Balanced Parentheses Problem Using Stack in DSA Python - Build from Scratch

Balanced Parentheses Problem Using Stack

📖 Scenario: Imagine you are building a simple code editor that checks if the parentheses in a line of code are balanced. Balanced parentheses mean every opening bracket has a matching closing bracket in the correct order.

🎯 Goal: You will create a program that uses a stack to check if a given string of parentheses is balanced or not.

📋 What You'll Learn

Create a stack using a list

Use a loop to process each character in the input string

Push opening brackets onto the stack

Pop from the stack when a matching closing bracket is found

Check if the stack is empty at the end to decide if parentheses are balanced

💡 Why This Matters

🌍 Real World

Checking balanced parentheses is important in code editors and compilers to ensure code syntax is correct.

💼 Career

Understanding stacks and balanced parentheses is a common interview question and a fundamental concept in programming and software development.

1

Create the input string of parentheses

Create a variable called parentheses and set it to the string '(()())'.

DSA Python

# Create the input string called parentheses
# Your code here

Hint

Use single quotes to create the string exactly as shown.

2

Create an empty stack

Create a variable called stack and set it to an empty list [].

DSA Python

parentheses = '(()())'
# Create an empty list called stack
# Your code here

Hint

A stack can be created using an empty list in Python.

3

Use a loop to check parentheses using the stack

Use a for loop with variable char to iterate over parentheses. Inside the loop, if char is '(', append it to stack. If char is ')', pop from stack if it is not empty. If stack is empty when trying to pop, set a variable balanced to False and break the loop. After the loop, if stack is empty and balanced is not False, set balanced to True, else set it to False.

DSA Python

parentheses = '(()())'
stack = []
# Use a for loop to check each character in parentheses
# Your code here

Hint

Use stack.append() to add and stack.pop() to remove the last item.

Check if stack is empty before popping to avoid errors.

4

Print if the parentheses are balanced

Print "Balanced" if balanced is True, else print "Not Balanced".

DSA Python

parentheses = '(()())'
stack = []
balanced = None
for char in parentheses:
    if char == '(':  
        stack.append(char)
    elif char == ')':
        if stack:
            stack.pop()
        else:
            balanced = False
            break
if balanced is None:
    balanced = len(stack) == 0
# Print "Balanced" if balanced is True, else print "Not Balanced"
# Your code here

Hint

Use an if statement to print the correct message based on the balanced variable.