Count number of ways to reach destination in a Maze

Given a maze with obstacles, count the number of paths to reach the rightmost-bottommost cell from the topmost-leftmost cell. A cell in the given maze has a value of -1 if it is a blockage or dead-end, else 0.

From a given cell, we are allowed to move to cells (i+1, j) and (i, j+1) only.

Examples: 

Input: maze[R][C] =  {{0,  0, 0, 0},
                      {0, -1, 0, 0},
                      {-1, 0, 0, 0},
                      {0,  0, 0, 0}};
Output: 4
There are four possible paths as shown in
below diagram.

Recursion:
When considering the cell (0,0) as the starting point, there is actually 1 way to reach it, which is by not making any move at all. This understanding helps us establish the base case for our recursive solution.

By observing the recursion tree, we can notice that there is repetition in the function calls. This is a common occurrence when multiple recursive calls are made.

We can have confidence in the recursive function’s ability to provide the number of unique paths from an intermediate point to the destination.

Since we can only move downward or to the right, we pass these options to the recursive function, trusting that it will determine the answer by traversing from these points to the destination.

By summing up the number of unique ways from the downward and rightward paths, we obtain the total number of unique ways.

4

Time Complexity:O(2 ^{n * m})

Space Complexity:O(n*m)

Memoization:

We can cache the overlapping subproblems to make it a efficient solution

4

Time Complexity:O(n*m) 

Space Complexity:O(n*m)

Tabulation:

Intuition:

1. We declare a 2-D matrix of size matrix[n+1][m+1]
2. We fill the matrix with -1 where the blocked cells we given.
3. Then we traverse through the matrix and put the sum of matrix[i-1][j] and matrix[i][j-1] if there doesn’t exist a -1.
4. Atlast we return the matrix[n][m] as our ans

Implementation:

1

Time Complexity: O(N*M)
Space Complexity: O(N*M) Since we are using a 2-D matrix

This problem is an extension of the below problem.

Backtracking | Set 2 (Rat in a Maze)

In this post, a different solution is discussed that can be used to solve the above Rat in a Maze problem also.
The idea is to modify the given grid[][] so that grid[i][j] contains count of paths to reach (i, j) from (0, 0) if (i, j) is not a blockage, else grid[i][j] remains -1. 

We can recursively compute grid[i][j] using below 
formula and finally return grid[R-1][C-1]
  // If current cell is a blockage
  if (maze[i][j] == -1)
      maze[i][j] = -1; //  Do not change
  // If we can reach maze[i][j] from maze[i-1][j]
  // then increment count.
  else if (maze[i-1][j] > 0)
      maze[i][j] = (maze[i][j] + maze[i-1][j]);
  // If we can reach maze[i][j] from maze[i][j-1]
  // then increment count.
  else if (maze[i][j-1] > 0)
      maze[i][j] = (maze[i][j] + maze[i][j-1]);

Below is the implementation of the above idea. 

4

Time Complexity: O(R x C)
Auxiliary Space: O(1)

This article is contributed by Roshni Agarwal. If you like neveropen and would like to contribute, you can also write an article using write.neveropen.co.za or mail your article to review-team@neveropen.co.za. See your article appearing on the neveropen main page and help other Geeks.