// If N=2 , means their must be 2 opening and 2 closing brackets
// you can not place a closing at 1st place
// you have two cases
// place a opening if no. of opening less then N
// place a closing if no. of closing less then opening ;
// we have to print the all possible combinations
#include<iostream>
using namespace std;
int noOfparenthese(int i,int o,int c , int n){
// base case
int t;
if (i==2*n)
{
// t=t+1;
return 1;
}
// cout<<"hello"<<endl;
// recursive case
if (o<n)
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