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

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  • Difficulty Level : Medium
  • Last Updated : 23 Jun, 2022

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++




// C++ 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: Dimensions 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 above functions
int main()
{
    int mat[2][3] = { {1, 2, 3}, {4, 5, 6} };
    printAllPaths(*mat, 2, 3);
    return 0;
}

Java




import java.util.*;
// Java program to Print all possible paths from
// top left to bottom right of a mXn matrix
public class MatrixTraversal
{
 
 
  
    private static void printMatrix(int mat[][], int m, int n,
                                    int i, int j, List<Integer> list)
    {
          //return if i or j crosses matrix size
        if(i > m || j > n)
            return;
        list.add(mat[i][j]);
        if(i == m && j == n){
            System.out.println(list);
        }
        printMatrix(mat, m, n, i+1, j, list);
        printMatrix(mat, m, n, i, j+1, list);
        list.remove(list.size()-1);
         
         
    }
     
    // Driver code
    public static void main(String[] args)
    {
        int m = 2;
        int n = 3;
        int mat[][] = { { 1, 2, 3 },
                        { 4, 5, 6 } };
          List<Integer> list = new ArrayList<>();
        printMatrix(mat, m-1, n-1, 0, 0, list);
    }
}
//This article is contributed by Harneet

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: Dimensions 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: Dimensions 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

Javascript




<script>
// Javascript 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: Dimensions 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) */
function printMatrix(mat,m,n,i,j,path,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 (let k = j + 1; k < n; k++)
            {
                path[idx + k - j] = mat[i][k];
            }
            for (let l = 0; l < idx + n - j; l++)
            {
                document.write(path[l] + " ");
            }
            document.write("<br>");
            return;
        }
          
        // Reached the right corner of the matrix we are left with
        // only the downward movement.
        if (j == n - 1)
        {
            for (let k = i + 1; k < m; k++)
            {
                path[idx + k - i] = mat[k][j];
            }
            for (let l = 0; l < idx + m - i; l++)
            {
                document.write(path[l] + " ");
            }
            document.write();
            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
let m = 2;
let n = 3;
let mat = [[ 1, 2, 3 ],
[ 4, 5, 6 ]];
let maxLengthOfPath = m + n - 1;
printMatrix(mat, m, n, 0, 0, new Array(maxLengthOfPath), 0);
 
 
// This code is contributed by ab2127
</script>

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

C++




// C++ program to Print all possible paths from 
// top left to bottom right of a mXn matrix
#include <bits/stdc++.h>
using namespace std;
 
vector<vector<int>> allPaths;
 
void findPathsUtil(vector<vector<int>> maze, int m,
                                 int n, int i, int j,
                          vector<int> path, int index)
{
     
    // If we reach the bottom of maze,
    // we can only move right
    if (i == m - 1)
    {
        for(int k = j; k < n; k++)
        {
             
            //path.append(maze[i][k])
            path[index + k - j] = maze[i][k];
        }
         
        // If we hit this block, it means one
        // path is completed. Add it to paths
        // list and print
        cout << "[" << path[0] << ", ";
        for(int z = 1; z < path.size() - 1; z++)
        {
            cout << path[z] << ", ";
        }
        cout << path[path.size() - 1] << "]" << endl;
        allPaths.push_back(path);
        return;
    }
         
    // If we reach to the right most
    // corner, we can only move down
    if (j == n - 1)
    {
        for(int k = i; k < m; k++)
        {
            path[index + 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
        cout << "[" << path[0] << ", ";
        for(int z = 1; z < path.size() - 1; z++)
        {
            cout << path[z] << ", ";
        }
        cout << path[path.size() - 1] << "]" << endl;
        allPaths.push_back(path);
        return;
    }
       
    // Add current element to the path list
    //path.append(maze[i][j])
    path[index] = maze[i][j];
       
    // Move down in y direction and call
    // findPathsUtil recursively
    findPathsUtil(maze, m, n, i + 1,
                  j, path, index + 1);
       
