Balanced Parentheses Problem Using Stack in DSA C - 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 correct order.

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

📋 What You'll Learn

Create a stack data structure using an array and an integer top pointer.

Use a configuration variable to store the input string of parentheses.

Implement the logic to push and pop from the stack to check for balanced parentheses.

Print Balanced if parentheses match correctly, otherwise print Not Balanced.

💡 Why This Matters

🌍 Real World

Checking balanced parentheses is important in code editors and compilers to ensure code syntax is correct.

💼 Career

Understanding stacks and balanced parentheses is a common interview question and a fundamental concept in programming.

1

Create the stack data structure

Create an integer array called stack of size 100 and an integer variable top initialized to -1.

DSA C

# Create the stack array and top variable
# Your code here

Hint

The stack is an array and top keeps track of the last filled position. Start top at -1 to show the stack is empty.

2

Add the input parentheses string

Create a character array input initialized with the string "(()())".

DSA C

int stack[100];
int top = -1;

// Create the input string
# Your code here

Hint

The input string contains parentheses to check. Use double quotes for the string.

3

Implement the balanced parentheses check logic

Write a for loop using i to iterate over input until the null character. Inside the loop, push '(' onto stack by increasing top when found. When ')' is found, pop from stack by decreasing top. If top is -1 when popping, break the loop early.

DSA C

int stack[100];
int top = -1;

char input[] = "(()())";

// Loop through input and use stack to check parentheses
# Your code here

Hint

Use top++ to push and top-- to pop. Check for unmatched closing parentheses by seeing if top is -1 before popping.

4

Print if parentheses are balanced or not

After the loop, print Balanced if top is -1, otherwise print Not Balanced.

DSA C

int stack[100];
int top = -1;

char input[] = "(()())";

for (int i = 0; input[i] != '\0'; i++) {
    if (input[i] == '(') {
        stack[++top] = '(';
    } else if (input[i] == ')') {
        if (top == -1) {
            break;
        } else {
            top--;
        }
    }
}

// Print result based on top value
# Your code here

Hint

If the stack is empty at the end (top == -1), parentheses are balanced. Otherwise, they are not.

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 expression

C. To reverse the order of parentheses in the expression

D. To store only the closing parentheses encountered

Solution

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

   A stack helps to remember the last opening parenthesis to match it with the next closing one.

2. Step 2: Identify why other options are incorrect

   Counting or reversing parentheses does not solve matching pairs; storing only closing parentheses misses the matching logic.

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 remembers last opening bracket for matching [OK]

Common Mistakes:

• Thinking stack counts parentheses instead of matching
• Using stack to store only closing parentheses
• Ignoring order of parentheses

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

easy

A. push(stack, '(');

B. stack.push('(');

C. stack->push('(');

D. push('(', stack);

Solution

1. Step 1: Recognize typical C stack push syntax

   In C, stack operations are usually functions like push(stack, element).

2. Step 2: Analyze each option

   push(stack, '('); matches the common C function call syntax. Options B and C use dot or arrow notation typical in C++ or OOP, not plain C. push('(', stack); has arguments reversed.

3. Final Answer:

   push(stack, '('); -> Option A

4. Quick Check:

   C stack push syntax = A [OK]

Hint: In C, use push(stack, element) function call [OK]

Common Mistakes:

• Using C++ style stack.push() in C
• Reversing arguments in push function
• Using arrow operator on non-pointer stack

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

char expr[] = "({[]})";
int result = isBalanced(expr);
printf("%d", result);

Assume isBalanced returns 1 if balanced, 0 otherwise.

medium

A. Runtime error

B. 0

C. Compilation error

D. 1

Solution

1. Step 1: Analyze the expression "({[]})"

   The expression has matching pairs: '(' with ')', '{' with '}', '[' with ']'. All are properly nested.

2. Step 2: Understand isBalanced function behavior

   Since the expression is balanced, isBalanced returns 1, so printf prints 1.

3. Final Answer:

   1 -> Option D

4. Quick Check:

   Balanced expression returns 1 = A [OK]

Hint: Balanced nested pairs return 1 from isBalanced [OK]

Common Mistakes:

• Assuming only () pairs are valid
• Confusing balanced with equal count
• Expecting 0 for nested but balanced

4. Identify the bug in this C function snippet for checking balanced parentheses:

int isBalanced(char* expr) {
  Stack s;
  for (int i = 0; expr[i] != '\0'; i++) {
    if (expr[i] == '(') push(&s, '(');
    else if (expr[i] == ')') {
      if (isEmpty(&s)) return 0;
      pop(&s);
    }
  }
  return isEmpty(&s);
}

medium

A. Return statement should be inside the loop

B. Stack s is not initialized before use

C. pop() is called without checking if stack is empty

D. The function does not check for other types of brackets

Solution

1. Step 1: Check stack initialization

   The stack variable 's' is declared but not initialized, so push/pop operations may fail.

2. Step 2: Verify other parts

   Other options are less critical: checking only () is a limitation but not a bug; pop is called after empty check; return outside loop is correct.

3. Final Answer:

   Stack s is not initialized before use -> Option B

4. Quick Check:

   Uninitialized stack causes errors = D [OK]

Hint: Always initialize stack before push/pop [OK]

Common Mistakes:

• Ignoring stack initialization
• Assuming pop safe without empty check
• Thinking return must be inside loop

5. Given the expression "[{()}](({}))", which of the following modifications to the isBalanced function will correctly handle all types of brackets: (), {}, and []?

hard

A. Ignore brackets inside quotes and only check outer brackets

B. Count total opening and closing brackets separately and compare counts

C. Push opening brackets and check matching closing brackets by popping and comparing types

D. Replace all brackets with parentheses before checking

Solution

1. Step 1: Understand bracket matching logic

   To handle multiple bracket types, push opening brackets and on closing bracket, pop and check if types match.

2. Step 2: Evaluate other options

   Counting brackets ignores order; ignoring brackets inside quotes is unrelated; replacing brackets loses type info.

3. Final Answer:

   Push opening brackets and check matching closing brackets by popping and comparing types -> Option C

4. Quick Check:

   Match types by push/pop for all brackets = C [OK]

Hint: Match each closing bracket with last opening of same type [OK]

Common Mistakes:

• Counting brackets without order check
• Ignoring bracket types
• Replacing brackets loses matching info