Find paths from corner cell to middle cell in maze

Difficulty Level : Hard
Last Updated : 10 Jun, 2021

Given a square maze containing positive numbers, find all paths from a corner cell (any of the extreme four corners) to the middle cell. We can move exactly n steps from a cell in 4 directions i.e. North, East, West and South where n is value of the cell,

We can move to mat[i+n][j], mat[i-n][j], mat[i][j+n], and mat[i][j-n] from a cell mat[i][j] where n is value of mat[i][j].

Example

Input:  9 x 9 maze
[ 3, 5, 4, 4, 7, 3, 4, 6, 3 ]
[ 6, 7, 5, 6, 6, 2, 6, 6, 2 ]
[ 3, 3, 4, 3, 2, 5, 4, 7, 2 ]
[ 6, 5, 5, 1, 2, 3, 6, 5, 6 ]
[ 3, 3, 4, 3, 0, 1, 4, 3, 4 ]
[ 3, 5, 4, 3, 2, 2, 3, 3, 5 ]
[ 3, 5, 4, 3, 2, 6, 4, 4, 3 ]
[ 3, 5, 1, 3, 7, 5, 3, 6, 4 ]
[ 6, 2, 4, 3, 4, 5, 4, 5, 1 ]

Output:
(0, 0) -> (0, 3) -> (0, 7) -> 
(6, 7) -> (6, 3) -> (3, 3) -> 
(3, 4) -> (5, 4) -> (5, 2) -> 
(1, 2) -> (1, 7) -> (7, 7) ->
(7, 1) -> (2, 1) -> (2, 4) -> 
(4, 4) -> MID

The idea is to use backtracking. We start with each corner cell of the maze and recursively checks if it leads to the solution or not. Following is the Backtracking algorithm –
If destination is reached

print the path

Else

Mark current cell as visited and add it to path array.
Move forward in all 4 allowed directions and recursively check if any of them leads to a solution.
If none of the above solutions work then mark this cell as not visitedand remove it from path array.

Below is the implementation of the above approach:

C++

// C++ program to find a path from corner cell to
// middle cell in maze containing positive numbers
#include <bits/stdc++.h>
using namespace std;
 
// Rows and columns in given maze
#define N 9
 
// check whether given cell is a valid cell or not.
bool isValid(set<pair<int, int> > visited,
             pair<int, int> pt)
{
    // check if cell is not visited yet to
    // avoid cycles (infinite loop) and its
    // row and column number is in range
    return (pt.first >= 0) && (pt.first  < N) &&
           (pt.second >= 0) && (pt.second < N) &&
           (visited.find(pt) == visited.end());
}
 
// Function to print path from source to middle coordinate
void printPath(list<pair<int, int> > path)
{
    for (auto it = path.begin(); it != path.end(); it++)
        cout << "(" << it->first << ", "
             << it->second << ") -> ";
 
    cout << "MID" << endl << endl;
}
 
// For searching in all 4 direction
int row[] = {-1, 1, 0, 0};
int col[] = { 0, 0, -1, 1};
 
// Coordinates of 4 corners of matrix
int _row[] = { 0, 0, N-1, N-1};
int _col[] = { 0, N-1, 0, N-1};
 
void findPathInMazeUtil(int maze[N][N],
                list<pair<int, int> > &path,
                set<pair<int, int> > &visited,
                pair<int, int> &curr)
{
    // If we have reached the destination cell.
    // print the complete path
    if (curr.first == N / 2 && curr.second == N / 2)
    {
        printPath(path);
        return;
    }
 
    // consider each direction
    for (int i = 0; i < 4; ++i)
    {
        // get value of current cell
        int n = maze[curr.first][curr.second];
 
        // We can move N cells in either of 4 directions
        int x = curr.first + row[i]*n;
        int y = curr.second + col[i]*n;
 
        // Constructs a pair object with its first element
        // set to x and its second element set to y
        pair<int, int> next = make_pair(x, y);
 
        // if valid pair
        if (isValid(visited, next))
        {
            // mark cell as visited
            visited.insert(next);
 
            // add cell to current path
            path.push_back(next);
 
            // recuse for next cell
            findPathInMazeUtil(maze, path, visited, next);
 
            // backtrack
            path.pop_back();
             
            // remove cell from current path
            visited.erase(next);
        }
    }
}
 
// Function to find a path from corner cell to
// middle cell in maze containing positive numbers
void findPathInMaze(int maze[N][N])
{
    // list to store complete path
    // from source to destination
    list<pair<int, int> > path;
 
    // to store cells already visisted in current path
    set<pair<int, int> > visited;
 
    // Consider each corners as the starting
    // point and search in maze
    for (int i = 0; i < 4; ++i)
    {
        int x = _row[i];
        int y = _col[i];
 
        // Constructs a pair object
        pair<int, int> pt = make_pair(x, y);
 
