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Count of ways to traverse a Matrix according to given conditions
  • Difficulty Level : Medium
  • Last Updated : 12 May, 2021

Given an integer N which represents an N x N Square Matrix, the task is to print the number of ways to move from top left to the bottom right of the Square Matrix by following the conditions:  

  • If the current position of the pointer is at the edges of the Square Matrix, then the next move can either be a vertical or a horizontal movement, any number of steps (Like a rook on the chessboard).
  • If the current position of the pointer is at the diagonals of the Square Matrix, then the next move should also lie on the diagonal. (Like a bishop on the chessboard).
  • If the current position of the pointer is at any other place than the above two, then the next possible step can be in any way like a knight moves on a chessboard but the row and column of the new position > row and column of the old position.

Examples: 

Input: N = 2 
Output:
Explanation: 
The three possible ways are: 
{0-0} – {0-1} – {1-1} 
{0-0} – {1-0} – {1-1} 
{0-0} – {1-1} 
Input:
Output: 18 
Explanation: 
The possible ways are: 
{0-0} – {2-1} – {2-2} 
{0-0} – {1-2} – {2-2} 
{0-0} – {0-1} – {2-2} 
{0-0} – {0-1} – {0-2} – {1-2} – {2-2} 
{0-0} – {0-1} – {0-2} – {2-2} 
{0-0} – {0-1} – {1-1} – {2-2} 
{0-0} – {0-1} – {2-1} – {2-2} 
{0-0} – {0-2} – {1-2} – {2-2} 
{0-0} – {0-2} – {2-2} 
{0-0} – {1-0} – {2-2} 
{0-0} – {1-0} – {1-1} – {2-2} 
{0-0} – {1-0} – {1-2} – {2-2} 
{0-0} – {1-0} – {2-0} – {2-1} – {2-2} 
{0-0} – {1-0} – {2-0} – {2-2} 
{0-0} – {2-0} – {2-1} – {2-2} 
{0-0} – {2-0} – {2-2} 
{0-0} – {1-1} – {2-2} 
{0-0} – {2-2} 

Approach: The idea is to recursively check for every possible case whether it reaches the end of the Square Matrix or not. To do this, a recursive function is defined which takes the current position and the last position as the input parameters. The following are the conditions of the recursive function:  

  • Base Case: The base case of the function is to check if the current pointer has reached the bottom right position in the Square Matrix. If reached, then the counter count which keeps the note of the total number of ways is incremented. If the last position cannot be reached, then the function should be stopped and shouldn’t be called for the next iteration. This is implemented as:
if (cr == er && cc == ec) {
    count++;
    return;
}

if (cr > er || cc > ec) {
   return;
}
  • Recursive Case: Given that three types of traversals are possible. Therefore, the recursive case is to recursively call the function by checking if the current position. If the current position is at the edges of the Square Matrix, then the pointer can be moved only horizontally or vertically. And for every subsequent vertical traversal, two more choices of horizontal and vertical traversals are possible. Therefore, in order to keep a track of the total number of ways, both of these are called recursively in a loop.
for (int i = 1; i <= er; i++) {
    if (cc == 0 || cr == 0 
       || cr == er || cc == ec) {
        chessboard1(cr, cc + i, er, ec);
    }
}

for (int i = 1; i <= er; i++) {
    if (cc == 0 || cr == 0 
       || cr == er || cc == ec) {
        chessboard1(cr + i, cc, er, ec);
    }
}
  • Apart from this, if the current pointer is at the diagonals, then the pointer can move diagonally. However, for every subsequent diagonal traversal, another set of subsequent diagonals are possible. Therefore, in order to keep a track of the total number of ways, a for loop is used to recursively call the function. 
     
for (int i = 1; i <= er; i++) {
    if (cr == cc || cr + cc == er) {
        chessboard1(cr + i, cc + i,
                    er, ec);
    }
}
  • If none of the above cases is satisfied, then the next position can be moved in such a way that:
    • new row is > old row.
    • new column > old column.
  • After executing the above function, the value stored in the count variable is the total number of ways to traverse the Square Matrix.

