Parenthesis in a recursive way (Python)

Question

def paren(s, cnt=0):
    if s == '':
        return True
    if s[0] == '(':
        return paren(s[1:], cnt + 1)
    elif s[0] == ')':
        return paren(s[1:], cnt - 1)
    return cnt == 0

So this code works for all cases if there is the same number of "(" and ")".
But for example it doesn’t work for "))(( ".
how can I modify the code for this to work that for every opening bracket there is a closing one, then it returns True.

Asked By: Moh 0189

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Source

Answer 1

def paren(s):
    _s = s.replace('()','')
    if not _s:
        return True
    elif _s==s:
        return False
    return paren(_s)

print(paren(')()('))

Answered By: islam abdelmoumen

Answer 2

Check if at any point c < 0, and fix the return for when s == ''

def paren(s, cnt=0):
    if c < 0: return False
    elif s == '': return c == 0
    elif s[0] == '(':
        return paren(s[1:], cnt + 1)
    elif s[0] == ')':
        return paren(s[1:], cnt - 1)
    # here, there's a non-parentheses character. I'll assume you want to ignore these
    # if they're impossible, just remove this line
    return parent(s[1:], cnt)

Answered By: Samathingamajig

Answer 3

To check that for every opening bracket there is a closing bracket, you can simply check if the final value of cnt is 0. If it is, then it means that there was an equal number of opening and closing brackets, so the string is balanced. Here is one way you could modify the code to do this:

def paren(s, cnt=0):
if s == '':
    # If the string is empty, return True if the count is 0,
    # otherwise return False
    return cnt == 0
if s[0] == '(':
    # If the first character is an opening bracket, increment the count
    return paren(s[1:], cnt + 1)
elif s[0] == ')':
    # If the first character is a closing bracket, decrement the count
    return paren(s[1:], cnt - 1)
# If the first character is neither an opening nor closing bracket,
# just recurse on the rest of the string
return paren(s[1:], cnt)

This code should work for the example you gave, "))(( ". When it reaches the first closing bracket, cnt will be decremented to -1. When it reaches the next closing bracket, cnt will be decremented again to -2. When it reaches the first opening bracket, cnt will be incremented to -1. Finally, when it reaches the last opening bracket, cnt will be incremented again to 0. When the string is empty, cnt will be 0, so the function will return True.

Answered By: 0xkratos

Parenthesis in a recursive way (Python)

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