Why Balanced Parentheses Problem Using Stack in DSA Typescript?

The Big Idea

What if you could instantly know if every bracket in your code matches perfectly without counting by hand?

The Scenario

Imagine you are checking if a math expression typed by hand has matching parentheses. You try to count opening and closing brackets manually by scanning the expression from left to right.

The Problem

Counting manually is slow and easy to mess up, especially when parentheses are nested or mixed with other types like curly braces or square brackets. You might miss a closing bracket or get confused about the order.

The Solution

Using a stack, you can push each opening bracket and pop it when you find a matching closing bracket. This way, you always check the most recent opening bracket first, ensuring correct order and pairing automatically.

Before vs After

✗ Before

function checkParentheses(expression: string): boolean {
  let count = 0;
  for (let char of expression) {
    if (char === '(') count++;
    else if (char === ')') count--;
  }
  return count === 0;
}

✓ After

function checkParentheses(expression: string): boolean {
  const stack: string[] = [];
  for (const char of expression) {
    if (char === '(') stack.push(char);
    else if (char === ')') {
      if (stack.length === 0) return false;
      stack.pop();
    }
  }
  return stack.length === 0;
}

What It Enables

This method lets you quickly and reliably verify complex nested parentheses and other bracket types in any expression.

Real Life Example

Code editors use this technique to highlight matching brackets and warn you if you forget to close one while writing code.

Key Takeaways

Manual counting fails with nested or mixed brackets.

Stack keeps track of the order of opening brackets.

Matching closing brackets pop from the stack to ensure correctness.

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