# HackerEarth: Batman and Tick-tack-toe

This is a basic implementaation problem but need to think critically and I used editorial + previous solved code to understand the logic

code:

```#include<iostream>
using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
char s[4][4];
int i,j;
bool ans=false;
for(i=0; i<4; i++)
{
for(j=0; j<4; j++)
{
cin>>s[i][j];
}
}
//for column match
for(i=0;i<2;i++)
{
for(j=0;j<4;j++)
{
if(s[i][j]=='x'&&s[i+1][j]=='x'&&s[i+2][j]=='.')
ans=true;
if(s[i][j]=='x'&&s[i+1][j]=='.'&&s[i+2][j]=='x')
ans=true;
if(s[i][j]=='.'&&s[i+1][j]=='x'&&s[i+2][j]=='x')
ans=true;
}
}
//for row match
for(i=0;i<4;i++)
{
for(j=0;j<2;j++)
{
if(s[i][j+1]=='x'&&s[i][j]=='x'&&s[i][j+2]=='.')
ans=true;
if(s[i][j+1]=='x'&&s[i][j]=='.'&&s[i][j+2]=='x')
ans=true;
if(s[i][j+1]=='.'&&s[i][j]=='x'&&s[i][j+2]=='x')
ans=true;
}
}
//for diagonal match
for(i=0;i<2;i++){
for(j=0;j<2;j++){
if(s[i][j]=='x'&&s[i+1][j+1]=='x'&&s[i+2][j+2]=='.')
ans=true;
if(s[i][j]=='x'&&s[i+1][j+1]=='.'&&s[i+2][j+2]=='x')
ans=true;
if(s[i][j]=='.'&&s[i+1][j+1]=='x'&&s[i+2][j+2]=='x')
ans=true;
}
}
//reverse diagonal match
for(i=0;i<2;i++)
{
for(j=2;j<4;j++)
{
if(s[i][j]=='x' && s[i+1][j-1]=='x'&&s[i+2][j-2]=='.')
ans=true;
if(s[i][j]=='x' && s[i+1][j-1]=='.'&&s[i+2][j-2]=='x')
ans=true;
if(s[i][j]=='.'&&s[i+1][j-1]=='x'&&s[i+2][j-2]=='x')
ans=true;
}
}
if(ans)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}
```

It would be a great help, if you support by sharing :)
Author: zakilive