Shortest path in 2d arrays

Question

*...*..D
.G..*.....
**...**.
.S....*.
........
...G**..
........
.G..*...

Here is 2d array where
S- Source
D-Destination
G-Point must be

Answer 1

To find the shortest distance from start point to all other points in the map, you can use a BFS. Sample code:

public void visit(String []map , Point start){
        int []x = {0,0,1,-1};//This represent 4 directions right, left, down , up
        int []y = {1,-1,0,0};
        LinkedList q = new LinkedList();
        q.add(start);
        int n = map.length;
        int m = map[0].length();
        int[][]dist = new int[n][m];
        for(int []a : dist){
            Arrays.fill(a,-1);
        }
        dist[start.x][start.y] = 0;
        while(!q.isEmpty()){
            Point p = q.removeFirst();
            for(int i = 0; i < 4; i++){
                int a = p.x + x[i];
                int b = p.y + y[i];
                if(a >= 0 && b >= 0 && a < n && b < m && dist[a][b] == -1 && map[a].charAt(b) != '*' ){
                    dist[a][b] = 1 + dist[p.x][p.y];
                    q.add(new Point(a,b));
                }
            }
        }
    }

The second path of the problem is actually a traveling salesman problem, so you need to convert from your original graph to a graph which only contains G,D and S points, with the weight of each edge in this graph is the shortest path between them in original path. From that onward, if the number of G is small (less than 17) you can use dynamic programming and bitmask to solve the problem.