    // Move down in y direction and
    // call findPathsUtil recursively
    findPathsUtil(maze, m, n, i, j + 1,
                        path, index + 1);
}
     
void findPaths(vector<vector<int>> maze,
                       int m, int n)
{
    vector<int> path(m + n - 1, 0);
    findPathsUtil(maze, m, n, 0, 0, path, 0);
}
 
// Driver Code
int main()
{
    vector<vector<int>> maze{ { 1, 2, 3 },
                              { 4, 5, 6 },
                              { 7, 8, 9 } };
    findPaths(maze, 3, 3);
     
    //print(allPaths)
    return 0;
}
 
// This code is contributed by divyeshrabadiya07

Java




// Java program to Print all possible paths from 
// top left to bottom right of a mXn matrix
import java.io.*;
import java.util.*;
class GFG
{
  static ArrayList<ArrayList<Integer>> allPaths =
    new ArrayList<ArrayList<Integer>>();
  static void findPathsUtil(ArrayList<ArrayList<Integer>> maze,
                            int m, int n, int i,int j,
                            ArrayList<Integer> path,int index)
  {
 
    // If we reach the bottom of maze,
    // we can only move right
    if(i == m - 1)
    {
      for(int k = j; k < n; k++)
      {
 
        // path.append(maze[i][k])
        path.set(index + k - j, maze.get(i).get(k));
 
      }
 
      // If we hit this block, it means one
      // path is completed. Add it to paths
      // list and print
      System.out.print("[" + path.get(0) + ", ");
      for(int z = 1; z < path.size() - 1; z++)
      {
        System.out.print(path.get(z) + ", ");
      }
      System.out.println(path.get(path.size() - 1) + "]");
      allPaths.add(path);
      return;
    }
 
    // If we reach to the right most
    // corner, we can only move down
    if(j == n - 1)
    {
      for(int k = i; k < m; k++)
      {
        path.set(index + k - i,maze.get(k).get(j));
      }
 
      // path.append(maze[j][k])
      // If we hit this block, it means one
      // path is completed. Add it to paths
      // list and print
      System.out.print("[" + path.get(0) + ", ");
      for(int z = 1; z < path.size() - 1; z++)
      {
        System.out.print(path.get(z) + ", ");
 
      }
      System.out.println(path.get(path.size() - 1) + "]");
      allPaths.add(path);
      return;
    }
 
    // Add current element to the path list
    //path.append(maze[i][j])
    path.set(index,maze.get(i).get(j));
 
    // Move down in y direction and call
    // findPathsUtil recursively
    findPathsUtil(maze, m, n, i + 1, j, path, index + 1);
 
    // Move down in y direction and
    // call findPathsUtil recursively
    findPathsUtil(maze, m, n, i, j + 1, path, index + 1);
 
  }
  static void findPaths(ArrayList<ArrayList<Integer>> maze,
                        int m, int n)
  {
    ArrayList<Integer> path = new ArrayList<Integer>();
    for(int i = 0; i < m + n - 1; i++)
    {
      path.add(0);
    }
    findPathsUtil(maze, m, n, 0, 0, path, 0);
  }
 
  // Driver code
  public static void main (String[] args)
  {
    ArrayList<ArrayList<Integer>> maze =
      new ArrayList<ArrayList<Integer>>();
    maze.add(new ArrayList<Integer>
             (Arrays.asList(1,2,3)));
    maze.add(new ArrayList<Integer>
             (Arrays.asList(4,5,6)));
    maze.add(new ArrayList<Integer>
             (Arrays.asList(7,8,9)));
 
    findPaths(maze, 3, 3);       
  }
}
 
// This code is contributed by avanitrachhadiya2155

Python3




# 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,index):
    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[index+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[index+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[index] = maze[i][j]
     
    # move down in y direction and call findPathsUtil recursively
    findPathsUtil(maze, m, n, i+1, j, path, index+1)
     
    # move down in y direction and call findPathsUtil recursively
    findPathsUtil(maze, m, n, i, j+1, path, index+1)
 
if __name__ == '__main__':
    maze = [[1,2,3],
            [4,5,6],
            [7,8,9]]
    findPaths(maze,3,3)
    #print(allPaths)

