Balanced Parentheses Problem Using Stack in DSA Typescript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main purpose of using a stack in the Balanced Parentheses Problem?

A stack helps keep track of opening parentheses so that each closing parenthesis can be matched correctly in the right order.

beginner

Which characters are typically checked in the Balanced Parentheses Problem?

The characters checked are opening brackets '(', '{', '[' and their corresponding closing brackets ')', '}', ']'.

beginner

What happens if a closing parenthesis does not match the top of the stack?

It means the parentheses are not balanced, so the expression is invalid.

beginner

Why do we check if the stack is empty at the end of the Balanced Parentheses check?

If the stack is empty, it means all opening parentheses were matched properly. If not, some opening parentheses were not closed.

beginner

In TypeScript, which data structure is commonly used to implement a stack for this problem?

An array is commonly used as a stack by using push() to add and pop() to remove elements.

What is the first step when you encounter an opening parenthesis in the string?

A Push it onto the stack

B Pop from the stack

C Ignore it

D Check if the stack is empty

When you see an opening parenthesis, you push it onto the stack to remember it for matching later.

What should you do when you find a closing parenthesis in the string?

A Push it onto the stack

B Pop the top element from the stack and check if it matches

C Ignore it

D Clear the stack

You pop the top element from the stack and check if it matches the closing parenthesis type.

If the stack is not empty after processing the entire string, what does it mean?

A Parentheses are balanced

B There are unmatched closing parentheses

C The string is empty

D There are unmatched opening parentheses

A non-empty stack means some opening parentheses were not matched by closing ones.

Which of these strings is balanced?

A (()))

B ({[)]}

C ({[]})

D ((())

Only '({[]})' has all parentheses correctly matched and nested.

What data structure property is essential for solving the Balanced Parentheses Problem?

A Last In First Out (LIFO)

B First In First Out (FIFO)

C Random Access

D Sorted Order

Stack uses Last In First Out (LIFO), which is needed to match the most recent opening parenthesis first.

Explain how a stack helps in checking if parentheses in a string are balanced.

Think about how you remember the last opening bracket you saw.

Describe the step-by-step process to validate balanced parentheses using a stack.

Imagine reading the string from left to right and using a stack to track.

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