|
| 1 | +/* Rat In a Maze |
| 2 | + --------------------------------------------------------------------- |
| 3 | + |
| 4 | +Problem: Given a maze(2D matrix) with obstacles, starting from (0,0) you have to |
| 5 | +reach (n-1, n-1). If you are currently on (x,y), you can move to (x+1,y) or (x,y+1). |
| 6 | +You can not move to the walls. |
| 7 | +Idea: Try all the possible paths to see if you can reach (n-1,n-1) |
| 8 | +
|
| 9 | +-------------------------------------------------------------------------------- |
| 10 | +
|
| 11 | +Input: |
| 12 | +---- |
| 13 | +0 denotes wall, 1 denotes free path |
| 14 | +two numbers n, m |
| 15 | +Next n lines contain m numbers (0 or 1) |
| 16 | +
|
| 17 | +Output: |
| 18 | +-------- |
| 19 | +Print 1 if rat can reach (n-1,m-1) |
| 20 | +Print 0 if it can not reach (n-1,m-1) |
| 21 | +
|
| 22 | +------------------------ |
| 23 | +
|
| 24 | +Test Case 1: |
| 25 | +
|
| 26 | +Input: |
| 27 | +5 5 |
| 28 | +1 0 1 0 1 |
| 29 | +1 1 1 1 1 |
| 30 | +0 1 0 1 0 |
| 31 | +1 0 0 1 1 |
| 32 | +1 1 1 0 1 |
| 33 | +
|
| 34 | +Output: |
| 35 | +
|
| 36 | +1 0 0 0 0 |
| 37 | +1 1 1 1 0 |
| 38 | +0 0 0 1 0 |
| 39 | +0 0 0 1 1 |
| 40 | +0 0 0 0 1 |
| 41 | +
|
| 42 | +*/ |
| 43 | + |
| 44 | + |
| 45 | +#include <iostream> |
| 46 | +using namespace std; |
| 47 | + |
| 48 | +bool isSafe(int **arr, int x, int y, int n){ |
| 49 | + if(x<n && y<n && arr[x][y] == 1){ |
| 50 | + return true; |
| 51 | + } |
| 52 | + return false; |
| 53 | +} |
| 54 | + |
| 55 | +bool ratinMaze(int **arr, int x, int y,int n, int** solArr){ |
| 56 | + |
| 57 | + if((x== (n-1)) && (y== (n-1))){ |
| 58 | + solArr[x][y]=1; |
| 59 | + return true; |
| 60 | + } |
| 61 | + |
| 62 | + if(isSafe(arr,x,y,n)){ |
| 63 | + solArr[x][y]==1; |
| 64 | + if(ratinMaze(arr,x+1,y,n,solArr)){ |
| 65 | + return true; |
| 66 | + } |
| 67 | + if(ratinMaze(arr,x,y+1,n,solArr)){ |
| 68 | + return true; |
| 69 | + } |
| 70 | + |
| 71 | + solArr[x][y]=0; //backtracking |
| 72 | + return false; |
| 73 | + } |
| 74 | + return false; |
| 75 | + |
| 76 | +} |
| 77 | + |
| 78 | +int main(){ |
| 79 | + |
| 80 | + int n; |
| 81 | + cin>>n; |
| 82 | + |
| 83 | + int** arr=new int*[n]; |
| 84 | + for(int i=0;i<n;i++){ |
| 85 | + arr[i]=new int[n]; |
| 86 | + } |
| 87 | + |
| 88 | + for(int i=0;i<n;i++){ |
| 89 | + for(int j=0;j<n;j++){ |
| 90 | + cin>>arr[i][j]; |
| 91 | + } |
| 92 | + } |
| 93 | + |
| 94 | + int** solArr=new int*[n]; |
| 95 | + for(int i=0;i<n;i++){ |
| 96 | + solArr[i]=new int[n]; |
| 97 | + for(int j=0;j<n;j++){ |
| 98 | + solArr[i][j]=0; |
| 99 | + } |
| 100 | + |
| 101 | + } |
| 102 | + |
| 103 | + if(ratinMaze(arr,0,0,n,solArr)){ |
| 104 | + for(int i=0;i<n;i++){ |
| 105 | + for(int j=0;j<n;j++){ |
| 106 | + cout<<solArr[i][j]<<" "; |
| 107 | + } cout<<endl; |
| 108 | + } |
| 109 | + |
| 110 | + } |
| 111 | + |
| 112 | + |
| 113 | + return 0; |
| 114 | +} |
| 115 | + |
| 116 | +/* Time Complexity: O(2n) |
| 117 | +Space Complexity: O(2n) |
| 118 | +*/ |
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