Balanced Parentheses Problem Using Stack in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to push an opening bracket onto the stack.

DSA Javascript

if (char === '(') {
  stack.[1](char);
}

Drag options to blanks, or click blank then click option'

Aunshift

Bpop

Cshift

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push removes elements instead of adding.

Using shift or unshift changes the wrong end of the array.

We use push() to add an element to the top of the stack.

2fill in blank

medium

Complete the code to check if the stack is empty before popping.

DSA Javascript

if (stack.[1] === 0) {
  return false;
}

Drag options to blanks, or click blank then click option'

Alength

Bsize

Ccount

Dempty

Attempts:

3 left

💡 Hint

Common Mistakes

Using size or count which are not valid properties for arrays in JavaScript.

Using empty which is not a property.

The length property gives the number of elements in an array.

3fill in blank

hard

Fix the error in popping the top element from the stack.

DSA Javascript

const top = stack.[1]();

Drag options to blanks, or click blank then click option'

Apop

Bpush

Cshift

Dunshift

Attempts:

3 left

💡 Hint

Common Mistakes

Using push which adds elements instead of removing.

Using shift or unshift which operate on the start of the array.

pop() removes and returns the last element from the array, which is the top of the stack.

4fill in blank

hard

Fill both blanks to check if the popped bracket matches the current closing bracket.

DSA Javascript

if (top [1] '(' && char [2] ')') {
  continue;
} else {
  return false;
}

Drag options to blanks, or click blank then click option'

A===

B!==

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using != or !== which check for inequality.

Using == which is less strict and can cause bugs.

We use strict equality (===) to compare values exactly.

5fill in blank

hard

Fill all three blanks to complete the function that checks balanced parentheses.

DSA Javascript

function isBalanced(str) {
  const stack = [];
  for (let i = 0; i < str.[1]; i++) {
    const char = str[i];
    if (char === '(') {
      stack.[2](char);
    } else if (char === ')') {
      if (stack.[3] === 0) {
        return false;
      }
      stack.pop();
    }
  }
  return stack.length === 0;
}

Drag options to blanks, or click blank then click option'

Alength

Bpush

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push to add elements.

Checking stack.pop() === 0 which is incorrect.

We loop using str.length, push opening brackets, and check stack.length before popping.

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