Print all valid combinations of n-pairs of parentheses

    For such recursive problems check if f(n) can be built from f(n-1)  , by inserting an instance of n = 1
    This is a hint for general recursive problem
   
    Example : n = 1 : ()
              n = 2 : ()() , (())
              n = 3 : ()()() , (())() , () (()) , (()()) , ((()))
             
    f(3) can be found from f(3-1) , i,.e  f(2) , by adding a single pair of parenthesis to f(3-1) inside every existing pair of parenthesis
   
    Eg. -  n = 2 : ()() -
                         (())()  // inserted after first left parenthesis
                         () (()) // inserted after second left parenthesis
                         ()()()  // inserted at beginning
                        
                    (()) -
                          (()())  // inserted after first left parenthesis
                          ((()))  // inserted after second left parenthesis
                          () (()) // inserted at beginning
     
      Since  () (()) occurs twice so use of set is required
       Placing a pair at end would reduce to previous case as already did
       eg -    (())()  inserted at end of (()) is same as inserted after first left parenthesis in ()() 


public   Set<String> generateParen(int remainder)
{
    //HashSet will automatically check that no same values are inserted in set
    Set<String> set = new HashSet<>();
    //base case
    if( remainder == 0 )
    {
        set.add("");
    }
   
    else
    {
        //recurse to f(n-1)
        Set<String> prev =  generateParen(remainder-1);
       
        for(Strring str : prev)
        {
            for( int i = 0 ; i < str.length() ; ++i)
            {
                //if a left parenthesis is found
                if(str.charAt(i) == '(')
                {
                    String s = insertParenAt(str, i );
                    set.add(s);
                }
            }
           
            //insert at beginning
            set.add("()" + str);
        }
       
   
    }

}

public String insertParenAt (String parent , int index)
{
    String left = parent.substring(0,index+1);
    String right = parent.substring(index+1 , parent.length());
    return left+ "()" + right ;
}