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Print all possible paths to escape out of a matrix from a given position using at most K moves

Given a matrix mat[][] of dimension N*M, a positive integer K and the source cell (X, Y), the task is to print all possible paths to move out of the matrix from the source cell (X, Y) by moving in all four directions in each move in at most K moves.

Examples:

Input: N = 2, M = 2, X = 1, Y = 1, K = 2
Output:
(1 1)(1 0)
(1 1)(1 2)(1 3)
(1 1)(1 2)(0 2)
(1 1)(0 1)
(1 1)(2 1)(2 0)
(1 1)(2 1)(3 1)

Input: N = 1, M = 1, X = 1, Y = 1, K = 2
Output:
(1 1)(1 0)
(1 1)(1 2)
(1 1)(0 1)
(1 1)(2 1)

Approach: The given problem can be solved by using Recursion and Backtracking. Follow the steps below to solve the given problem:

  • Initialize an array, say, arrayOfMoves[] that stores all the moves moving from the source cell to the out of the matrix.
  • Define a recursive function, say printAllmoves(N, M, moves, X, Y, arrayOfMoves), and perform the following steps:
    • Base Case:
      • If the value of the moves is non-negative and the current cell (X, Y) is out of the matrix, then print all the moves stored in the ArrayOfMoves[].
      • If the value of moves is less than 0 then return from the function.
    • Insert the current cell (X, Y) in the array arrayOfMoves[].
    • Recursively call the function in all the four directions of the current cell (X, Y) by decrementing the value of moves by 1
    • If the size of the array arrayOfMoves[] is greater than 1, then remove the last cell inserted for the Backtracking steps.
  • Call the function printAllmoves(N, M, moves, X, Y, arrayOfMoves) to print all possible moves.

Below is the implementation of the above approach: 

C++

// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
 
// Function to print all the paths
// that are outside of the matrix
void printAllmoves(
  int n, int m, int x, int y,
  int moves, vector<pair<int,int>> ArrayOfMoves)
{
   
  // Base Case
  if (x <= 0 || y <= 0 || x >= n + 1
      || y >= m + 1 && moves >= 0) {
 
    // Add the last position
    ArrayOfMoves.push_back({x, y});
 
    // Traverse the pairs
    for (auto ob : ArrayOfMoves) {
 
      // Print all the paths
      cout<<"("<<ob.first<<" "<<ob.second<<")";
    }
 
    cout<<endl;
 
    // Backtracking Steps
    if(ArrayOfMoves.size() > 1)
      ArrayOfMoves.pop_back();
 
    return;
  }
 
  // If no moves remain
  if (moves <= 0) {
    return;
  }
 
  // Add the current position
  // in the list
  ArrayOfMoves.push_back({x, y});
 
  // Recursive function Call
  // in all the four directions
  printAllmoves(n, m, x, y - 1, moves - 1,
                ArrayOfMoves);
  printAllmoves(n, m, x, y + 1, moves - 1,
                ArrayOfMoves);
  printAllmoves(n, m, x - 1, y, moves - 1,
                ArrayOfMoves);
  printAllmoves(n, m, x + 1, y, moves - 1,
                ArrayOfMoves);
 
  // Backtracking Steps
  if (ArrayOfMoves.size() > 1) {
    ArrayOfMoves.pop_back();
  }
}
 
// Driver Code
int main()
{
  int N = 2, M = 2;
  int X = 1;
  int Y = 1;
  int K = 2;
  vector<pair<int,int>> ArrayOfMoves;
 
  // Function Call
  printAllmoves(N, M, X, Y, K,
                ArrayOfMoves);
}
 
// This code is contributed by ipg2016107.