C#




// C# program to Print all possible paths from 
// top left to bottom right of a mXn matrix
using System;
using System.Collections.Generic;
class GFG
{
   
    static List<List<int>> allPaths = new List<List<int>>();
  
    static void findPathsUtil(List<List<int>> maze,
                              int m, int n, int i,
                              int j, List<int> path,
                              int index)
    {
          
        // If we reach the bottom of maze,
        // we can only move right
        if (i == m - 1)
        {
            for(int k = j; k < n; k++)
            {
                  
                // path.append(maze[i][k])
                path[index + k - j] = maze[i][k];
            }
              
            // If we hit this block, it means one
            // path is completed. Add it to paths
            // list and print
            Console.Write( "[" + path[0] + ", ");
            for(int z = 1; z < path.Count - 1; z++)
            {
                Console.Write(path[z] + ", ");
            }
            Console.WriteLine(path[path.Count - 1] + "]");
            allPaths.Add(path);
            return;
        }
              
        // If we reach to the right most
        // corner, we can only move down
        if (j == n - 1)
        {
            for(int k = i; k < m; k++)
            {
                path[index + 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
            Console.Write( "[" + path[0] + ", ");
            for(int z = 1; z < path.Count - 1; z++)
            {
                Console.Write(path[z] + ", ");
            }
            Console.WriteLine(path[path.Count - 1] + "]");
            allPaths.Add(path); 
            return;
        }
            
        // Add current element to the path list
        //path.append(maze[i][j])
        path[index] = maze[i][j];
            
        // Move down in y direction and call
        // findPathsUtil recursively
        findPathsUtil(maze, m, n, i + 1,
                      j, path, index + 1);
            
        // Move down in y direction and
        // call findPathsUtil recursively
        findPathsUtil(maze, m, n, i, j + 1,
                            path, index + 1);
    }
          
    static void findPaths(List<List<int>> maze, int m, int n)
    {
        List<int> path = new List<int>();
        for(int i = 0; i < m + n - 1; i++)
        {
            path.Add(0);
        }
        findPathsUtil(maze, m, n, 0, 0, path, 0);
    }
 
  // Driver code
  static void Main()
  {
    List<List<int>> maze = new List<List<int>>();
    maze.Add(new List<int> { 1, 2, 3 });
    maze.Add(new List<int> { 4, 5, 6 });
    maze.Add(new List<int> { 7, 8, 9 });
 
    findPaths(maze, 3, 3);
  }
}
 
// This code is contributed by divyesh072019

Javascript




<script>
// Javascript program to Print all possible paths from 
// top left to bottom right of a mXn matrix
let allPaths = [];
 
function findPathsUtil(maze, m, n, i, j, path, index)
{
     
    // If we reach the bottom of maze,
    // we can only move right
    if (i == m - 1)
    {
        for(let k = j; k < n; k++)
        {
             
            //path.append(maze[i][k])
            path[index + k - j] = maze[i][k];
        }
         
        // If we hit this block, it means one
        // path is completed. Add it to paths
        // list and print
        document.write("[" + path[0] + ", ");
        for(let z = 1; z < path.length - 1; z++)
        {
            document.write(path[z] + ", ");
        }
        document.write(path[path.length - 1] + "]" + "<br>");
        allPaths.push(path);
        return;
    }
         
    // If we reach to the right most
    // corner, we can only move down
    if (j == n - 1)
    {
        for(let k = i; k < m; k++)
        {
            path[index + 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
        document.write("[" + path[0] + ", ");
        for(let z = 1; z < path.length - 1; z++)
        {
            document.write(path[z] + ", ");
        }
        document.write(path[path.length - 1] + "]" + "<br>");
        allPaths.push(path);
        return;
    }
       
    // Add current element to the path list
    // path.append(maze[i][j])
    path[index] = maze[i][j];
       
    // Move down in y direction and call
    // findPathsUtil recursively
    findPathsUtil(maze, m, n, i + 1,
                  j, path, index + 1);
       
