Balanced Parentheses Problem Using Stack in DSA Javascript - 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 grows as the input string gets longer.

How does the number of steps change when we have more characters to check?

Scenario Under Consideration

Analyze the time complexity of the following 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;
}
    

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 input string gets longer, the number of steps grows directly with the number of characters.

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 steps increase evenly as the input size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to check balanced parentheses grows in a straight line with the input size.

Common Mistake

[X] Wrong: "The time grows faster because of the stack operations inside the loop."

[OK] Correct: Each stack operation (push/pop) takes constant time, so they don't add extra loops or slow down the process beyond the single pass.

Interview Connect

Understanding this helps you explain how simple data structures like stacks can efficiently solve common problems, a skill valued in many coding challenges.

Self-Check

"What if we had to check for multiple types of brackets like (), {}, and []? How would the time complexity change?"

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