Java

// Java program for the above approach
 
import java.io.*;
import java.util.*;
 
public class GFG {
 
    // Class for the pairs
    static class Pair {
        int a;
        int b;
        Pair(int a, int b)
        {
            this.a = a;
            this.b = b;
        }
    }
 
    // Function to print all the paths
    // that are outside of the matrix
    static void printAllmoves(
        int n, int m, int x, int y,
        int moves, ArrayList<Pair> ArrayOfMoves)
    {
        // Base Case
        if (x <= 0 || y <= 0 || x >= n + 1
            || y >= m + 1 && moves >= 0) {
 
            // Add the last position
            ArrayOfMoves.add(new Pair(x, y));
 
            // Traverse the pairs
            for (Pair ob : ArrayOfMoves) {
 
                // Print all the paths
                System.out.print("(" + ob.a
                                 + " " + ob.b
                                 + ")");
            }
 
            System.out.println();
 
            // Backtracking Steps
            if (ArrayOfMoves.size() > 1)
                ArrayOfMoves.remove(
                    ArrayOfMoves.size() - 1);
 
            return;
        }
 
        // If no moves remain
        if (moves <= 0) {
            return;
        }
 
        // Add the current position
        // in the list
        ArrayOfMoves.add(new Pair(x, y));
 
        // Recursive function Call
        // in all the four directions
        printAllmoves(n, m, x, y - 1, moves - 1,
                      ArrayOfMoves);
        printAllmoves(n, m, x, y + 1, moves - 1,
                      ArrayOfMoves);
        printAllmoves(n, m, x - 1, y, moves - 1,
                      ArrayOfMoves);
        printAllmoves(n, m, x + 1, y, moves - 1,
                      ArrayOfMoves);
 
        // Backtracking Steps
        if (ArrayOfMoves.size() > 1) {
            ArrayOfMoves.remove(
                ArrayOfMoves.size() - 1);
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 2, M = 2;
        int X = 1;
        int Y = 1;
        int K = 2;
        ArrayList<Pair> ArrayOfMoves
            = new ArrayList<>();
 
        // Function Call
        printAllmoves(N, M, X, Y, K,
                      ArrayOfMoves);
    }
}

Python3

# Python program for the above approach
 
# Function to print all the paths
# that are outside of the matrix
def printAllmoves(n,m,x,y, moves,ArrayOfMoves):
   
  # Base Case
  if (x <= 0 or y <= 0 or x >= n + 1 or y >= m + 1 and moves >= 0):
     
    # Add the last position
    ArrayOfMoves.append([x, y])
 
    # Traverse the pairs
    for ob in ArrayOfMoves:
       
      # Print all the paths
      print("(",ob[0],ob[1],")",end="")
 
    print("\n",end = "")
 
    # Backtracking Steps
    if(len(ArrayOfMoves) > 1):
      ArrayOfMoves.pop()
 
    return
 
  # If no moves remain
  if (moves <= 0):
    return
 
  # Add the current position
  # in the list
  ArrayOfMoves.append([x, y])
 
  # Recursive function Call
  # in all the four directions
  printAllmoves(n, m, x, y - 1, moves - 1,ArrayOfMoves)
  printAllmoves(n, m, x, y + 1, moves - 1,ArrayOfMoves)
  printAllmoves(n, m, x - 1, y, moves - 1,ArrayOfMoves)
  printAllmoves(n, m, x + 1, y, moves - 1,ArrayOfMoves)
 
  # Backtracking Steps
  if (len(ArrayOfMoves) > 1):
    ArrayOfMoves.pop()
 
# Driver Code
if __name__ == '__main__':
  N = 2
  M = 2
  X = 1
  Y = 1
  K = 2
  ArrayOfMoves = []
 
  # Function Call
  printAllmoves(N, M, X, Y, K,ArrayOfMoves)
 
  # This code is contributed by SURENDRA_GANGWAR.

C#

using System;
using System.Collections;
public class GFG {
    class Pair {
        public int a;
        public int b;
        public Pair(int a, int b)
        {
            this.a = a;
            this.b = b;
        }
    }
 
    // Function to print all the paths
    // that are outside of the matrix
    static void printAllmoves(int n, int m, int x, int y,
                              int moves,
                              ArrayList ArrayOfMoves)
    {
        // Base Case
        if (x <= 0 || y <= 0 || x >= n + 1
            || y >= m + 1 && moves >= 0) {
 
