Balanced Parentheses Problem Using Stack in DSA Rust - 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 program's work increase when the input size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn is_balanced(s: &str) -> bool {
    let mut stack = Vec::new();
    for ch in s.chars() {
        if ch == '(' || ch == '{' || ch == '[' {
            stack.push(ch);
        } else {
            if stack.is_empty() { return false; }
            let top = stack.pop().unwrap();
            if !matches!(top, '(' if ch == ')' | '{' if ch == '}' | '[' if ch == ']') {
                return false;
            }
        }
    }
    stack.is_empty()
}
    

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 in the input string once.
• How many times: Exactly once for each character, so n times for input length n.

How Execution Grows With Input

As the input string gets longer, the program checks each character once, so the work grows directly with input size.

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 operations grows linearly as the input size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to check balanced parentheses grows in direct proportion to the input length.

Common Mistake

[X] Wrong: "The time grows faster because of nested checks or multiple passes over the string."

[OK] Correct: The code only loops once through the string, and stack operations are fast, so time grows linearly, not faster.

Interview Connect

Understanding this linear time check is a key skill for many coding problems involving stacks and strings. It shows you can reason about efficient solutions.

Self-Check

"What if we changed the input to include other characters besides parentheses? How would the time complexity change?"

Practice

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

easy

A. To keep track of opening parentheses and match them with closing ones

B. To count the total number of parentheses in the string

C. To reverse the order of parentheses in the string

D. To sort the parentheses alphabetically

Solution

1. Step 1: Understand the role of stack in parentheses matching

   A stack stores opening parentheses as they appear to check if each closing parenthesis matches the last opened one.

2. Step 2: Explain why other options are incorrect

   Counting, reversing, or sorting parentheses does not help in checking if they are balanced.

3. Final Answer:

   To keep track of opening parentheses and match them with closing ones -> Option A

4. Quick Check:

   Stack tracks openings for matching = B [OK]

Hint: Stack stores openings to match closings quickly [OK]

Common Mistakes:

• Thinking stack counts parentheses instead of matching
• Confusing stack with queue for this problem
• Assuming stack sorts parentheses

2. Which of the following Rust code snippets correctly pushes an opening parenthesis '(' onto a stack named stack?

easy

A. stack.insert('(');

B. stack.push('(');

C. stack.add('(');

D. stack.append('(');

Solution

1. Step 1: Recall Rust Vec method to add elements

   In Rust, Vec uses push() to add elements at the end.

2. Step 2: Verify other methods are invalid for Vec

   add(), insert() require different parameters or do not exist; append() merges vectors, not single elements.

3. Final Answer:

   stack.push('('); -> Option B

4. Quick Check:

   Vec push method = push() [OK]

Hint: Use push() to add items to Rust Vec stack [OK]

Common Mistakes:

• Using append() to add single element
• Confusing insert() with push()
• Using add() which is not a Vec method

3. What will be the output of the following Rust code snippet that checks if parentheses are balanced?

fn is_balanced(s: &str) -> bool {
    let mut stack = Vec::new();
    for ch in s.chars() {
        if ch == '(' {
            stack.push(ch);
        } else if ch == ')' {
            if stack.pop().is_none() {
                return false;
            }
        }
    }
    stack.is_empty()
}

fn main() {
    println!("{}", is_balanced("(()())"));
}

medium

A. false

B. Runtime panic

C. Compilation error

D. true

Solution

1. Step 1: Trace the input string "(()())" character by character

   Process '(', push ['(']; '(', push ['(','(']; ')', pop ['(']; '(', push ['(','(']; ')', pop ['(']; ')', pop []. All match.

2. Step 2: Check final stack state

   Stack is NOT empty after processing all 6 characters because the input string "(()())" has 6 characters but the last character is ')', so the stack ends empty. Wait, let's count carefully:
   Characters: '(' push ['(']
   '(' push ['(','(']
   ')' pop ['(']
   '(' push ['(','(']
   ')' pop ['(']
   ')' pop []
   Stack is empty after processing all characters, so stack.is_empty() returns true.

3. Final Answer:

   true -> Option D

4. Quick Check:

   Matched pairs and empty stack = true [OK]

Hint: Count openings and closings; unmatched means false [OK]

Common Mistakes:

• Mis-counting the number of parentheses
• Ignoring stack emptiness at end
• Confusing pop() returning None with Some

4. Identify the bug in this Rust function that checks balanced parentheses:

fn check_balance(s: &str) -> bool {
    let mut stack = Vec::new();
    for ch in s.chars() {
        if ch == '(' {
            stack.push(ch);
        } else if ch == ')' {
            stack.pop();
        }
    }
    stack.is_empty()
}

medium

A. It uses a wrong data structure instead of Vec

B. It pushes closing parentheses instead of opening

C. It does not check if pop() returns None before popping

D. It returns true even if stack is not empty

Solution

1. Step 1: Analyze pop() usage on empty stack

   The function calls stack.pop() without checking if stack is empty, so popping from empty stack returns None but is ignored.

2. Step 2: Explain consequence of missing check

   This causes false positives when closing parentheses appear without matching openings, as function never returns false early.

3. Final Answer:

   It does not check if pop() returns None before popping -> Option C

4. Quick Check:

   Missing pop() None check causes errors [OK]

Hint: Always check pop() result before proceeding [OK]

Common Mistakes:

• Ignoring pop() returning None on empty stack
• Pushing wrong characters
• Assuming stack emptiness check is enough

5. Given the Rust function below, what will be the output when checking the string "{[(())]}"? The function supports '()', '[]', and '{}' pairs.

fn is_balanced(s: &str) -> bool {
    let mut stack = Vec::new();
    for ch in s.chars() {
        if ch == '(' || ch == '[' || ch == '{' {
            stack.push(ch);
        } else {
            if let Some(top) = stack.pop() {
                if (ch == ')' && top != '(') ||
                   (ch == ']' && top != '[') ||
                   (ch == '}' && top != '{') {
                    return false;
                }
            } else {
                return false;
            }
        }
    }
    stack.is_empty()
}

fn main() {
    println!("{}", is_balanced("{[(())]}"));
}

hard

A. true

B. false

C. Compilation error

D. Runtime panic

Solution

1. Step 1: Trace the string "{[(())]}" step-by-step

   Push '{', '[', '(', '('. Then pop for ')', ')', ']', '}'. Each closing matches the last opening correctly.

2. Step 2: Check stack emptiness after processing

   All pairs matched, stack is empty, so function returns true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   All pairs matched and stack empty = true [OK]

Hint: Match each closing with last opening in stack [OK]

Common Mistakes:

• Not handling multiple bracket types
• Ignoring order of matching pairs
• Forgetting to check stack empty at end