Below is the implementation of the above approach:
 



C++




// C++ program to count the number
// of ways to traverse the given
// N x N Square Matrix recursively
#include<bits/stdc++.h>
using namespace std;
   
// Variable to store the
// total number of ways to
// traverse the Square Matrix
     
// Function to recursively
// count the total number
// of ways to traverse the
// given N x N Square Matrix
void chessboard1(int cr, int cc,
                 int er, int ec,
                 int& count)
{
 
  // If the last index has been
  // reached, then the count is
  // incremented and returned
  if (cr == er && cc == ec)
  {
    count++;
    return;
  }
 
  // If the last index cannot
  // be reached
  if (cr > er || cc > ec)
  {
    return;
  }
 
  // If the current position is
  // neither on the edges nor
  // on the diagonal, then the
  // pointer moves like a knight
  chessboard1(cr + 2, cc + 1,
              er, ec,count);
  chessboard1(cr + 1, cc + 2,
              er, ec,count);
 
  // If the pointer is on the
  // edges of the Square Matrix
  // next move can be horizontal
  // or vertical
 
  // This for loop is used to include
  // all the horizontal traversal cases
  // recursively
  for (int i = 1; i <= er; i++)
  {
    if (cc == 0 || cr == 0||
        cr == er || cc == ec)
    {
      chessboard1(cr, cc + i,
                  er, ec,count);
    }
  }
 
  // This for loop is used to include
  // all the vertical traversal cases
  // recursively
  for (int i = 1; i <= er; i++)
  {
    if (cc == 0 || cr == 0||
        cr == er || cc == ec)
    {
      chessboard1(cr + i, cc,
                  er, ec,count);
    }
  }
 
  // If the pointer is on the
  // diagonal of the Square Matrix
  // next move is also diagonal.
  // For loop is used to include
  // all the diagonal traversal cases.
  for (int i = 1; i <= er; i++)
  {
    if (cr == cc || cr + cc == er)
    {
      chessboard1(cr + i,  cc + i, 
                  er, ec, count);
    }
  }
}
   
// Driver code
int main()
{
  int N = 3;
  int count=0;
  chessboard1(0, 0, N - 1, N - 1,count);
  cout<<count;
}
 
// This code is contributed by chahattekwani71

Java




// Java program to count the number
// of ways to traverse the given
// N x N Square Matrix recursively
 
public class GFG {
 
    // Variable to store the total number
    // of ways to traverse the Square Matrix
    static int count = 0;
 
    // Function to recursively
    // count the total number
    // of ways to traverse the
    // given N x N Square Matrix
    public static void chessboard1(
        int cr, int cc,
        int er, int ec)
    {
 
        // If the last index has been reached, then
        // the count is incremented and returned
        if (cr == er && cc == ec) {
            count++;
            return;
        }
 
        // If the last index cannot be reached
        if (cr > er || cc > ec) {
            return;
        }
 
        // If the current position is neither
        // on the edges nor on the diagonal,
        // then the pointer moves
        // like a knight
        chessboard1(cr + 2, cc + 1, er, ec);
        chessboard1(cr + 1, cc + 2, er, ec);
 
        // If the pointer is on the
        // edges of the Square Matrix
        // next move can be horizontal or vertical
 
        // This for loop is used to include all the
        // horizontal traversal cases recursively
        for (int i = 1; i <= er; i++) {
            if (cc == 0 || cr == 0
                || cr == er || cc == ec) {
                chessboard1(cr, cc + i, er, ec);
            }
        }
 
        // This for loop is used to include all the
        // vertical traversal cases recursively
        for (int i = 1; i <= er; i++) {
            if (cc == 0 || cr == 0
                || cr == er || cc == ec) {
                chessboard1(cr + i, cc, er, ec);
            }
        }
 
        // If the pointer is on the
        // diagonal of the Square Matrix
        // next move is also diagonal.
        // For loop is used to include
        // all the diagonal traversal cases.
        for (int i = 1; i <= er; i++) {
            if (cr == cc
                || cr + cc == er) {
 
                chessboard1(cr + i,
                            cc + i,
                            er, ec);
            }
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        int N = 3;
        chessboard1(0, 0, N - 1, N - 1);
        System.out.println(count);
    }
}