    // Move down in y direction and
    // call findPathsUtil recursively
    findPathsUtil(maze, m, n, i, j + 1,
                        path, index + 1);
}
     
function findPaths(maze, m, n)
{
    let path = new Array(m + n - 1).fill(0);
    findPathsUtil(maze, m, n, 0, 0, path, 0);
}
 
// Driver Code
let maze = [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ];
findPaths(maze, 3, 3)
 
// This code is contributed by Saurabh Jaiswal
</script>

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]

Note all the above approach take some extra time and space for solving the problem ,we can simply use backtracking algorithm to solve problem in optimized manner 

C++




#include<bits/stdc++.h>
using namespace std;
 
// function to display the path
void display(vector<int> &ans) {
  for(auto i :ans ) {
    cout<<i <<" ";
  }
  cout<<endl;
}
 
// a function which check whether our step is safe or not
bool issafe(int r,int c,vector<vector<int>>& visited,int n,int m) {
  return (r < n and c <m and visited[r] !=-1 );  // return true if all values satisfied else false
}
 
 
void FindPaths(vector<vector<int>> &grid,int r,int c, int n,int m,vector<int> &ans) {
  // when we hit the last cell we reach to destination then directly push the path
  if(r == n-1 and c == m-1) {
    ans.push_back(grid[r]);
    display(ans);  // function to display the path stored in ans vector
    ans.pop_back(); // pop back because we need to backtrack to explore more path
    return ;
  
   
  // we will store the current value in ch and mark the visited place as -1
  int ch = grid[r];
  
  ans.push_back(ch); // push the path in ans array
  grid[r] = -1;  // mark the visited place with -1
   
  // if is it safe to take next downward step then take it
  if(issafe(r+1,c,grid,n,m)) {
    FindPaths(grid,r+1,c,n,m,ans);
  }
   
  // if is it safe to take next rightward step then take it
  if(issafe(r,c+1,grid,n,m)) {
    FindPaths(grid,r,c+1,n,m,ans);
  }
   
  // backtracking step we need to make values original so to we can visit it by some another path
  grid[r] = ch;
   
  // remove the current path element we explore
  ans.pop_back();
  return ;
}
 
int main() {
      int n = 3 ,m =3;
      vector<vector<int> >grid{ {1,2,3},{4,5,6},{7,8,9}};
      vector<int>ans ; // it will store the path which we have covered
       
      FindPaths(grid,0,0,n,m,ans); // here 0,0 are initial position to start with
    return 0;
}

Java




import java.util.*;
 
class Main {
    public static void main(String[] args)
    {
        int n = 3, m = 3;
        int[][] grid
            = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
        ArrayList<Integer> ans
            = new ArrayList<>(); // it will store the path
                                 // which we have covered
 
        FindPaths(grid, 0, 0, n, m,
                  ans); // here 0,0 are initial position to
                        // start with
    }
    // function to display the path
    public static void display(ArrayList<Integer> ans)
    {
        for (int i : ans) {
            System.out.print(i + " ");
        }
        System.out.println();
    }
    // a function which check whether our step is safe or
    // not
    public static boolean
    issafe(int r, int c, int[][] visited, int n, int m)
    {
        return (r < n && c < m && visited[r] != -1);
    }
 
    public static void FindPaths(int[][] grid, int r, int c,
                                 int n, int m,
                                 ArrayList<Integer> ans)
    {
        // when we hit the last cell we reach to destination
        // then directly push the path
        if (r == n - 1 && c == m - 1) {
            ans.add(grid[r]);
            display(ans); // function to display the path
                          // stored in ans vector
            ans.remove(
                ans.size()
                - 1); // remove last element because we need
                      // to backtrack to explore more path
            return;
        }
        // we will store the current value in ch and mark
        // the visited place as -1
        int ch = grid[r];
 
        ans.add(ch); // add the path in ans array
        grid[r] = -1; // mark the visited place with -1
 
