algorithm - Python program to check matching of simple parentheses

Question

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

I have seen examples here using the stack command which I haven't encountered yet. So I attempted a different approach. Can anyone tell me where I am going wrong?

def matched(str):
    ope = []
    clo = []
    for i in range(0,len(str)):
        l = str[i]
        if l == "(":
            ope = ope + ["("]
        else:
            if l == ")":
                clo = clo  + [")"]
            else:
                return(ope, clo)
    if len(ope)==len(clo):
        return True
    else:
        return False

The idea is to pile up "(" and ")" into two separate lists and then compare the length of the lists. I also had another version where I had appended the lists ope and clo with the relevant I which held either ( or ) respectively.

Answer 1

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 2

This checks whether parentheses are properly matched, not just whether there is an equal number of opening and closing parentheses. We use a list as a stack and push onto it when we encounter opening parentheses and pop from it when we encounter closing parentheses.

The main problem with your solution is that it only counts the number of parentheses but does not match them. One way of keeping track of the current depth of nesting is by pushing opening parentheses onto a stack and popping them from the stack when we encounter a closing parenthesis.

def do_parentheses_match(input_string):
    s = []
    balanced = True
    index = 0
    while index < len(input_string) and balanced:
        token = input_string[index]
        if token == "(":
            s.append(token)
        elif token == ")":
            if len(s) == 0:
                balanced = False
            else:
                s.pop()

        index += 1

    return balanced and len(s) == 0

Answer 3

My solution here works for brackets, parentheses & braces

openList = ["[","{","("]
closeList = ["]","}",")"]
def balance(myStr):
    stack= []
    for i in myStr:
        if i in openList:
            stack.append(i)
        elif i in closeList:
            pos = closeList.index(i)
            if ((len(stack) > 0) and (openList[pos] == stack[len(stack)-1])):
                stack.pop()
            else:
                return "Unbalanced"
    if len(stack) == 0:
        return "Balanced"
print balance("{[()](){}}") 

Answer 4

Most blatant error done by you is:

    if l == ")":
        clo = clo  + [")"]
    else:
        return(ope, clo)  # here

By using return, you exit from function when first char not equal to "(" or ")" is encountered. Also some indentation is off.

Minimal change which allows your code to run (although it won't give correct answers for all possible input strings) is:

def matched(str):
    ope = []
    clo = []
    for i in range(0,len(str)):
        l = str[i]
        if l == "(":
            ope = ope + ["("]
        elif l == ")":
            clo = clo  + [")"]
    if len(ope)==len(clo):
        return True
    else:
        return False

Answer 5

The problem with your approach is that you don't consider the order. Following line would pass: ))) (((. I'd suggest to keep the count of open and closed parenthesis:

• counter starts from 0
• every ( symbol increment counter
• every ) symbol decrements counter
• if at any moment counter is negative it is an error
• if at the end of the line counter is 0 - string has matching parenthesis

Answer 6

a = "((a+b)*c)+(b*a))"

li = list(a)
result = []

for i in range(0, len(a)):

    if a[i] == "(":
        result.append(i)
    elif a[i] == ")":
        if len(result) > 0:
            result.pop()
        else:
            li.pop(i)

for i in range(0, len(result)):
    li.pop(result[i])

print("".join(li))

Answer 7

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

Answer 8

In case u also need to find the position of the first mismatching bracket from left u can use the below code which also cover certain edge cases:

def isBalanced(expr):
    opening=set('([{')
    new=set(')]}{[(')
    match=set([ ('(',')'), ('[',']'), ('{','}') ])
    stack=[]
    stackcount=[]
    for i,char in enumerate(expr,1):
        if char not in new:
            continue
        elif char in opening:
            stack.append(char)
            stackcount.append(i)
        else:
            if len(stack)==0:
                print(i)
                return False
            lastOpen=stack.pop()
            lastindex=stackcount.pop()
            if (lastOpen, char) not in match:
                print (i)
                return False
    length=len(stack)
    if length!=0:
      elem=stackcount[0]
      print (elem)
    return length==0
string =input()
ans=isBalanced(string)
if ans==True:
    print("Success")

