Balanced Parentheses Problem Using Stack in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Typescript

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

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

Apop

Bshift

Cpush

Dunshift

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 Typescript

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 array properties in TypeScript.

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 the code to pop the top element from the stack.

DSA Typescript

const top = stack.[1]();

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

Ashift

Bpush

Cunshift

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using push which adds instead of removes.

Using shift or unshift which affect 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 Typescript

if (top !== [1]) {
  return false;
}

const pairs = { ')': '(', [2]: '[' };

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

A(

B{

C]

D}

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up opening and closing brackets.

Using wrong bracket types in pairs.

We compare top with '(' for ')' and define pairs with ']' for '['.

5fill in blank

hard

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

DSA Typescript

function isBalanced(s: string): boolean {
  const stack: string[] = [];
  const pairs: { [key: string]: string } = { ')': '(', ']': '[1]', '[2]': '{' };

  for (const char of s) {
    if (char === '(' || char === '[' || char === '[3]') {
      stack.push(char);
    } else {
      if (stack.length === 0) return false;
      const top = stack.pop();
      if (top !== pairs[char]) return false;
    }
  }
  return stack.length === 0;
}

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

A[

B{

C}

D(

Attempts:

3 left

💡 Hint

Common Mistakes

Confusing opening and closing brackets in pairs.

Missing '{' in the opening brackets check.

Pairs map closing brackets to openings: ']' to '[', '}' to '{'. Opening brackets include '{'.

Practice

1. What is the core idea behind using a stack to validate balanced parentheses?

easy

A. Count total opening and closing brackets and compare

B. Push opening brackets onto the stack; when a closing bracket appears, check if the top matches

C. Sort the string and compare first and last characters

D. Use a counter that increments for '(' and decrements for ')'

Solution

1. Step 1: Stack tracks unmatched opens

   Opening brackets are pushed onto the stack. Each closing bracket must match the most recent unmatched opening bracket - the stack top.

2. Step 2: Why stack, not counter

   A counter works for a single bracket type but fails for mixed types: '([)]' has equal counts but is invalid. Stack tracks TYPE and ORDER.

3. Final Answer:

   Push open, match against stack top on close. -> Option B

4. Quick Check:

   Stack: push '(' '[' '{'. On close, pop and verify match. [OK]

Hint: Stack: push opens, pop+match on close. Counter alone fails for mixed bracket types. [OK]

Common Mistakes:

• Using a counter which fails for '([)]'
• Not checking bracket TYPE when matching
• Forgetting to check stack is empty at the end

2. What should be returned when a closing bracket is encountered but the stack is empty?

easy

A. true - the string might still be balanced

B. false - there is no matching opening bracket

C. Push the closing bracket and continue

D. Skip it and continue scanning

Solution

1. Step 1: Empty stack on close

   If we see ')' but the stack is empty, there is no unmatched opening bracket to pair with it. The string is invalid.

2. Step 2: Return false immediately

   No need to continue scanning - return false right away.

3. Final Answer:

   Return false - unmatched closing bracket. -> Option B

4. Quick Check:

   Close on empty stack = unmatched = false. [OK]

Hint: Empty stack + closing bracket = instant false. No match possible. [OK]

Common Mistakes:

• Continuing to scan (wastes time, wrong)
• Pushing closing bracket (corrupts the stack)
• Returning true (incorrect)

3. Is '([)]' a valid balanced parentheses string?

medium

A. Yes - it has equal counts of each bracket type

B. No - the brackets are nested incorrectly

C. Yes - all brackets are closed

D. Yes - opening and closing counts match

Solution

1. Step 1: Trace the stack

   Push '(': stack=['(']. Push '[': stack=['(','[']. See ')': stack top is '[' ≠ ')' - MISMATCH. Return false.

2. Step 2: Why counting fails

   Count approach: 1 '(', 1 '[', 1 ')', 1 ']' - all equal. But stack reveals the TYPE mismatch.

3. Final Answer:

   Invalid - '[' is at top when ')' arrives. -> Option B

4. Quick Check:

   Stack trace: push '(' push '[' → ')' arrives → top='[' ≠ ')' → false. [OK]

Hint: Trace stack: '([)]' → push '(' push '[' → ')' mismatches '[' → false. [OK]

Common Mistakes:

• Counting brackets and saying valid
• Not checking bracket type on pop
• Thinking equal counts = balanced

4. What is the time and space complexity of the stack-based balanced parentheses solution?

medium

A. Time O(n²), Space O(1)

B. Time O(n log n), Space O(n)

C. Time O(n), Space O(n)

D. Time O(n), Space O(1)

Solution

1. Step 1: Time

   Single pass through string: O(n). Each character: one push or one pop - O(1) each. Total: O(n).

2. Step 2: Space

   Stack can hold up to n/2 characters in worst case '((((...)))'. Space: O(n).

3. Final Answer:

   Time O(n), Space O(n). -> Option C

4. Quick Check:

   One pass = O(n). Stack = O(n) worst case. [OK]

Hint: Balanced parentheses: O(n) time (one pass), O(n) space (stack). [OK]

Common Mistakes:

• Saying O(1) space - stack can grow to O(n)
• Saying O(n²) - only one pass needed
• Confusing with sorting-based approaches

5. Which TypeScript code correctly implements the balanced parentheses check?

hard

A. return s.split('').filter(c => '({['.includes(c)).length === s.split('').filter(c => ')}]'.includes(c)).length;

B. const stack: string[] = []; const map: Record = {')':'(', ']':'[', '}':'{'}; for (const c of s) { if ('([{'.includes(c)) stack.push(c); else if (stack.pop() !== map[c]) return false; } return stack.length === 0;

C. let count = 0; for (const c of s) { if (c === '(') count++; else count--; if (count < 0) return false; } return count === 0;

D. return s.length % 2 === 0;

Solution

1. Step 1: Map for bracket pairs

   map pairs each closing bracket to its expected opening. On close: pop stack and check it matches map[c].

2. Step 2: Final check

   stack.length === 0 ensures no unmatched opening brackets remain.

3. Why A fails

   Only counts, doesn't check types or order.

4. Why C fails

   Counter only works for '()' - fails for '([)]' or mixed types.

5. Final Answer:

   Option B - map + stack with type checking. -> Option B

6. Quick Check:

   Map + stack + empty check at end = correct solution. [OK]

Hint: Use map {')':'(', ']':'[', '}':'{'} + stack. Pop on close, check match. Return stack.length===0. [OK]

Common Mistakes:

• Using count only (fails mixed brackets)
• Not checking stack empty at end (unmatched opens)
• Not using a map for bracket pairs