        // if is it safe to take next downward step then
        // take it
        if (issafe(r + 1, c, grid, n, m)) {
            FindPaths(grid, r + 1, c, n, m, ans);
        }
        // if is it safe to take next rightward step then
        // take it
        if (issafe(r, c + 1, grid, n, m)) {
            FindPaths(grid, r, c + 1, n, m, ans);
        }
        // backtracking step we need to make values original
        // so to we can visit it by some another path
        grid[r] = ch;
        // remove the current path element we explore
        ans.remove(ans.size() - 1);
        return;
    }
}
 
// This code is contributed by Tapesh(tapeshdua420)

Python3




# code
class Solution:
     
    def __init__(self):
        self.mapping = {}
     
    def printAllPaths(self, M, m, n):
        if not self.mapping.get((m,n)):
            if m == 1 and n == 1:
                return [M[m-1][n-1]]
            else:
                res = []
                if n > 1:
                    a = self.printAllPaths(M, m, n-1)
                    for i in a:
                        if not isinstance(i, list):
                            i = [i]
                        res.append(i+[M[m-1][n-1]])
                if m > 1:
                    b = self.printAllPaths(M, m-1, n)
                    for i in b:
                        if not isinstance(i, list):
                            i = [i]
                        res.append(i+[M[m-1][n-1]])
            self.mapping[(m,n)] = res
        return self.mapping.get((m,n))
 
M = [[1, 2, 3], [4, 5, 6], [7,8,9]]
m, n = len(M), len(M[0])
a = Solution()
res = a.printAllPaths(M, m, n)
for i in res:
    print(i)

C#




// C# program for above problem
using System;
using System.Collections.Generic;
 
class Program {
 
    // function to display the path
    static void display(List<int> ans)
    {
        foreach(var i in ans)
        {
            Console.Write(i + " ");
             
        }
 
        Console.WriteLine();
    }
 
    // a function which check whether our step is safe or not
    static bool issafe(int r, int c, int[][] visited, int n, int m)
    {
        return (r < n && c < m && visited[r] != -1); // return true if all values satisfied else false
    }
 
    static void FindPaths(int[][] grid, int r, int c, int n, int m, List<int> ans)
    {
        // when we hit the last cell we reach to destination then directly push the path
        if (r == n - 1 && c == m - 1) {
            ans.Add(grid[r]);
            display(ans); // function to display the path stored in ans vector
            ans.RemoveAt(ans.Count - 1); // remove last element because we need to backtrack to explore more path
            return;
        }
 
        int ch = grid[r];  // we will store the current value in ch and mark the visited place as -1
 
        ans.Add(ch); // add the path in ans array
 
        grid[r] = -1;  // mark the visited place with -1
 
        // if is it safe to take next downward step then take it
        if (issafe(r + 1, c, grid, n, m)) {
            FindPaths(grid, r + 1, c, n, m, ans);
        }
 
        // if is it safe to take next rightward step then take it
        if (issafe(r, c + 1, grid, n, m)) {
            FindPaths(grid, r, c + 1, n, m, ans);
        }
 
        // backtracking step we need to make values original so to we can visit it by some another path
        grid[r] = ch;
         
        // remove the current path element we explore
        ans.RemoveAt(ans.Count - 1);
    }
 
    static void Main()
    {
        int n = 3, m = 3;
        int[][] grid = new int[n][];
         
        grid[0] = new[] { 1, 2, 3 };
        grid[1] = new[] { 4, 5, 6 };
        grid[2] = new[] { 7, 8, 9 };
         
        List<int> ans = new List<int>(); // it will store the path which we have covered
 
         
        FindPaths(grid, 0, 0, n, m, ans); // here 0,0 are initial position to start with
    }
}
 
// This code is contributed by Tapesh(tapeshdua420)

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 

So by these method you can optimized your code.

TC- O(2^n*m)   , SC – O(n) 

Another Approach (Iterative) :

1. In this approach we will use BFS (breadth first search) to find all possible paths.

2. We will make a queue which contains the following information : 

    a)  Vector that stores the path up to a certain cell.

    b)  coordinates of the cell.

3. We will start from the top-left cell and push cell value and coordinates in the queue.

4. We will keep on exploring right and down cell (if possible) until queue is not empty 

   and push them in the queue by updating the current cell vector.

