|
3 | 3 | * - Given a NxN grid, find whether rat in cell (0,0) can reach target(N-1,N-1); |
4 | 4 | * - In grid "0" represent blocked cell and "1" represent empty cell |
5 | 5 | * |
| 6 | + * Reference for this problem: |
| 7 | + * - https://www.geeksforgeeks.org/rat-in-a-maze-backtracking-2/ |
| 8 | + * |
6 | 9 | */ |
7 | 10 |
|
| 11 | +// Helper function to find if current cell is out of boundary |
8 | 12 | const outOfBoundary = (grid, currentRow, currentColumn) => { |
9 | | - if (currentRow < 0 || currentColumn < 0 || currentRow >= grid.length || currentColumn >= grid[0].length) return true |
10 | | - else return false |
| 13 | + if (currentRow < 0 || currentColumn < 0 || currentRow >= grid.length || currentColumn >= grid[0].length) { |
| 14 | + return true |
| 15 | + } else { |
| 16 | + return false |
| 17 | + } |
11 | 18 | } |
12 | 19 |
|
13 | | -const isPossible = (grid, currentRow, currentColumn) => { |
14 | | - if (outOfBoundary(grid, currentRow, currentColumn)) return false |
15 | | - |
16 | | - if (grid[currentRow][currentColumn] === 0) return false |
17 | | - |
18 | | - if (currentRow === targetRow && currentColumn === targetColumn) { |
19 | | - return true |
20 | | - } |
21 | | - |
22 | | - const directions = [ |
23 | | - [1, 0], |
24 | | - [0, 1], |
25 | | - [-1, 0], |
26 | | - [0, -1] |
27 | | - ] |
28 | | - |
29 | | - for (let i = 0; i < directions.length; i++) { |
30 | | - const nextRow = currentRow + directions[i][0]; const nextColumn = currentColumn + directions[i][1] |
31 | | - grid[currentRow][currentColumn] = 0 |
32 | | - if (isPossible(grid, nextRow, nextColumn)) return true |
33 | | - grid[currentRow][currentColumn] = 1 |
34 | | - } |
35 | | - return false |
| 20 | +const printPath = (grid, currentRow, currentColumn, path) => { |
| 21 | + // If cell is out of boundary, we can't proceed |
| 22 | + if (outOfBoundary(grid, currentRow, currentColumn)) return false |
| 23 | + |
| 24 | + // If cell is blocked then you can't go ahead |
| 25 | + if (grid[currentRow][currentColumn] === 0) return false |
| 26 | + |
| 27 | + // If we reached target cell, then print path |
| 28 | + if (currentRow === targetRow && currentColumn === targetColumn) { |
| 29 | + console.log(path) |
| 30 | + return true |
| 31 | + } |
| 32 | + |
| 33 | + // R,L,D,U are directions `Right, Left, Down, Up` |
| 34 | + const directions = [ |
| 35 | + [1, 0, 'D'], |
| 36 | + [-1, 0, 'U'], |
| 37 | + [0, 1, 'R'], |
| 38 | + [0, -1, 'L'] |
| 39 | + ] |
| 40 | + |
| 41 | + for (let i = 0; i < directions.length; i++) { |
| 42 | + const nextRow = currentRow + directions[i][0] |
| 43 | + const nextColumn = currentColumn + directions[i][1] |
| 44 | + const updatedPath = path + directions[i][2] |
| 45 | + |
| 46 | + grid[currentRow][currentColumn] = 0 |
| 47 | + if (printPath(grid, nextRow, nextColumn, updatedPath)) return true |
| 48 | + grid[currentRow][currentColumn] = 1 |
| 49 | + } |
| 50 | + return false |
36 | 51 | } |
37 | 52 |
|
38 | 53 | // Driver Code |
39 | 54 |
|
40 | 55 | const grid = [ |
41 | | - [1, 1, 1, 1], |
42 | | - [1, 0, 0, 1], |
43 | | - [0, 0, 1, 0], |
44 | | - [1, 1, 0, 1] |
| 56 | + [1, 1, 1, 1], |
| 57 | + [1, 0, 0, 1], |
| 58 | + [0, 0, 1, 1], |
| 59 | + [1, 1, 0, 1] |
45 | 60 | ] |
46 | 61 |
|
47 | 62 | const targetRow = grid.length - 1 |
48 | 63 | const targetColumn = grid[0].length - 1 |
49 | 64 |
|
50 | | -if (isPossible(grid, 0, 0)) { |
51 | | - console.log('Possible') |
52 | | -} else { |
53 | | - console.log('Not Possible') |
54 | | -} |
| 65 | +// Variation 2 : print a possible path to reach from (0, 0) to (N-1, N-1) |
| 66 | +// If there is no path possible then it will print "Not Possible" |
| 67 | +!printPath(grid, 0, 0, '') && console.log('Not Possible') |
0 commit comments