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Print all possible paths from top left to bottom right of a mXn matrix

The problem is to print all the possible paths from top left to bottom right of a mXn matrix with the constraints that from each cell you can either move only to right or down.

Examples :

Input : 1 2 3
        4 5 6
Output : 1 4 5 6
         1 2 5 6
         1 2 3 6

Input : 1 2 
        3 4
Output : 1 2 4
         1 3 4

The algorithm is a simple recursive algorithm, from each cell first print all paths by going down and then print all paths by going right. Do this recursively for each cell encountered.

Following are implementation of the above algorithm.

C++

// CPP program to Print all possible paths from 
// top left to bottom right of a mXn matrix
#include<iostream>
  
using namespace std;
  
/* mat:  Pointer to the starting of mXn matrix
   i, j: Current position of the robot (For the first call use 0,0)
   m, n: Dimentions of given the matrix
   pi:   Next index to be filed in path array
   *path[0..pi-1]: The path traversed by robot till now (Array to hold the
                  path need to have space for at least m+n elements) */
void printAllPathsUtil(int *mat, int i, int j, int m, int n, int *path, int pi)
{
    // Reached the bottom of the matrix so we are left with
    // only option to move right
    if (i == m - 1)
    {
        for (int k = j; k < n; k++)
            path[pi + k - j] = *((mat + i*n) + k);
        for (int l = 0; l < pi + n - j; l++)
            cout << path[l] << " ";
        cout << endl;
        return;
    }
  
    // Reached the right corner of the matrix we are left with
    // only the downward movement.
    if (j == n - 1)
    {
        for (int k = i; k < m; k++)
            path[pi + k - i] = *((mat + k*n) + j);
        for (int l = 0; l < pi + m - i; l++)
            cout << path[l] << " ";
        cout << endl;
        return;
    }
  
    // Add the current cell to the path being generated
    path[pi] = *((mat + i*n) + j);
  
    // Print all the paths that are possible after moving down
    printAllPathsUtil(mat, i+1, j, m, n, path, pi + 1);
  
    // Print all the paths that are possible after moving right
    printAllPathsUtil(mat, i, j+1, m, n, path, pi + 1);
  
    // Print all the paths that are possible after moving diagonal
    // printAllPathsUtil(mat, i+1, j+1, m, n, path, pi + 1);
}
  
// The main function that prints all paths from top left to bottom right
// in a matrix 'mat' of size mXn
void printAllPaths(int *mat, int m, int n)
{
    int *path = new int[m+n];
    printAllPathsUtil(mat, 0, 0, m, n, path, 0);
}
  
// Driver program to test abve functions
int main()
{
    int mat[2][3] = { {1, 2, 3}, {4, 5, 6} };
    printAllPaths(*mat, 2, 3);
    return 0;
}

Java

// Java program to Print all possible paths from 
// top left to bottom right of a mXn matrix
public class MatrixTraversal 
{
  
  
    /* mat:  Pointer to the starting of mXn matrix
   i, j: Current position of the robot (For the first call use 0,0)
   m, n: Dimentions of given the matrix
   pi:   Next index to be filed in path array
   *path[0..pi-1]: The path traversed by robot till now (Array to hold the
                  path need to have space for at least m+n elements) */
    private static void printMatrix(int mat[][], int m, int n,
                                    int i, int j, int path[], int idx)
    {
        path[idx] = mat[i][j];
          
         // Reached the bottom of the matrix so we are left with
        // only option to move right
        if (i == m - 1
        {
            for (int k = j + 1; k < n; k++)
            {
                path[idx + k - j] = mat[i][k];
            }
            for (int l = 0; l < idx + n - j; l++) 
            {
                System.out.print(path[l] + " ");
            }
            System.out.println();
            return;
        }
          
        // Reached the right corner of the matrix we are left with
        // only the downward movement.
        if (j == n - 1
        {
            for (int k = i + 1; k < m; k++) 
            {
                path[idx + k - i] = mat[k][j];
            }
            for (int l = 0; l < idx + m - i; l++) 
            {
                System.out.print(path[l] + " ");
            }
            System.out.println();
            return;
        }
        // Print all the paths that are possible after moving down
        printMatrix(mat, m, n, i + 1, j, path, idx + 1);
  
         // Print all the paths that are possible after moving right
        printMatrix(mat, m, n, i, j + 1, path, idx + 1);
    }
      