            // Add the last position
            ArrayOfMoves.Add(new Pair(x, y));
 
            // Traverse the pairs
            foreach (Pair ob in ArrayOfMoves) {
 
                // Print all the paths
                Console.Write("(" + ob.a + " " + ob.b
                                  + ")");
            }
 
            Console.WriteLine();
 
            // Backtracking Steps
            if (ArrayOfMoves.Count > 1)
                ArrayOfMoves.Remove(ArrayOfMoves.Count
                                    - 1);
 
            return;
        }
 
        // If no moves remain
        if (moves <= 0) {
            return;
        }
 
        // Add the current position
        // in the list
        ArrayOfMoves.Add(new Pair(x, y));
 
        // Recursive function Call
        // in all the four directions
        printAllmoves(n, m, x, y - 1, moves - 1,
                      ArrayOfMoves);
        printAllmoves(n, m, x, y + 1, moves - 1,
                      ArrayOfMoves);
        printAllmoves(n, m, x - 1, y, moves - 1,
                      ArrayOfMoves);
        printAllmoves(n, m, x + 1, y, moves - 1,
                      ArrayOfMoves);
 
        // Backtracking Steps
        if (ArrayOfMoves.Count > 1) {
            ArrayOfMoves.Remove(ArrayOfMoves.Count - 1);
        }
    }
    static public void Main()
    {
        int N = 2, M = 2;
        int X = 1;
        int Y = 1;
        int K = 2;
        ArrayList ArrayOfMoves = new ArrayList();
 
        // Function Call
        printAllmoves(N, M, X, Y, K, ArrayOfMoves);
    }
}
 
// This code is contributed by maddler.

Javascript

<script>
// Javascript program for the above approach
 
    // Class for the pairs
     class Pair {
     
        constructor(a , b) {
            this.a = a;
            this.b = b;
        }
    }
function ob(item)
{
 
    // Print all the paths
    document.write("(" + item.a + " " + item.b + ")");
}
 
    // Function to print all the paths
    // that are outside of the matrix
    function printAllmoves(n , m , x , y , moves,  ArrayOfMoves)
    {
        // Base Case
        if (x <= 0 || y <= 0 || x >= n + 1 || y >= m + 1 && moves >= 0) {
 
            // Add the last position
            ArrayOfMoves.push(new Pair(x, y));
 
            // Traverse the pairs
            ArrayOfMoves.forEach (ob);
 
            document.write("<br/>");
 
            // Backtracking Steps
            if (ArrayOfMoves.length > 1)
                ArrayOfMoves.pop(ArrayOfMoves.length - 1);
 
            return;
        }
 
        // If no moves remain
        if (moves <= 0) {
            return;
        }
 
        // Add the current position
        // in the list
        ArrayOfMoves.push(new Pair(x, y));
 
        // Recursive function Call
        // in all the four directions
        printAllmoves(n, m, x, y - 1, moves - 1, ArrayOfMoves);
        printAllmoves(n, m, x, y + 1, moves - 1, ArrayOfMoves);
        printAllmoves(n, m, x - 1, y, moves - 1, ArrayOfMoves);
        printAllmoves(n, m, x + 1, y, moves - 1, ArrayOfMoves);
 
        // Backtracking Steps
        if (ArrayOfMoves.length > 1) {
            ArrayOfMoves.pop(ArrayOfMoves.length - 1);
        }
    }
 
    // Driver Code
        var N = 2, M = 2;
        var X = 1;
        var Y = 1;
        var K = 2;
        var ArrayOfMoves =new Array();
 
        // Function Call
        printAllmoves(N, M, X, Y, K, ArrayOfMoves);
 
// This code is contributed by gauravrajput1
</script>

Output: 

(1 1)(1 0)
(1 1)(1 2)(1 3)
(1 1)(1 2)(0 2)
(1 1)(0 1)
(1 1)(2 1)(2 0)
(1 1)(2 1)(3 1)

 

Time Complexity: O(4N)
Auxiliary Space: O(4N)


 


Reffered: https://www.geeksforgeeks.org

Matrix

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