C#




// C# program to count the number
// of ways to traverse the given
// N x N Square Matrix recursively
using System;
 
class GFG{
 
// Variable to store the total number
// of ways to traverse the Square Matrix
static int count = 0;
 
// Function to recursively
// count the total number
// of ways to traverse the
// given N x N Square Matrix
public static void chessboard1(int cr, int cc,
                               int er, int ec)
{
     
    // If the last index has been reached, then
    // the count is incremented and returned
    if (cr == er && cc == ec)
    {
        count++;
        return;
    }
 
    // If the last index cannot be reached
    if (cr > er || cc > ec)
    {
        return;
    }
 
    // If the current position is neither
    // on the edges nor on the diagonal,
    // then the pointer moves
    // like a knight
    chessboard1(cr + 2, cc + 1, er, ec);
    chessboard1(cr + 1, cc + 2, er, ec);
 
    // If the pointer is on the edges
    // of the Square Matrix next move
    // can be horizontal or vertical
 
    // This for loop is used to include all the
    // horizontal traversal cases recursively
    for(int i = 1; i <= er; i++)
    {
        if (cc == 0 || cr == 0 ||
           cr == er || cc == ec)
        {
            chessboard1(cr, cc + i, er, ec);
        }
    }
     
    // This for loop is used to include all the
    // vertical traversal cases recursively
    for(int i = 1; i <= er; i++)
    {
        if (cc == 0 || cr == 0 ||
           cr == er || cc == ec)
        {
            chessboard1(cr + i, cc, er, ec);
        }
    }
     
    // If the pointer is on the
    // diagonal of the Square Matrix
    // next move is also diagonal.
    // For loop is used to include
    // all the diagonal traversal cases.
    for(int i = 1; i <= er; i++)
    {
        if (cr == cc || cr + cc == er)
        {
            chessboard1(cr + i, cc + i,
                        er, ec);
        }
    }
}
 
// Driver code
public static void Main(string[] args)
{
    int N = 3;
     
    chessboard1(0, 0, N - 1, N - 1);
     
    Console.Write(count);
}
}
 
// This code is contributed by rutvik_56

Javascript




<script>
 
// Javascript program to count the number
// of ways to traverse the given
// N x N Square Matrix recursively
 
   
// Variable to store the
// total number of ways to
// traverse the Square Matrix
let count = 0
     
// Function to recursively
// count the total number
// of ways to traverse the
// given N x N Square Matrix
function chessboard1(cr, cc, er, ec)
{
 
  // If the last index has been
  // reached, then the count is
  // incremented and returned
  if (cr == er && cc == ec)
  {
    count++;
    return;
  }
 
  // If the last index cannot
  // be reached
  if (cr > er || cc > ec)
  {
    return;
  }
 
  // If the current position is
  // neither on the edges nor
  // on the diagonal, then the
  // pointer moves like a knight
  chessboard1(cr + 2, cc + 1,
              er, ec);
  chessboard1(cr + 1, cc + 2,
              er, ec);
 
  // If the pointer is on the
  // edges of the Square Matrix
  // next move can be horizontal
  // or vertical
 
  // This for loop is used to include
  // all the horizontal traversal cases
  // recursively
  for (let i = 1; i <= er; i++)
  {
    if (cc == 0 || cr == 0||
        cr == er || cc == ec)
    {
      chessboard1(cr, cc + i, er, ec);
    }
  }
 
  // This for loop is used to include
  // all the vertical traversal cases
  // recursively
  for (let i = 1; i <= er; i++)
  {
    if (cc == 0 || cr == 0||
        cr == er || cc == ec)
    {
      chessboard1(cr + i, cc,er, ec);
    }
  }
 
  // If the pointer is on the
  // diagonal of the Square Matrix
  // next move is also diagonal.
  // For loop is used to include
  // all the diagonal traversal cases.
  for (let i = 1; i <= er; i++)
  {
    if (cr == cc || cr + cc == er)
    {
      chessboard1(cr + i, cc + i, er, ec);
    }
  }
}
   
 
// Driver Code
let N = 3;
chessboard1(0, 0, N - 1, N - 1,count);
document.write(count);
 
// This code is contributed by jana_sayantan.
</script>

Output :

18

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