All possiblities of balanced Parenthesis

print all balanced parenthesis of size "n".
[ex:- for 1: (), for 2: ()(), (()); like this]

Printing balance paranthesis, need dynamically updating your options.
Such as initially only openeing brace can be choosen and closing braces are not present.
You choose one opening brace and add into your result, this will decrease the number of opening
brace but increase the closing brace count by one.

In normal permuations program, we try with every element in set with every other element.
So, here are only two choices given one opening brance and closing brace with n numbers.
So, everytime we try to choose, we check for the opening brace count available and closing brance count available.

I am posting my program here,


#include
using namespace std;
static int permcnt;
void print( char out[], int cnt)
{
//Print all the index from 0 to cnt-1 of out
permcnt++;
}

void generate ( char in[], int i, int opara,int cpara)
{
int j;
if ( opara == 0 && cpara == 0)
{
print(in, i);
return;
}
for ( j = 0; j < 2;j++)
{
if ( j == 0 && opara != 0 )
{
in[i] = '{';
generate(in,i+1,opara-1,cpara+1);
}
else if ( j == 1 && cpara != 0 )
{
in[i] = '}';
generate(in,i+1,opara,cpara-1);
}
}
}
int main()
{
int num;
cout<<"Enter the size:";
cin>>num;
char *in = new char [num*2];
in[0] = '{';
generate(in,1,num-1,1);
//Print permcnt to know total permutations
return 0;
}