Determine if brackets are balanced. Not getting expected output

Question

# Stack Data Structure.

I just started leaning Data structures and algorithms. After learning Stack I wrote this algorithm for finding that brackets are balanced or not. Bot I don't know why I'm not getting correct output.

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

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

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

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

    def peek(self):
        if not self.is_empty():
            return self.items[-1]
    
    def get_stack(self):
        return self.items 

I wrote this is_match function which is going to check that given two values are equal or not.

def is_match(a,b):
    if a == b :
        return True
    else:
        return False

Here is the main code started for determining that given barckets are balanced or not.

# DETERMINE IF BRACKETS ARE BALANCED

def is_paren_balanced(paren_string):
    s = Stack()
    is_balanced = True
    index = 0

    while index < len(paren_string) and is_balanced:
        paren = paren_string[index]
        if paren in "([{":
            s.push(paren)
        else:
            if s.is_empty():
                is_balanced = False
                break
            else:
                top = s.pop()
                if not is_match(top, paren):
                    is_balanced = False
                    break
        index += 1

    if s.is_empty() and is_balanced:
        return True
    else:
        return False

I called out is_paren_balanced function and passed some parameter for testing my code, as I have passed balanced bracket but still it is returning output as False.

result = is_paren_balanced("()")
print(result)

Answer 1

Your is_match function is wrong, it should match each opening parenthesis to the corresponding closing one.

def is_match(a,b):
    if (a == "(" and b == ")") or (a == "[" and b == "]") or (a == "{" and b == "}") :
        return True
    else:
        return False

You can write the if statement in cleaner fashion, but that gives you a basic idea.