Balanced Parentheses Problem Using Stack in DSA Java - 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 the string size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public boolean isBalanced(String s) {
    Stack<Character> stack = new Stack<>();
    for (int i = 0; i < s.length(); i++) {
        char c = s.charAt(i);
        if (c == '(' || c == '{' || c == '[') {
            stack.push(c);
        } else {
            if (stack.isEmpty() || !matches(stack.pop(), c)) {
                return false;
            }
        }
    }
    return stack.isEmpty();
}

private boolean matches(char open, char close) {
    return (open == '(' && close == ')') ||
           (open == '{' && close == '}') ||
           (open == '[' && close == ']');
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through each character of the string once.
• How many times: Exactly once for each character, so n times where n is string length.

How Execution Grows With Input

As the string gets longer, the number of steps grows directly with its length.

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 in a straight line as input size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to check balanced parentheses grows linearly with the string length.

Common Mistake

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

[OK] Correct: Each character is processed once, and stack operations are simple and fast, so they don't add extra loops or repeated work.

Interview Connect

Understanding this linear time check helps you solve many problems involving matching pairs efficiently, a common skill interviewers look for.

Self-Check

"What if the input string contains only opening parentheses? How would the time complexity change?"

Practice

1. What does the Balanced Parentheses problem using a stack check for in a string?

easy

A. Whether every opening bracket has a matching closing bracket in correct order

B. If the string contains only alphabets

C. If the string length is even

D. Whether the string contains any digits

Solution

1. Step 1: Understand the problem goal

   The Balanced Parentheses problem checks if all opening brackets like '(' have a matching closing bracket ')' in the correct order.

2. Step 2: Role of stack in checking

   A stack helps track opening brackets and ensures each is closed properly in sequence.

3. Final Answer:

   Whether every opening bracket has a matching closing bracket in correct order -> Option A

4. Quick Check:

   Balanced parentheses = matching pairs [OK]

Hint: Match each '(' with ')' using stack push/pop [OK]

Common Mistakes:

• Thinking it checks string length
• Ignoring order of brackets
• Checking only for digits or alphabets

2. Which of the following Java code snippets correctly initializes a stack to check balanced parentheses?

easy

A. Stack<Character> stack = new Stack<>();

B. Stack stack = new Stack<Character>();

C. Stack<Character> stack = Stack();

D. Stack<Character> stack = new Stack<Character>;

Solution

1. Step 1: Check correct Stack initialization syntax

   In Java, to create a stack of Characters, use: Stack<Character> stack = new Stack<>(); which uses diamond operator for type inference.

2. Step 2: Identify errors in other options

   Stack stack = new Stack<Character>(); misses generic type on right side; Stack<Character> stack = Stack(); misses 'new' keyword; Stack<Character> stack = new Stack<Character>; misses parentheses after constructor.

3. Final Answer:

   Stack<Character> stack = new Stack<>(); -> Option A

4. Quick Check:

   Correct Java stack init = Stack<Character> stack = new Stack<>(); [OK]

Hint: Use diamond operator <> with new Stack<>() [OK]

Common Mistakes:

• Omitting 'new' keyword
• Missing parentheses after constructor
• Incorrect generic type syntax

3. What will be the output of the following Java code snippet?

String s = "({[]})";
System.out.println(isBalanced(s));

public static boolean isBalanced(String s) {
    Stack<Character> stack = new Stack<>();
    for (char c : s.toCharArray()) {
        if (c == '(' || c == '{' || c == '[') {
            stack.push(c);
        } else {
            if (stack.isEmpty()) return false;
            char top = stack.pop();
            if ((c == ')' && top != '(') ||
                (c == '}' && top != '{') ||
                (c == ']' && top != '[')) {
                return false;
            }
        }
    }
    return stack.isEmpty();
}

medium

A. Compilation error

B. false

C. true

D. Runtime exception

Solution

1. Step 1: Trace the input string "({[]})" character by character

   Push '(' '{' '[' onto stack; then pop matching pairs for ']' '}' ')'. Stack ends empty.

2. Step 2: Check final stack state

   Stack is empty after processing all characters, so return true.

3. Final Answer:

   true -> Option C

4. Quick Check:

   All brackets matched = true [OK]

Hint: Stack empty at end means balanced [OK]

Common Mistakes:

• Not checking stack empty at end
• Mismatching brackets order
• Ignoring empty stack before pop

4. Identify the error in this Java method for checking balanced parentheses:

public static boolean checkBalanced(String s) {
    Stack<Character> stack = new Stack<>();
    for (char c : s.toCharArray()) {
        if (c == '(' || c == '{' || c == '[') {
            stack.push(c);
        } else {
            char top = stack.pop();
            if ((c == ')' && top != '(') ||
                (c == '}' && top != '{') ||
                (c == ']' && top != '[')) {
                return false;
            }
        }
    }
    return stack.isEmpty();
}

medium

A. For loop syntax is incorrect

B. Wrong comparison operators used

C. Stack should store Strings not Characters

D. Missing check if stack is empty before pop

Solution

1. Step 1: Analyze stack pop operation

   Pop is called without checking if stack is empty, which can cause runtime exception if closing bracket appears first.

2. Step 2: Identify missing empty check

   Must check stack.isEmpty() before pop to avoid EmptyStackException.

3. Final Answer:

   Missing check if stack is empty before pop -> Option D

4. Quick Check:

   Always check stack empty before pop [OK]

Hint: Check stack empty before pop to avoid errors [OK]

Common Mistakes:

• Ignoring empty stack before pop
• Assuming input always starts with opening bracket
• Not handling runtime exceptions

5. Given the string "[{()}]({[]})", which of the following modifications to the balanced parentheses algorithm will correctly handle this input and return true?

hard

A. Use two stacks, one for opening and one for closing brackets separately

B. Use a stack and check matching pairs for all bracket types, ensuring stack is empty at end

C. Only check if the string length is even and contains brackets

D. Ignore curly braces '{}' and only check parentheses '()'

Solution

1. Step 1: Understand input complexity

   The string contains multiple types of brackets nested and sequentially combined.

2. Step 2: Apply stack-based matching for all bracket types

   Push opening brackets, pop and match closing brackets; ensure stack empty at end for balanced check.

3. Step 3: Evaluate other options

   Length check or ignoring bracket types will fail; two stacks unnecessary and incorrect.

4. Final Answer:

   Use a stack and check matching pairs for all bracket types, ensuring stack is empty at end -> Option B

5. Quick Check:

   Stack matching all brackets = true [OK]

Hint: Match all bracket types with one stack, check empty at end [OK]

Common Mistakes:

• Ignoring some bracket types
• Relying on string length only
• Using multiple stacks unnecessarily