Given a N X N matrix (M) filled with 1, 0, 2, 3. The task is to find whether there is a path possible from source to the destination while traversing through blank cells only. You can traverse up, down, right, and left.

Given a N X N matrix (M) filled with 1, 0, 2, 3. The task is to find whether there is a path possible from source to the destination while traversing through blank cells only. You can traverse up, down, right, and left.

1. A value of cell 1 means Source.
2. A value of cell 2 means Destination.
3. A value of cell 3 means Blank cell.

A value of cell 0 means Blank Wall.

Note: there are only a single source and a single destination

For each test case in a new line print 1 if the path exists from source to destination else print 0.

#include <bits/stdc++.h>

using namespace std;

#define N 21

int check(int i,int j,int n,int a[N][N])

{

// if i and j outoff index or a[i][j] is zero then return 0

    if(i<0||j<0||i>=n||j>=n||a[i][j]==0) 

    return 0;

// if a[i][j] is destination when its value is 2

    if(a[i][j]==2)

    return 1;

    a[i][j]=0;

// here return all combination of  check value with or logic

   return  check(i,j+1,n,a) || check(i,j-1,n,a)||check(i-1,j,n,a)||check(i+1,j,n,a);

    

}

int main() {

//code

int t;

cin>>t;

while(t--)

{

    int n,x,y,d,m,l;

    cin>>n;

    int a[N][N];

    

    for(int i=0;i<n;++i)

    {

        for(int j=0;j<n;++j)

        {

            cin>>a[i][j];

// finding the source index when a[i][j] is 1

            if(a[i][j]==1)

            {

                x=i,y=j;

            }

            

        }

    }

    

    cout<<check(x,y,n,a)<<endl;

   

}

return 0;

}