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

Balanced Parentheses Problem Using Stack

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

🎯 Goal: Build a program that uses a stack to check if a string of parentheses is balanced or not.

📋 What You'll Learn

Create a string variable with a sequence of parentheses

Create an empty array to use as a stack

Use a loop to process each character in the string

Push opening parentheses onto the stack

Pop from the stack when a matching closing parenthesis is found

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

Print 'Balanced' if parentheses match correctly, otherwise print 'Not Balanced'

💡 Why This Matters

🌍 Real World

Checking balanced parentheses is important in code editors, compilers, and calculators to ensure expressions are correct.

💼 Career

Understanding stack usage and string processing is fundamental for software development, especially in parsing and syntax checking.

1

Create the parentheses string

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

DSA Javascript

# Create a string variable called parentheses with value '(()())'
# Your code here

Hint

Use const to create the string variable with the exact value.

2

Create an empty stack

Create an empty array called stack to use as a stack.

DSA Javascript

const parentheses = '(()())';
# Create an empty array called stack
# Your code here

Hint

Use [] to create an empty array for the stack.

3

Process each character with stack logic

Use a for loop with variable char to iterate over parentheses. Inside the loop, if char is '(', push it onto stack. If char is ')', pop from stack only if stack is not empty; otherwise, set a variable isBalanced to false and break the loop. Initialize isBalanced as true before the loop.

DSA Javascript

const parentheses = '(()())';
const stack = [];
// Initialize isBalanced as true
// Use a for loop to check each character
// Push '(' to stack
// Pop from stack for ')', set isBalanced false and break if stack empty
# Your code here

Hint

Use a for...of loop to check each character. Use stack.push() and stack.pop() to manage the stack.

4

Print if parentheses are balanced

After the loop, check if isBalanced is true and stack is empty. If both conditions are true, print 'Balanced'. Otherwise, print 'Not Balanced'.

DSA Javascript

const parentheses = '(()())';
const stack = [];
let isBalanced = true;
for (const char of parentheses) {
    if (char === '(') {
        stack.push(char);
    } else if (char === ')') {
        if (stack.length === 0) {
            isBalanced = false;
            break;
        } else {
            stack.pop();
        }
    }
}
// Print 'Balanced' if isBalanced is true and stack is empty, else print 'Not Balanced'
# Your code here

Hint

Use console.log() to print the result based on the conditions.

Practice

1. What is the main purpose of using a stack to solve the Balanced Parentheses Problem?

easy

A. To sort the parentheses in order

B. To count the total number of parentheses in the string

C. To reverse the string of parentheses

D. To keep track of opening parentheses and match them with closing ones

Solution

1. Step 1: Understand the role of stack in matching parentheses

   A stack helps store opening parentheses as they appear, so we can check if each closing parenthesis matches the last opened one.

2. Step 2: Explain why other options are incorrect

   Counting or reversing parentheses does not help check if they are balanced. Sorting changes order and breaks the original sequence.

3. Final Answer:

   To keep track of opening parentheses and match them with closing ones -> Option D

4. Quick Check:

   Stack tracks openings for matching [OK]

Hint: Stack stores openings to match closings quickly [OK]

Common Mistakes:

• Thinking stack counts parentheses instead of matching
• Assuming reversing string solves balance
• Confusing sorting with balancing

2. Which of the following is the correct way to push an element onto a stack in JavaScript?

easy

A. stack.add(element);

B. stack.push(element);

C. stack.insert(element);

D. stack.append(element);

Solution

1. Step 1: Recall JavaScript array method for adding elements at the end

   JavaScript arrays use push() to add elements to the end, which simulates stack push.

2. Step 2: Identify incorrect methods

   add(), insert(), and append() are not valid JavaScript array methods.

3. Final Answer:

   stack.push(element); -> Option B

4. Quick Check:

   Use push() to add to stack [OK]

Hint: Use push() to add elements to stack in JavaScript [OK]

Common Mistakes:

• Using add() or append() which are not JavaScript array methods
• Confusing push() with pop()
• Trying to use insert() which doesn't exist

3. What will be the output of the following JavaScript code snippet?

function isBalanced(s) {
  const stack = [];
  for (const char of s) {
    if (char === '(') stack.push(char);
    else if (char === ')') {
      if (stack.length === 0) return false;
      stack.pop();
    }
  }
  return stack.length === 0;
}

console.log(isBalanced('(()())'));
console.log(isBalanced('(()'));
console.log(isBalanced(')('));

medium

A. true, true, false

B. false, true, false

C. true, false, false

D. false, false, true

Solution

1. Step 1: Trace the function for each input string

   For '(()())': all parentheses match, stack ends empty, returns true.
   For '(()': one '(' remains unmatched, returns false.
   For ')(': first char is ')', stack empty, returns false.

2. Step 2: Confirm outputs match true, false, false

   Outputs are true, false, false respectively.

3. Final Answer:

   true, false, false -> Option C

4. Quick Check:

   Stack empty at end means balanced [OK]

Hint: Stack empty at end means balanced parentheses [OK]

Common Mistakes:

• Assuming unmatched openings are balanced
• Ignoring early unmatched closing parentheses
• Confusing true/false outputs

4. The following code is intended to check balanced parentheses but has a bug. What is the bug?

function checkBalance(str) {
  const stack = [];
  for (let i = 0; i < str.length; i++) {
    if (str[i] === '(') {
      stack.push('(');
    } else {
      if (stack.length === 0) return false;
      stack.pop();
    }
  }
  return true;
}

medium

A. It returns true even if stack is not empty after loop

B. It pushes closing parentheses instead of opening

C. It does not check for closing parentheses

D. It uses wrong loop syntax

Solution

1. Step 1: Analyze the return statement after loop

   The function returns true without checking if stack is empty, so leftover '(' are ignored.

2. Step 2: Explain why this causes incorrect results

   If there are unmatched '(' left in stack, string is unbalanced but function returns true incorrectly.

3. Final Answer:

   It returns true even if stack is not empty after loop -> Option A

4. Quick Check:

   Check stack.length === 0 before return true [OK]

Hint: Always check stack empty before returning true [OK]

Common Mistakes:

• Forgetting to check stack length after loop
• Returning true too early
• Ignoring unmatched opening parentheses

5. Given the string s = '{[()]}', which modification is needed in the stack-based balanced parentheses algorithm to correctly check this string?

hard

A. Add support for matching pairs of '()', '{}', and '[]' with stack push/pop logic

B. Ignore all characters except '(' and ')' during stack operations

C. Use two separate stacks for different types of brackets

D. Reverse the string before checking with the existing algorithm

Solution

1. Step 1: Understand the input string contains multiple bracket types

   The string has '()', '{}', and '[]' brackets, so the algorithm must handle all types.

2. Step 2: Explain how to handle multiple bracket types with one stack

   Push opening brackets onto stack, and on closing bracket, check if it matches the last opening bracket type before popping.

3. Final Answer:

   Add support for matching pairs of '()', '{}', and '[]' with stack push/pop logic -> Option A

4. Quick Check:

   One stack handles all bracket types [OK]

Hint: Match each closing bracket with last opening bracket type [OK]

Common Mistakes:

• Ignoring bracket types other than parentheses
• Using multiple stacks unnecessarily
• Reversing string instead of matching pairs