Parentheses pairing ({}[]()<>) issue

Question

I want to be able to pair up all parentheses in a string, if they aren't paired then then they get their index number and False. It seems like it is repeating some values over and over, i.e cl == pop[1]. I have tried to see where the problem is but I can't see it no matter how hard I try. So I'm asking if anyone help me to locate the error and maybe even improve my code ;)

def check_parentheses(string):
    pending = 0
    brackets = []
    '''Checks if parens are paired, otherwise they are bad.'''
    parenstack = collections.deque()
    for ch in string:
        if ch in lrmap:
            try:
                cl = string.index(ch, pending)
                pending = cl + 1

            except:
                cl = False

        if ch in lparens:
            parenstack.append([ch, cl])
            print parenstack

        elif ch in rparens:
            try:
                pop = parenstack.pop()

                if lrmap[pop[0]] != ch:
                    print 'wrong type of parenthesis popped from stack',\
                    pop[0], ch, pop[1], cl

                    brackets.append([pop[1], False])
                    brackets.append([cl, False])
                else:
                    brackets.append([pop[1], cl])

            except IndexError:
                print 'no opening parenthesis left in stack'
                brackets.append([cl, False])

    # if we are not out of opening parentheses, we have a mismatch
    for p in parenstack:
        brackets.append([p[1],False])
    return brackets

Answer 1

The below code will display the missing parentheses and the no of times missing in the given string.

from collections import Counter

def find_missing(str):
    stack1 = []
    stack2 = []
    result = []
    res_dict = {}
    open_set = '<[{('
    closed_set = '>]})'
    a = list(str)
    for i in a:
        if i in open_set:
            stack1.append(i)
        elif i in closed_set:
            stack2.append(i)
    dict1 = Counter(stack1)
    dict2 = Counter(stack2)
    print(dict1)
    print(dict2)
    for i in open_set:
        if dict1[i] > dict2[closed_set[open_set.index(i)]]:
            res_dict[closed_set[open_set.index(i)]] = dict1[i] - dict2[closed_set[open_set.index(i)]]
            result.append(closed_set[open_set.index(i)])
    for i in closed_set:
        if dict2[i] > dict1[open_set[closed_set.index(i)]]:
            res_dict[open_set[closed_set.index(i)]] = dict2[i] - dict1[open_set[closed_set.index(i)]]
            result.append(open_set[closed_set.index(i)])
    return res_dict
    # return result

if __name__ == '__main__':
    str1 = '{This ((()bracket {[function]} <<going> crazy}'
    x = find_missing(str1)
    if len(x) > 0:
        print("Imbalanced")
        print(x)
    else:
        print("Balanced")

Answer 2

BRACES = { '(': ')', '[': ']', '{': '}' }

def group_check(s):
    stack = []
    for b in s:
        c = BRACES.get(b)
        if c:
            stack.append(c)
        elif not stack or stack.pop() != b:
            return False
    return not stack

Answer 3

First we will scan the string from left to right, and every time we see an opening parenthesis we push it to a stack, because we want the last opening parenthesis to be closed first. (Remember the FILO structure of a stack!) Then, when we see a closing parenthesis we check whether the last opened one is the corresponding closing match, by popping an element from the stack. If it’s a valid match, then we proceed forward, if not return false. Code: https://gist.github.com/i143code/51962bfb1bd5925f75007d4dcbcf7f55