# Find paths from corner cell to middle cell in maze

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```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

1. print the path

Else

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

Below is its C++ and Java implementation

C++ Implementation

 `// C++ program to find a path from corner cell to ` `// middle cell in maze containing positive numbers ` `#include ` `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 > 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 > 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}; ` ` `  `// Cordinates 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 > &path, ` `                ``set > &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 contaning positive numbers ` `void` `findPathInMaze(``int` `maze[N][N]) ` `{ ` `    ``// list to store complete path ` `    ``// from source to destination ` `    ``list > path; ` ` `  `    ``// to store cells already visisted in current path ` `    ``set > 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 "``); ` `        ``} ` ` `  `        ``// down call ` `        ``if` `(i+k "``); ` `        ``} ` ` `  `        ``// 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; ` `    ``} ` `                         `  `} `

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
```

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Improved By : raghavbansal011

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