Balanced Parentheses Problem Using Stack in DSA Typescript - Time & Space Complexity

Time Complexity: Balanced Parentheses Problem Using Stack

O(n)

Understanding Time Complexity

We want to know how the time needed to check balanced parentheses changes as the input string grows.

How does the number of steps grow when the string gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function isBalanced(s: string): boolean {
  const stack: string[] = [];
  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;
}
    

This code checks if parentheses in a string are balanced using a stack.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through each character in the string once.
• How many times: Exactly once for each character in the input string.

How Execution Grows With Input

As the string gets longer, the code checks each character one by one.

Input Size (n)	Approx. Operations
10	About 10 checks and stack operations
100	About 100 checks and stack operations
1000	About 1000 checks and stack operations

Pattern observation: The number of steps grows directly with the input size.

Final Time Complexity

Time Complexity: O(n)

This means the time to check balanced parentheses grows in a straight line with the string length.

Common Mistake

[X] Wrong: "The stack operations inside the loop make this O(n²) because push and pop are slow."

[OK] Correct: Push and pop on a stack are very fast and happen in constant time, so they don't multiply the loop's work.

Interview Connect

Understanding this helps you explain how to efficiently check matching pairs, a common problem in coding interviews.

Self-Check

"What if we added other types of brackets like {} and []? How would the time complexity change?"

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