzle that requires a combination of breadth-first search, depth-first search, backtracking, and recursion to solve efficiently. In this blog post, we will explore the “Rat in a Maze” problem in detail and provide you with a step-by-step guide to approaching and solving this problem on Leetcode.

## Understanding the Problem

In the “Rat in a Maze” problem, we are given a maze represented as a two-dimensional array where 0s represent walls, and 1s represent open paths. The rat starts at the top-left corner of the maze and needs to reach the bottom-right corner. The rat can move in four directions – up, down, left, and right – but cannot pass through walls.

The task is to find a path from the starting point to the destination point in the maze. If a path exists, you need to return a sequence of movements that the rat needs to make to reach the destination. If no path is possible, you should return an empty sequence.

## Approach to Solving the Problem

To solve the “Rat in a Maze” problem efficiently, we can use a backtracking algorithm. Here’s a step-by-step guide to approaching this problem:

### 1. Define the Helper Function

First, we need to define a helper function that will recursively explore all possible paths in the maze. This function will take the current position of the rat, the current path, and the visited cells as input.

### 2. Base Case

Next, we define the base case for our recursive function. The base case will check if the rat has reached the destination cell (bottom-right corner). If the rat has reached the destination, we return the path.

### 3. Recursive Case

In the recursive case, we need to explore all four possible directions from the current cell (up, down, left, and right). We will check if each direction is valid (i.e., within the bounds of the maze and not a wall). If a valid move is found, we mark the cell as visited and recursively call the function with the new position and path.

### 4. Backtrack

If none of the directions lead to the destination, we need to backtrack by marking the current cell as unvisited and remove it from the path.

### 5. Call the Helper Function

Finally, we call the helper function with the starting position, an empty path, and a set to keep track of visited cells.

## Pseudocode

Here’s a pseudocode representation of the approach described above:

“`
function findPath(maze):
// Define helper function
function backtrackHelper(row, col, path, visited):
// Base case
if (row, col) is destination:
return path
// Recursive case
for direction in directions:
if valid move:
mark cell as visited
path.append(direction)
result = backtrackHelper(new row, new col, path, visited)
if result is not empty:
return result
// Backtrack
mark cell as unvisited
path.pop()
return ”

``````// Call helper function
return backtrackHelper(starting row, starting col, empty path, empty visited set)
``````

“`

## Complexity Analysis

• Time Complexity: The time complexity of this approach is O(3^(n*m)), where n and m are the dimensions of the maze. This is because at each cell, we have three possible directions to explore (up, down, left, right) except for the cell we came from.
• Space Complexity: The space complexity is O(n*m) to store the visited cells and the path.

## Implementation in Python

Let’s look at a sample implementation of the “Rat in a Maze” problem in Python:

“`python
def findPath(maze):
def backtrackHelper(row, col, path, visited):
if (row, col) == (destinationRow, destinationCol):
return path

``````    for dr, dc in directions:
new_row, new_col = row + dr, col + dc
if 0 <= new_row < len(maze) and 0 <= new_col < len(maze[0]) and maze[new_row][new_col] == 1 and (new_row, new_col) not in visited:
path.append((dr, dc))
result = backtrackHelper(new_row, new_col, path, visited)
if result:
return result

visited.remove((new_row, new_col))
path.pop()

return []

# Start from top-left corner and end at bottom-right corner
sourceRow, sourceCol = 0, 0
destinationRow, destinationCol = len(maze) - 1, len(maze[0]) - 1
directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]

return backtrackHelper(sourceRow, sourceCol, [], {(sourceRow, sourceCol)})
``````

“`

## FAQs:

### Q1: What is the “Rat in a Maze” problem?

A1: The “Rat in a Maze” problem is a classic algorithmic problem where a rat needs to find a path from the starting point to the destination point in a maze with walls.

### Q2: What is the approach to solving the “Rat in a Maze” problem?

A2: The problem can be solved efficiently using a backtracking algorithm that explores all possible paths in the maze.

### Q3: What is the time complexity of solving the “Rat in a Maze” problem?

A3: The time complexity of the backtracking approach is O(3^(n*m)), where n and m are the dimensions of the maze.

### Q4: What is the space complexity of the backtracking approach?

A4: The space complexity is O(n*m) to store the visited cells and the path during the recursive traversal.

### Q5: Can the “Rat in a Maze” problem be solved iteratively?

A5: While the problem is commonly solved using recursion and backtracking, it can also be solved iteratively using a stack to simulate the recursive calls.

In conclusion, the “Rat in a Maze” problem is an interesting challenge that tests your understanding of graph traversal algorithms. By breaking down the problem into smaller steps and following the backtracking approach outlined in this article, you can efficiently navigate through the maze and find the optimal path for the rat. Happy coding!