Question

I am trying to solve one of the “classic” dynamic programming problems. The problem is – Given a number as input , generate possible nested conditions.

Edit: as pointed out by temp below , i am going to first try and sort this out using recursion and then try with dynamic programming.

ie. if n = 3

O/p
((()))
()()()
(())()
()(())
(()())

My approach to the problem is based on two rules.

1. Add a “(” brace if max number (
   here 3 ) is not reached and number
   of left braces is less than or equal
   to number of right braces.
2. Add a right brace only if max number
   is
   not reached and number of right
   braces is less than number of left
   braces.

In theory they sound right ,but it falls face flat on the below source. Please excuse the hard-coding.

Edit : I have made some modifications and moved an inch closer to the solution 🙂

 #include <iostream>
#include <vector>
#include <string>

using namespace std;

void printPar(int l,int r,string s)
{
    if(l > 3 || r > 3 || r >l)
        return;



    if(l==3 && r==3)
    {
        cout<<s<<endl;
        return;
    }
    else
    {

         if((l<3))
         {
            s+="<";
            l = l+1;
            printPar(l,r,s);
         }
          if(r<3 && r < l)
         {
             s+=">";
             r = r+1;
            printPar(l,r,s);
         }
       //  cout<<"Exiting "<<l<<" & "<<r<<" "<<s<<endl;
    }
}




int main()
{
    string s;
    printPar(0,0,s);
    return 0;
}

Debug:

<<<>>>
<<<>>>
<<><>>
<<><>>
<><<>>
<><<>>
<><><>
<><><>

I understand why there are duplicate values in the list. i.e once the function is called using recursion and ends up following the next branch on execution. The second print is due to the function it self falling on the second branch. Is there any way to handle this ? I really do not want to go the global-set route.

Also , in my head this code should print (())() – Yet it does not 🙁

Can some one please point out the error ?

Thanks!

I know the condition needs some tweaking , but i have been staring at this endlessly. Halp!

Answer 1

Right now, your solution doesn’t take advantage of dynamic programming. If you want to use DP here, you’ll need to think about the problem recursively. Fortunately, there’s a great recursive formulation to this problem:

1. There is only one way to make balanced parentheses from no parentheses, which is to have no parentheses at all.
2. If you have n + 1 pairs of parentheses to balance, then you can generate all balanced parentheses as follows: for all i from 0 to n, construct every string of the form (X)Y, where X is a string of i balanced parentheses and Y is a string of n – i balanced parentheses.

The beauty of this setup is that to compute all strings of n + 1 balanced parentheses, you only need to know how to make balanced parentheses for 0, 1, 2, …, n + 1. Consequently, you can solve this problem by iterative constructing solutions for n = 0, 1, 2, …, etc. and reusing the results you produced at earlier steps.

Hope this helps!