        // mark cell as visited
        visited.insert(pt);
 
        // add cell to current path
        path.push_back(pt);
 
        findPathInMazeUtil(maze, path, visited, pt);
 
        // backtrack
        path.pop_back();
 
        // remove cell from current path
        visited.erase(pt);
    }
}
 
int main()
{
    int maze[N][N] =
    {
        { 3, 5, 4, 4, 7, 3, 4, 6, 3 },
        { 6, 7, 5, 6, 6, 2, 6, 6, 2 },
        { 3, 3, 4, 3, 2, 5, 4, 7, 2 },
        { 6, 5, 5, 1, 2, 3, 6, 5, 6 },
        { 3, 3, 4, 3, 0, 1, 4, 3, 4 },
        { 3, 5, 4, 3, 2, 2, 3, 3, 5 },
        { 3, 5, 4, 3, 2, 6, 4, 4, 3 },
        { 3, 5, 1, 3, 7, 5, 3, 6, 4 },
        { 6, 2, 4, 3, 4, 5, 4, 5, 1 }
    };
 
    findPathInMaze(maze);
 
    return 0;
}

Output :

(0, 0) -> (0, 3) -> (0, 7) -> 
(6, 7) -> (6, 3) -> (3, 3) -> 
(3, 4) -> (5, 4) -> (5, 2) -> 
(1, 2) -> (1, 7) -> (7, 7) ->
(7, 1) -> (2, 1) -> (2, 4) -> 
(4, 4) -> MID

A better approach:

Java

// Java program to find a path from corner cell to
// middle cell in maze containing positive numbers
import java.io.*;
 
class GFG {
    public static void main (String[] args) {
 
        // Creating the maze
        int[][] maze = {
            { 3, 5, 4, 4, 7, 3, 4, 6, 3 },
            { 6, 7, 5, 6, 6, 2, 6, 6, 2 },
            { 3, 3, 4, 3, 2, 5, 4, 7, 2 },
            { 6, 5, 5, 1, 2, 3, 6, 5, 6 },
            { 3, 3, 4, 3, 0, 1, 4, 3, 4 },
            { 3, 5, 4, 3, 2, 2, 3, 3, 5 },
            { 3, 5, 4, 3, 2, 6, 4, 4, 3 },
            { 3, 5, 1, 3, 7, 5, 3, 6, 4 },
            { 6, 2, 4, 3, 4, 5, 4, 5, 1 }
        };
         
        // Calling the printPath function
        printPath(maze,0,0,"");
    }
     
    public static void printPath(int[][] maze, int i, int j, String ans){
 
        // If we reach the center cell
        if (i == maze.length/2 && j==maze.length/2){
 
            // Make the final answer, Print the
                // final answer and Return
            ans += "("+i+", "+j+") -> MID";
            System.out.println(ans);
            return;
        }
         
        // If the element at the current position
            // in maze is 0, simply Return as it has
            // been visited before.
        if (maze[i][j]==0){
            return;
        }
         
        // If element is non-zero, then note
            // the element in variable 'k'
        int k = maze[i][j];
         
        // Mark the cell visited by making the
            // element 0. Don't worry, the element
            // is safe in 'k'
        maze[i][j]=0;
         
        // Make recursive calls in all 4
            // directions pro-actively i.e. if the next
            // cell lies in maze or not. Right call
        if (j+k<maze.length){
            printPath(maze, i, j+k, ans+"("+i+", "+j+") -> ");
        }
 
        // down call
        if (i+k<maze.length){
            printPath(maze, i+k, j, ans+"("+i+", "+j+") -> ");
        }
 
        // left call
        if (j-k>0){
            printPath(maze, i, j-k, ans+"("+i+", "+j+") -> ");
        }
 
        // up call
        if (i-k>0){
            printPath(maze, i-k, j, ans+"("+i+", "+j+") -> ");
        }
         
        // Unmark the visited cell by substituting
            // its original value from 'k'
        maze[i][j] = k;
    }
                         
}

Python3

# Python program to find a path from corner cell to
# middle cell in maze containing positive numbers
def printPath(maze, i, j, ans):
 
    # If we reach the center cell
    if (i == len(maze) // 2 and j == len(maze) // 2):
 
        # Make the final answer, Prthe
        # final answer and Return
        ans += "(" + str(i) + ", " + str(j) + ") -> MID";
        print(ans);
        return;
     
    # If the element at the current position
    # in maze is 0, simply Return as it has
    # been visited before.
    if (maze[i][j] == 0):
        return;
     
    # If element is non-zero, then note
    # the element in variable 'k'
    k = maze[i][j];
 
    # Mark the cell visited by making the
    # element 0. Don't worry, the element
    # is safe in 'k'
    maze[i][j] = 0;
 
    # Make recursive calls in all 4
    # directions pro-actively i.e. if the next
    # cell lies in maze or not. Right call
    if (j + k < len(maze)):
        printPath(maze, i, j + k, ans + "(" + str(i) + ", " + str(j) + ") -> ");
     
