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); 
    } 
} 

Python3

# Python3 program to Print all possible paths from 
# top left to bottom right of a mXn matrix 
  
''' 
/* 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) */ 
'''
def printAllPathsUtil(mat, i, j, m, n, path, pi): 
  
    # Reached the bottom of the matrix  
    # so we are left with only option to move right 
    if (i == m - 1): 
        for k in range(j, n): 
            path[pi + k - j] = mat[i][k] 
  
        for l in range(pi + n - j): 
            print(path[l], end = " ") 
        print() 
        return
  
    # Reached the right corner of the matrix  
    # we are left with only the downward movement. 
    if (j == n - 1): 
  
        for k in range(i, m): 
            path[pi + k - i] = mat[k][j] 
  
        for l in range(pi + m - i): 
            print(path[l], end = " ") 
        print() 
        return
  
    # Add the current cell  
    # to the path being generated 
    path[pi] = mat[i][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 
def printAllPaths(mat, m, n): 
  
    path = [0 for i in range(m + n)] 
    printAllPathsUtil(mat, 0, 0, m, n, path, 0) 
  
# Driver Code 
mat = [[1, 2, 3],  
       [4, 5, 6]] 
  
printAllPaths(mat, 2, 3) 
  
# This code is contributed by Mohit Kumar 

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.

This article is contributed by Hariprasad NG. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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]

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My Personal Notes

C++

Java

Python3

C#

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