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: 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
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 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[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[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[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[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<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 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<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,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<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 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[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[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[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[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<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 |
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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