Simple ruby solution with recursion

• This code follows your thought process when you are asked to generate the answer manually by writing it down on a piece of paper. We keep appending parenthesis following two main rules. For each parenthesis that we want to insert we check if

1. we have utilized all the open parenthesis (= n). If we have, we continue by only appending ')' to our solution.
2. if we haven't then
1) if each '(' is matched with a ')' , we continue by appending a '('. We cannot add a ')' here as that would break the valid parenthesis rule.
2) if not, we have the option to add either a '(' or ')'. We follow through by appending each of the options and continuing towards generating the result.

Finally, we have a valid result when we have utilized all the ')' . We save the result and return.
Below is the code.

``````def generate_parenthesis(n)
result = []
insert_p(result,'', n, 0, 0)
result
end

def insert_p(res, ele, n, num_open_p, num_closed_p)
if num_closed_p == n
res.push(ele)
return
end
if (num_open_p == n)
insert_p(res, ele + ')', n, num_open_p, num_closed_p+1)
else
if (num_open_p == num_closed_p)
insert_p(res, ele + '(', n, num_open_p+1, num_closed_p)
else
insert_p(res, ele + '(', n, num_open_p+1, num_closed_p)
insert_p(res, ele + ')', n, num_open_p, num_closed_p+1)
end
end
end
``````

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