    # down call
    if (i + k < len(maze)):
        printPath(maze, i + k, j, ans + "(" + str(i) + ", " + str(j) + ") -> ");
     
    # left call
    if (j - k > 0):
        printPath(maze, i, j - k, ans + "(" + str(i) + ", " + str(j) + ") -> ");
     
    # up call
    if (i - k > 0):
        printPath(maze, i - k, j, ans + "(" + str(i) + ", " + str(j) + ") -> ");
     
    # Unmark the visited cell by substituting
    # its original value from 'k'
    maze[i][j] = k;
 
    # Driver code
if __name__ == '__main__':
 
    # Creating the maze
    maze = [[ 3, 5, 4, 4, 7, 3, 4, 6, 3 ],[ 6, 7, 5, 6, 6, 2, 6, 6, 2 ],[ 3, 3, 4, 3, 2, 5, 4, 7, 2 ],
            [ 6, 5, 5, 1, 2, 3, 6, 5, 6 ],[ 3, 3, 4, 3, 0, 1, 4, 3, 4 ],[ 3, 5, 4, 3, 2, 2, 3, 3, 5 ],
            [ 3, 5, 4, 3, 2, 6, 4, 4, 3 ],[ 3, 5, 1, 3, 7, 5, 3, 6, 4 ],[ 6, 2, 4, 3, 4, 5, 4, 5, 1 ]] ;
 
    # Calling the printPath function
    printPath(maze, 0, 0, "");
 
# This code contributed by gauravrajput1

C#

// C# program to find a path from corner
// cell to middle cell in maze containing
// positive numbers
using System;
 
class GFG{
 
// Driver Code   
public static void Main(String[] args)
{
     
    // Creating the maze
    int[,] maze = {
        { 3, 5, 4, 4, 7, 3, 4, 6, 3 },
        { 6, 7, 5, 6, 6, 2, 6, 6, 2 },
        { 3, 3, 4, 3, 2, 5, 4, 7, 2 },
        { 6, 5, 5, 1, 2, 3, 6, 5, 6 },
        { 3, 3, 4, 3, 0, 1, 4, 3, 4 },
        { 3, 5, 4, 3, 2, 2, 3, 3, 5 },
        { 3, 5, 4, 3, 2, 6, 4, 4, 3 },
        { 3, 5, 1, 3, 7, 5, 3, 6, 4 },
        { 6, 2, 4, 3, 4, 5, 4, 5, 1 }
    };
     
    // Calling the printPath function
    printPath(maze, 0, 0, "");
}
 
public static void printPath(int[,] maze, int i,
                             int j, String ans)
{
     
    // If we reach the center cell
    if (i == maze.GetLength(0) / 2 &&
        j == maze.GetLength(1) / 2)
    {
         
        // Make the readonly answer, Print the
        // readonly answer and Return
        ans += "(" + i + ", " + j + ") -> MID";
        Console.WriteLine(ans);
        return;
    }
     
    // If the element at the current position
    // in maze is 0, simply Return as it has
    // been visited before.
    if (maze[i, j] == 0)
    {
        return;
    }
     
    // If element is non-zero, then note
    // the element in variable 'k'
    int k = maze[i, j];
     
    // Mark the cell visited by making the
    // element 0. Don't worry, the element
    // is safe in 'k'
    maze[i, j] = 0;
     
    // Make recursive calls in all 4
    // directions pro-actively i.e. if the next
    // cell lies in maze or not. Right call
    if (j + k < maze.GetLength(1))
    {
        printPath(maze, i, j + k,
                  ans + "(" + i +
                  ", " + j + ") -> ");
    }
 
    // Down call
    if (i + k < maze.GetLength(0))
    {
        printPath(maze, i + k, j,
                  ans + "(" + i +
                  ", " + j + ") -> ");
    }
 
    // Left call
    if (j - k > 0)
    {
        printPath(maze, i, j - k,
                  ans + "(" + i +
                  ", " + j + ") -> ");
    }
 
    // Up call
    if (i - k > 0)
    {
        printPath(maze, i - k, j,
                  ans + "(" + i +
                  ", " + j + ") -> ");
    }
     
    // Unmark the visited cell by substituting
    // its original value from 'k'
    maze[i, j] = k;
}
}
 
// This code is contributed by gauravrajput1

Output :

(0, 0) -> (0, 3) -> (0, 7) -> 
(6, 7) -> (6, 3) -> (3, 3) ->
(3, 4) -> (5, 4) -> (5, 2) ->
(1, 2) -> (1, 7) -> (7, 7) -> 
(7, 1) -> (2, 1) -> (2, 4) -> 
(4, 4) -> MID

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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