Generate Parenthesis - where am I going wrong

Question

So here's the problem. For a given n we need to generate all valid parenthesis.

For e.g, for n=2 the possible output will be ()(), (()))

So I tried solving using recursion and backtracking.

I take two variables, open_bracket and close_bracket along with a ds to store the parenthesis.

Here's my recursive diagram looks like

So with that I tried this:

function generatePArenthesis(n, open_bracket=0, close_bracket=0, ds=[]){
    if(open_bracket === n && close_bracket === n){
    console.log(ds.toString());
    return
  }
  
  if(open_bracket < n){
    ds.push('(');
    generatePArenthesis(n, open_bracket+ 1, close_bracket, ds);
  }
  
  if(close_bracket<open_bracket){
    ds.push(')');
    generatePArenthesis(n, open_bracket, close_bracket +1, ds);
  }
  
}

generatePArenthesis(2)

when I dry run this program for n=2 it does seem to produce the above recursive diagram. But doesn't give me the correct output. Where am I making mistake?

MBo · Accepted Answer · 2023-12-28 08:38:28Z

1

Restore ds after recursive calls:

function generatePArenthesis(n, open_bracket=0, close_bracket=0, ds=[]){
    if(open_bracket === n && close_bracket === n){
    console.log(ds.toString());
    return
  }
  
  if(open_bracket < n){
    ds.push('(');
    generatePArenthesis(n, open_bracket+ 1, close_bracket, ds);
    ds.pop();
  }
  
  if(close_bracket<open_bracket){
    ds.push(')');
    generatePArenthesis(n, open_bracket, close_bracket +1, ds);
        ds.pop();
  }
  
}

generatePArenthesis(3)

answered Dec 28, 2023 at 8:38

MBo

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4 Comments

ABGR Over a year ago

can you please explain why as in my recursive tree, I'm pushing into the existing content without clearing it and it still produces the output correctly.

MBo Over a year ago

When you first made (,(,),) in left big branch, ds contains 4 items and it is used in the right branch, so you add ( not to (,) as you wanted but to (,(,),)

MBo Over a year ago

You are using single instance of ds everywhere. You can create new copies in recursive calls (I don't know JS enough, but something like generatePArenthesis(n, open_bracket+ 1, close_bracket, ds +'(')

ABGR Over a year ago

That’s right. Like this […ds, ‘(‘]

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