Python program to check matching of simple parentheses

Question

I am a Python newbie and I came across this exercise of checking whether or not the simple brackets \"(\", \")\" in a given string are matched evenly.

I have seen ex

Answer 1

You can do this in a couple of lines using accumulate (from itertools). The idea is to compute a cumulative parenthesis level going through the string with opening parentheses counting as level+1 and closing parentheses counting as level-1. If, at any point, the accumulated level falls below zero then there is an extra closing parenthesis. If the final level is not zero, then there is a missing closing parenthesis:

from itertools import accumulate
def matched(s):
    levels = list(accumulate((c=="(")-(c==")") for c in s))
    return all( level >= 0 for level in levels) and levels[-1] == 0

Answer 2

def parenthesis_check(parenthesis):
    chars = []
    matches = {')':'(',']':'[','}':'{'}
    for i in parenthesis:
        if i in matches:
            if chars.pop() != matches[i]:
                return False
        else:
            chars.append(i)
    return chars == []        

Answer 3

here's another way to solve it by having a counter that tracks how many open parentheses that are difference at this very moment. this should take care all of the cases.

def matched(str):
    diffCounter = 0
    length = len(str)
    for i in range(length):
        if str[i] == '(':
            diffCounter += 1
        elif str[i] == ')':
            diffCounter -= 1
    if diffCounter == 0:
        return True
    else:
        return False

Answer 4

foo1="()()())("  

def bracket(foo1):
    count = 0
    for i in foo1:
        if i == "(":
           count += 1
        else:
           if count==0 and i ==")":
               return False
           count -= 1

   if count == 0:
       return True
   else:
       return False

bracket(foo1)

Answer 5

A very slightly more elegant way to do this is below. It cleans up the for loop and replaces the lists with a simple counter variable. It also returns false if the counter drops below zero so that matched(")(") will return False.

def matched(str):
    count = 0
    for i in str:
        if i == "(":
            count += 1
        elif i == ")":
            count -= 1
        if count < 0:
            return False
    return count == 0

Answer 6

this code works fine

def matched(s):
  p_list=[]
  for i in range(0,len(s)):
    if s[i] =='(':
      p_list.append('(')
    elif s[i] ==')' :
      if not p_list:
        return False
      else:
        p_list.pop()
  if not p_list:
    return True
  else:
    return False