Balanced Parentheses Run. Do anyone know answer for this? I am getting wrong answer for two hidden cases.

Question

Parentheses are balanced, if all opening parentheses have their corresponding closing parentheses.

Given an expression as input, we need to find out whether the parentheses are balanced or not.
For example, "(x+y)*(z-2*(6))" is balanced, while "7-(3(2*9))4) (1" is not balanced.

The problem can be solved using a stack.
Push each opening parenthesis to the stack and pop the last inserted opening parenthesis whenever a closing parenthesis is encountered.
If the closing bracket does not correspond to the opening bracket, then stop and say that the brackets are not balanced.
Also, after checking all the parentheses, we need to check the stack to be empty -- if it's not empty, then the parentheses are not balanced.

Implement the balanced() function to return True if the parentheses in the given expression are balanced, and False if not.

Sample Input:
(a( ) eee) )

Sample Output:
False

Answer 1

Share your code and we can likely find cases it'll misbehave on.

One case that breaks for a lot of people is:
)

Make sure your code correctly indicates "False" for a single closing bracket. Your script shouldn't throw an exception or say "True".

Answer 2

Tejaswin, the following is a bit simpler and based on the observation that the stack of only "(" characters is nothing but a glorified counter.

def balanced(expression):
count = 0
for char in expression:
if char == "(":
count += 1
elif char == ")":
if count == 0:
return False
count -= 1
return count == 0

print(balanced(input()))

Answer 3

Tejaswin wrote, "I have attached the code below. ..."

Response:

This case should return True but instead you get False:
(1+1)

I changed one line below and commented it to pass all test cases:
def balanced(expression):
stack = []
for char in expression:
if char in ["("]:
stack.append(char)
elif char == ")": # replace else with this and your code should work.
if not stack:
return False
current_char = stack.pop()
if current_char == '(':
if char != ")":
return False

if stack:
return False
return True

print(balanced(input()))

Answer 4

Guys my code is failing in only one test case.
https://code.sololearn.com/coV3HUJR26xe/?ref=app

Answer 5

DENIZ MUSTAFAOGLU, ")(" should not pass but your code would. "()" should pass. There are an infinite number of other cases like "())(" which should fail too when your function returns True.

Answer 6

Test case 1 fail but test case 2 passed where is the mistake

#your code goes here

class Stack:
def __init__(self):
self.items = []

def isEmpty(self):
return self.items == []

def push(self, item):
self.items.append(item)

def pop(self):
return self.items.pop()

def peek(self):
return self.items[len(self.items)-1]

def size(self):
return len(self.items)

def balanced(expression):
a = Stack()
for i in expression:
if i == "(": a.push(i)
for i in expression:
if i == ")":
if a.size() == 0:
return False
else:
a.pop()
if a.size() == 0:
return True
else:
return False

print(balanced(input()))

Answer 7

## Correcr answer
def balanced(expression):
count = 0
for char in expression:
if char == "(":
count += 1
elif char == ")":
if count == 0:
return False
count -= 1
return count == 0

print(balanced(input()))

Answer 8

def balanced(expression):
#your code goes here
listt = []
for i in expression:
if i == '(' or i == ')':
listt.insert(0, i)
right = 0
left = 0
for j in listt:
if j == '(':
right+=1
if j == ')':
left+=1
if right==left and listt[0] == ')' and listt[-1] == '(':
return True
else:
return False

print(balanced(input()))

Answer 9

#Correct answer

def balanced(expression):
count = 0
for char in expression:
if char == "(":
count += 1
elif char == ")":
if count == 0:
return False
count -= 1
return count == 0

print(balanced(input()))

Answer 10

Why does not work for Test Case 7?

def balanced(expression):
a = expression.count("(")
b = expression.count(")")
if a == b:
return True
else:
return False

print(balanced(input()))

Answer 11

#Correct answer

def balanced(expression):
#your code goes here
l=[]
a=b=0
for i in expression:
if i=="(":
l.insert(0,i)
a+=1

elif i=="(":
l.insert(0,i)
b+=1

if l.pop(0)==")" and a==b:
return True
else:
return False

print(balanced(input()), end='')

Answer 12

The problem can be solved using a stack.

class Stack:
def __init__(self):
self.items = []

def isEmpty(self):
return self.items == []

def push(self, item):
self.items.append(item)

def pop(self):
return self.items.pop()

def peek(self):
return self.items[len(self.items)-1]

def size(self):
return len(self.items)

def balanced(expression):
a = Stack()
for char in expression:
if char == "(":
a.push(char)
elif char == ")":
if a.size() == 0:
return False
a.pop()
return a.size() == 0

print(balanced(input()))

Answer 13

n=input()
c=[]
for x in n:
if x=='(' or x==')':
c.append(x)

if len(c)==0 or len(c)%2==1 or c.count("(")!=c.count(")"):
print("False")
else:
while len(c)>2:
for y in c:
if y==")":
if c.index(y)==0:
print("False")
break
else:
c.remove(y)
c.remove(c[c.index(y)-1])
if c==[')', '(']:
print("False")
else:
print("True")

Answer 14

I have attached the code below. Check this. It is having some problem in two test cases..

def balanced(expression):
stack = []

for char in expression:
if char in ["("]:

stack.append(char)
else:

if not stack:
return False
current_char = stack.pop()
if current_char == '(':
if char != ")":
return False

if stack:
return False
return True

print(balanced(input()))