5. If we reach the last cell then we have got one answer and we will print the answer vector.

C++




// c++ implementation for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// this structure stores information
// about a particular cell i.e
// path upto that cell and cell's
// coordinates
struct info {
    vector<int> path;
    int i;
    int j;
};
 
void printAllPaths(vector<vector<int> >& maze)
{
    int n = maze.size();
    int m = maze[0].size();
 
    queue<info> q;
    // pushing top-left cell into the queue
    q.push({ { maze[0][0] }, 0, 0 });
 
    while (!q.empty()) {
        info p = q.front();
        q.pop();
 
        // if we reached the bottom-right cell
        // i.e the destination then print the path
        if (p.i == n - 1 && p.j == m - 1) {
            for (auto x : p.path)
                cout << x << " ";
 
            cout << "\n";
        }
 
        // if we are in the last row
        // then only right movement is possible
        else if (p.i == n - 1) {
            vector<int> temp = p.path;
            // updating the current path
            temp.push_back(maze[p.i][p.j + 1]);
 
            q.push({ temp, p.i, p.j + 1 });
        }
 
        // if we are in the last column
        // then only down movement is possible
        else if (p.j == m - 1) {
            vector<int> temp = p.path;
            // updating the current path
            temp.push_back(maze[p.i + 1][p.j]);
 
            q.push({ temp, p.i + 1, p.j });
        }
 
        // else both right and down movement
        // are possible
        else { // right movement
            vector<int> temp = p.path;
            // updating the current path
            temp.push_back(maze[p.i][p.j + 1]);
 
            q.push({ temp, p.i, p.j + 1 });
 
            // down movement
            temp.pop_back();
            // updating the current path
            temp.push_back(maze[p.i + 1][p.j]);
 
            q.push({ temp, p.i + 1, p.j });
        }
    }
}
 
// Driver Code
int main()
{
    vector<vector<int> > maze{ { 1, 2, 3 },
                               { 4, 5, 6 },
                               { 7, 8, 9 } };
 
    printAllPaths(maze);
 
    return 0;
}

Python3




# Python implementation for the above approach
from collections import deque
 
# this structure stores information
# about a particular cell i.e
# path upto that cell and cell's
# coordinates
 
 
class info:
    def __init__(self, path, i, j):
        self.path = path
        self.i = i
        self.j = j
 
 
def printAllPaths(maze):
    n = len(maze)
    m = len(maze[0])
 
    q = deque()
    # pushing top-left cell into the queue
    q.append(info([maze[0][0]], 0, 0))
 
    while len(q) > 0:
        p = q.popleft()
 
        # if we reached the bottom-right cell
        # i.e the destination then print the path
        if p.i == n - 1 and p.j == m - 1:
            for x in p.path:
                print(x, end=" ")
            print()
 
        # if we are in the last row
        # then only right movement is possible
        elif p.i == n-1:
            temp = p.path[:]
            # updating the current path
            temp.append(maze[p.i][p.j+1])
            q.append(info(temp, p.i, p.j+1))
 
        # if we are in the last column
        # then only down movement is possible
        elif p.j == m-1:
            temp = p.path[:]
            # updating the current path
            temp.append(maze[p.i+1][p.j])
            q.append(info(temp, p.i+1, p.j))
 
        # else both right and down movement
        # are possible
        else# right movement
            temp = p.path[:]
            # updating the current path
            temp.append(maze[p.i][p.j + 1])
            q.append(info(temp, p.i, p.j + 1))
 
            # down movement
            temp = temp[:-1]
 
            # updating the current path
            temp.append(maze[p.i + 1][p.j])
            q.append(info(temp, p.i + 1, p.j))
 
 
# Driver Code
maze = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
printAllPaths(maze)
 
# This code is contributed by Tapesh(tapeshdua420)

Output

1 2 3 6 9 
1 2 5 6 9 
1 2 5 8 9 
1 4 5 6 9 
1 4 5 8 9 
1 4 7 8 9 

  


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