Answer 9

if "(" ,")" these two characters are not present then we don't want to return true or false just return no matching found. if matching found i just checking the count of both characters are same then return true, else return false

def matched(str):
   count1=0
   count2=1
   for i in str:
       if i =="(":
           count1+=1:
       elif i==")":
           count2+=1:
       else:
           print "no matching found for (,)"
   if count1==count2:
        return True
   else:
        return False

Answer 10

Simplest of all , though all of you guys have done good:

def wellbracketed(s):
    left=[]
    right=[]
    for i in range(0,len(s)):``
        if s[i]=='(':
            left=left+['(']
        elif s[i]==')':
            if len(left)!=0:
                right=right+[')']
        else:
            return False

    return(len(left)==len(right))

Answer 11

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 12

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 13

An alternative to check for balanced nested parentheses:

def is_balanced(query: str) -> bool:
    # Alternative: re.sub(r"[^()]", "", query)
    query = "".join(i for i in query if i in {"(", ")"})
    while "()" in query:
        query = query.replace("()", "")
    return not query


for stmt in [
    "(()()()())",    # True
    "(((())))",      # True
    "(()((())()))",  # True
    "((((((())",     # False
    "()))",          # False
    "(()()))(()",    # False
    "foo",           # True
    "a or (b and (c or d)",  # False
    "a or (b and (c or d))"  # True
    "a or (b and (c or (d and e)))",  # True
]:
    print(stmt)
    print("Balanced:", is_balanced(stmt))
    print()

It works by:

1. Removing everything but parentheses
2. Recursively remove innermost parentheses pairs
3. If you're left with anything besides the empty string, the statement is not balanced. Otherwise, it is.

Answer 14

if the parenthesis sequence is not an issue (strings like )( ) this code is faster :

def matched_parenthesis(s):
    return s.count('(') == s.count(')')

Tested with 15KB string, it is ~20μs v.s. 1ms iterating over the whole string.

And for me the order is not an issue as the underlying protocol guaranties that the string is well-formed.

Answer 15

input_str = "{[()](){}}"
strblance=""

for i in input_str:

    if not strblance:
        strblance = strblance+i
    elif (i is '}' and strblance[len(strblance)-1] is '{') \
        or ( i is']'and strblance[len(strblance)-1] is '[') \
        or ( i is ')'and strblance[len(strblance)-1] is '('):
            strblance = strblance[:len(strblance)-1]
    else:
        strblance = strblance+i

if not strblance:

    print ("balanced")

else:

    print ("Not balanced")

Answer 16

More advanced example in which you additionally need to check a matching of square brackets '[]' and braces '{}' pars.

string = '([]{})'

def group_match(string):

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

    list_ = []

    for index, item in enumerate(string):
        if item in d.values():
            list_.append(item)

        elif (item in d.keys()) and (d.get(item) in list_):
            list_.pop()

    return len(list_) == 0

Answer 17

The simplest code ever!!

def checkpar(x):
    while len(''.join([e for e in x if e in "()"]).split('()'))>1: x=''.join(x.split('()'))
    return not x

Answer 18

you can check this code.
This code don't use stack operations.

def matched(s):
  count = 0
  for i in s:
    if i is "(":
      count += 1
    elif i is ")":
      if count != 0:
        count -= 1
      else:
        return (False)

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

Answer 19

    #function to check if number of closing brackets is equal to the number of opening brackets
    #this function also checks if the closing bracket appears after the opening bracket
    def matched(str1):
        if str1.count(")")== str1.count("("):
            p1=str1.find("(")
            p2=str1.find(")")
            if p2 >= p1:
                str1=str1[p1+1:p2]+ str1[p2+1:]
                if str1.count(")")>0 and str1.count("(")>0:
                    matched(str1)
                return True
            else:
                return False
        else:
            return False

    matched(str1)

