How to find all the possible increasing order paths in a multidimensional array?

Question

I wanted to know the Java code to find all the possible paths of a multidimensional array. We should start from the [0][0] node and move forward. The thing to keep in mind that the paths must be in an increasing order irrespective of the elements order. We should not traverse if the next element is equal to (or less then) the current one.

Example Array:

3   5   1
6   7   4
8   2   9

Result:

3,5,6,7,8
3,5,6,7,9
3,5,7,8
3,5,7,9
3,6,7,8
3,6,7,9
3,7,8
3,7,9

Answer 1

You should build a tree, beginning with your start point. In your example array, starting at the top left, the tree might look like this:

     3
   / \ \ 
  5  6  7 
 / \  \  \
7   6  ..  ..

Once you have that tree, you just have to step down each path in the tree.

Although it may look intimidating, you can break this up into two parts:

• Build the tree.
• Follow the paths.

---

• First, try to come up with a way to get your starting point to print. That's fairly easy.

• Once you've made it that far, print the neighbors of your starting point that are > its value.

• Then, do the same thing for each neighbor!

Answer 2

I think the following code does what you describe. It start checking with the first node (0,0). For every node that is checked, a vector of neighbors is created. The neighbors are the nodes that are eligible to be a continuation to the path (i.e. neighboring nodes with higher value in the table). Then for each neighbor, the path is cloned, and the new neighbor is checked. This continues until the checked node has no eligible neighbors, at which point the path is printed and the algorithm terminates.

Try this:

import java.util.Arrays;
import java.util.Vector;

class Main {
  class Coords {
    int x;
    int y;
    Coords(int x, int y) {
      this.x = x;
      this.y = y;
    }
  }
  int [][] array = { {3,5,1},{6,7,4},{8,2,9}};
  Vector<Coords> getNeighbors(Coords coords) {
     int x = coords.x;
     int y = coords.y;
     Vector<Coords> result = new Vector<Coords>();
     if (x < array.length - 1) {
        if (array[x + 1][y] >= array[x][y])
          result.add(new Coords(x + 1, y));
     }
     if (x > 0) {
        if (array[x - 1][y] >= array[x][y])
          result.add(new Coords(x - 1, y));
     }
     if (y < array[x].length - 1) {
        if (array[x][y + 1] >= array[x][y])
          result.add(new Coords(x, y + 1));
     }
     if (y > 0) {
        if (array[x][y - 1] >= array[x][y])
          result.add(new Coords(x, y - 1));
     }
     if (x < (array.length - 1 ) &&  (y < array[x].length - 1)) {
        if (array[x + 1][y + 1] >= array[x][y])
          result.add(new Coords(x + 1, y + 1));
     }
     if (x < (array.length - 1 ) &&  (y > 0)) {
        if (array[x + 1][y - 1] >= array[x][y])
          result.add(new Coords(x + 1, y - 1));
     }
     if (x > 0 &&  (y < array[x].length - 1)) {
        if (array[x - 1][y + 1] >= array[x][y])
          result.add(new Coords(x - 1, y + 1));
     }
     if (x > 0 &&  y > 0) {
        if (array[x -1][y - 1] >= array[x][y])
          result.add(new Coords(x - 1, y - 1));
     }
     return result;
  }
  void checkNode(Vector<Integer> path, Coords coords) {
    path.add(array[coords.x][coords.y]);
    Vector<Coords> neighbors = getNeighbors(coords);
    if (neighbors.size() == 0) {
       for (Integer i : path) {
         System.out.print(i+"\t");
       }
       System.out.println();
    }
    for (Coords c : neighbors) {
      Vector<Integer> newpath = (Vector<Integer>) path.clone();
      checkNode(newpath, c);
    }
  }
  Main() {
    System.out.println ("Array: " + Arrays.deepToString(array));
    checkNode(new Vector<Integer>(),new Coords(0,0));
  }

  public static void main(String args[]) {
    new Main();
  }
}

Output:

Array: [[3, 5, 1], [6, 7, 4], [8, 2, 9]]
3   6   8   
3   6   7   9   
3   6   7   8   
3   5   7   9   
3   5   7   8   
3   5   6   8   
3   5   6   7   9   
3   5   6   7   8   
3   7   9   
3   7   8   

It also gives me the path 3,6,8, which is not in your sample output