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

• Difficulty Level : Medium
• Last Updated : 21 Jul, 2021

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` `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 = { {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`

## Javascript

 ``
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.

## C++

 `// C++ program to Print all possible paths from ``// top left to bottom right of a mXn matrix``#include ``using` `namespace` `std;` `vector> allPaths;` `void` `findPathsUtil(vector> maze, ``int` `m,``                                 ``int` `n, ``int` `i, ``int` `j,``                          ``vector<``int``> path, ``int` `indx)``{``    ` `    ``// 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[indx + 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 << ``", "``;``        ``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[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``        ``cout << ``"["` `<< path << ``", "``;``        ``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[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);``}``    ` `void` `findPaths(vector> 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> 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> allPaths =``    ``new` `ArrayList>();``  ``static` `void` `findPathsUtil(ArrayList> maze,``                            ``int` `m, ``int` `n, ``int` `i,``int` `j,``                            ``ArrayList path,``int` `indx)``  ``{` `    ``// 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(indx + 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(indx + 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(indx,maze.get(i).get(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``);` `  ``}``  ``static` `void` `findPaths(ArrayList> maze,``                        ``int` `m, ``int` `n)``  ``{``    ``ArrayList path = ``new` `ArrayList();``    ``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> maze =``      ``new` `ArrayList>();``    ``maze.add(``new` `ArrayList``             ``(Arrays.asList(``1``,``2``,``3``)));``    ``maze.add(``new` `ArrayList``             ``(Arrays.asList(``4``,``5``,``6``)));``    ``maze.add(``new` `ArrayList``             ``(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,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)`

## 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> allPaths = ``new` `List>();`` ` `    ``static` `void` `findPathsUtil(List> maze,``                              ``int` `m, ``int` `n, ``int` `i,``                              ``int` `j, List<``int``> path,``                              ``int` `indx)``    ``{``         ` `        ``// 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[indx + 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 + ``", "``);``            ``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[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``            ``Console.Write( ``"["` `+ path + ``", "``);``            ``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[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);``    ``}``         ` `    ``static` `void` `findPaths(List> 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> maze = ``new` `List>();``    ``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`
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``using` `namespace` `std;` `// function to display the path``void` `display(vector<``int``> &ans) {``  ``for``(``auto` `i :ans ) {``    ``cout<>& visited,``int` `n,``int` `m) {``  ``return` `(r < n and c > &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 >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;``}`
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 ``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 >& maze)``{``    ``int` `n = maze.size();``    ``int` `m = maze.size();` `    ``queue q;``    ``// pushing top-left cell into the queue``    ``q.push({ { maze }, 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 > maze{ { 1, 2, 3 },``                               ``{ 4, 5, 6 },``                               ``{ 7, 8, 9 } };` `    ``printAllPaths(maze);` `    ``return` `0;``}`
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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