Balanced Parentheses Problem Using Stack in DSA Rust - Build from Scratch

Balanced Parentheses Problem Using Stack

📖 Scenario: Imagine you are building a simple code editor that checks if the parentheses in a line of code are balanced. Balanced parentheses mean every opening bracket has a matching closing bracket in the right order.

🎯 Goal: You will create a Rust program that uses a stack to check if a given string of parentheses is balanced or not.

📋 What You'll Learn

Create a string variable with a specific parentheses expression

Create a stack using a vector to help check parentheses

Write a loop to process each character in the string

Use conditions to push or pop from the stack based on parentheses

Print whether the parentheses are balanced or not

💡 Why This Matters

🌍 Real World

Checking balanced parentheses is important in code editors, compilers, and calculators to ensure expressions are correct.

💼 Career

Understanding stacks and balanced parentheses helps in software development roles involving parsing, syntax checking, and compiler design.

1

Create the parentheses string

Create a string variable called expression and set it to the exact value "(()())".

DSA Rust

// Create a string variable called expression with value "(()())"
// Your code here

Hint

Use let expression = "(()())"; to create the string.

2

Create an empty stack using a vector

Add a mutable vector called stack to hold characters. Initialize it as empty with Vec::new().

DSA Rust

fn main() {
    let expression = "(()())";
    // Create an empty mutable vector called stack
    // Your code here
}

Hint

Use let mut stack: Vec = Vec::new(); to create an empty stack.

3

Check each character and update the stack

Write a for loop using ch to go through each character in expression.chars(). Inside the loop, if ch is '(', push it to stack. If ch is ')', pop from stack if it is not empty.

DSA Rust

fn main() {
    let expression = "(()())";
    let mut stack: Vec = Vec::new();
    // Loop through each character in expression
    // Your code here
}

Hint

Use for ch in expression.chars() and inside check if ch == '(' to push and else if ch == ')' to pop if stack is not empty.

4

Print if parentheses are balanced

After the loop, print "Balanced" if stack.is_empty() is true, else print "Not Balanced".

DSA Rust

fn main() {
    let expression = "(()())";
    let mut stack: Vec = Vec::new();
    for ch in expression.chars() {
        if ch == '(' {
            stack.push(ch);
        } else if ch == ')' {
            if !stack.is_empty() {
                stack.pop();
            }
        }
    }
    // Print "Balanced" if stack is empty, else "Not Balanced"
    // Your code here
}

Hint

Use if stack.is_empty() { println!("Balanced"); } else { println!("Not Balanced"); } to print the result.

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