    // Driver code
    public static void main(String[] args) 
    {
        int m = 2;
        int n = 3;
        int mat[][] = { { 1, 2, 3 }, 
                        { 4, 5, 6 } };
        int maxLengthOfPath = m + n - 1;
        printMatrix(mat, m, n, 0, 0, new int[maxLengthOfPath], 0);
    }
}

/div>

C#

// C# program to Print all possible paths from 
// top left to bottom right of a mXn matrix
using System;
      
public class MatrixTraversal 
{
  
  
    /* mat: Pointer to the starting of mXn matrix
i, j: Current position of the robot (For the first call use 0,0)
m, n: Dimentions of given the matrix
pi: Next index to be filed in path array
*path[0..pi-1]: The path traversed by robot till now (Array to hold the
                path need to have space for at least m+n elements) */
    private static void printMatrix(int [,]mat, int m, int n,
                                    int i, int j, int []path, int idx)
    {
        path[idx] = mat[i,j];
          
        // Reached the bottom of the matrix so we are left with
        // only option to move right
        if (i == m - 1) 
        {
            for (int k = j + 1; k < n; k++)
            {
                path[idx + k - j] = mat[i,k];
            }
            for (int l = 0; l < idx + n - j; l++) 
            {
                Console.Write(path[l] + " ");
            }
            Console.WriteLine();
            return;
        }
          
        // Reached the right corner of the matrix we are left with
        // only the downward movement.
        if (j == n - 1) 
        {
            for (int k = i + 1; k < m; k++) 
            {
                path[idx + k - i] = mat[k,j];
            }
            for (int l = 0; l < idx + m - i; l++) 
            {
                Console.Write(path[l] + " ");
            }
            Console.WriteLine();
            return;
        }
          
        // Print all the paths that are possible after moving down
        printMatrix(mat, m, n, i + 1, j, path, idx + 1);
  
        // Print all the paths that are possible after moving right
        printMatrix(mat, m, n, i, j + 1, path, idx + 1);
    }
      
    // Driver code
    public static void Main(String[] args) 
    {
        int m = 2;
        int n = 3;
        int [,]mat = { { 1, 2, 3 }, 
                        { 4, 5, 6 } };
        int maxLengthOfPath = m + n - 1;
        printMatrix(mat, m, n, 0, 0, new int[maxLengthOfPath], 0);
    }
}
  
// This code contributed by Rajput-Ji


Output:



1 4 5 6
1 2 5 6
1 2 3 6

Note that in the above code, the last line of printAllPathsUtil() is commented, If we uncomment this line, we get all the paths from the top left to bottom right of a nXm matrix if the diagonal movements are also allowed. And also if moving to some of the cells are not permitted then the same code can be improved by passing the restriction array to the above function and that is left as an exercise.

Python implementation

# Python3 program to Print all possible paths from 
# top left to bottom right of a mXn matrix
allPaths = []
def findPaths(maze,m,n):
    path = [0 for d in range(m+n-1)]
    findPathsUtil(maze,m,n,0,0,path,0)
      
def findPathsUtil(maze,m,n,i,j,path,indx):
    global allPaths
    # if we reach the bottom of maze, we can only move right
    if i==m-1:
        for k in range(j,n):
            #path.append(maze[i][k])
            path[indx+k-j] = maze[i][k]
        # if we hit this block, it means one path is completed. 
        # Add it to paths list and print
        print(path)
        allPaths.append(path)
        return
    # if we reach to the right most corner, we can only move down
    if j == n-1:
        for k in range(i,m):
            path[indx+k-i] = maze[k][j]
          #path.append(maze[j][k])
        # if we hit this block, it means one path is completed. 
        # Add it to paths list and print
        print(path)
        allPaths.append(path)
        return
      
    # add current element to the path list
    #path.append(maze[i][j])
    path[indx] = maze[i][j]
      
    # move down in y direction and call findPathsUtil recursively
    findPathsUtil(maze, m, n, i+1, j, path, indx+1)
      
    # move down in y direction and call findPathsUtil recursively
    findPathsUtil(maze, m, n, i, j+1, path, indx+1)
  
if __name__ == '__main__':
    maze = [[1,2,3],
            [4,5,6],
            [7,8,9]]
    findPaths(maze,3,3)
    #print(allPaths)

Output:

[1, 4, 7, 8, 9]
[1, 4, 5, 8, 9]
[1, 4, 5, 6, 9]
[1, 2, 5, 8, 9]
[1, 2, 5, 6, 9]
[1, 2, 3, 6, 9]


This article is attributed to GeeksforGeeks.org

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