Answer 20

parenthesis_String = input("Enter your parenthesis string")
parenthesis_List = []
for p in parenthesis_String:
    parenthesis_List.append(p)
print(parenthesis_List)

if len(parenthesis_List)%2 != 0:
    print("Not Balanced Wrong number of input")

for p1 in parenthesis_List:
    last_parenthesis = parenthesis_List.pop()
    print(last_parenthesis)
    if (p1 == '{' and last_parenthesis == '}' or p1 == '[' and last_parenthesis == ']' or p1 == '(' and last_parenthesis == ')'):
        print("Balanced")
    else:
        print("Not balanced")

Answer 21

A little different one.

expression = '{(){({)}}'
brackets = '[](){}'
stack  = []
balanced = False
for e in expression:
    if e in brackets and stack: # Popping from the stack if it is closing bracket
        if stack [-1] == brackets[brackets.index(e)-1]:
            stack.pop()
            balanced = True
            continue # it will go to the new iteration skipping the next if below
    if e in brackets: # Push to stack if new bracket in the expression
        stack .append(e)
        balanced = False
balanced = 'Balanced' if balanced and not stack  else 'Unbalanced'
print(balanced, stack)

Answer 22

just modified Henry Prickett-Morgan's code a little bit to handle it more sensibly, namely taking into account that the number of "(" matches that of ")" but string starts with ")" or ends with "(" which are apparently not right.

def ValidParenthesis(s):
    count = 0
    if s[0] == ')' or s[-1] == '(':
      return False
    else:
      for c in s:
        if c == '(':
          count += 1
        elif c == ')':
          count -= 1
        else:
          continue
      return count == 0

Answer 23

The best way to understand this snippet is to follow along with all kind of scenarios.

in_data = ['{','[','(']
out_data = ['}',']',')']

def check_match(statements):
    stack = []
    for ch in statements:
        if ch in in_data:
            stack.append(ch)
        if ch in out_data:
            last = None
            if stack: 
                last = stack.pop()
            if last is '{' and ch is '}':
                continue
            elif last is '[' and ch is ']':
                continue
            elif last is '(' and ch is ')':
                continue
            else:
                return False
    if len(stack) > 0:
        return False
    else:
        return True

print(check_match("{www[eee}ee)eee"))
print(check_match("(ee)(eee[eeew]www)"))
print(check_match("(ss(ss[{ss}]zs)zss)"))
print(check_match("([{[[]]}])"))

Answer 24

def matched(str):
    braces = {"{": "}", "(": ")", "[": "]"}
    stack = []
    for c in str:
        if c in braces.keys():
            stack.append(c)
        elif c in braces.values():
            if not stack:
                return False
            last_brace = stack.pop()
            if braces[last_brace] != c:
                return False
    if stack:
        return False
    return True

print(matched("()"))
>> True
print(matched("(}"))
>> False
print(matched("}{"))
>> False
print(matched("}"))
>> False
print(matched("{"))
>> False
print(matched("(ff{fgg} [gg]h)"))
>> True

Answer 25

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 26

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 27

Although I'm not proposing a fix to your implementation, I suggest a cleaner and more pythonic version of the @kreld solution:

def check_parentheses(expr):
    s = []
    for c in expr:
        if c in '(':
            s.append(c)
        elif c in ')':
            if not len(s):
                break
            else:
                s.pop()
    else:
        return not len(s)
    return False

# test -----------------------------------------------------------------
test_expr = [')(', '(()', '())', '(', ')', '((', '))', '(()())', '(())',
             '()', '()(())']
for i, t in enumerate(test_expr, 1):
    print '%i\t%s\t%s' % (i, t, check_parentheses(t))

# output ---------------------------------------------------------------
1       )(      False
2       (()     False
3       ())     False
4       (       False
5       )       False
6       ((      False
7       ))      False
8       (()())  True
9       (())    True
10      ()      True
11      ()(())  True