🔍 Mastering Parentheses: A Guide to Validating Strings 📝

Introduction:
Welcome to our 🌟 awesome blog where we'll unravel the mystery behind one of the fundamental problems in programming - validating parentheses in a string. Brace yourselves as we journey through this fascinating problem and emerge victorious with our solution! 💪

The Problem:
Let's dive into the heart of the matter! We are given a string s containing only the characters '(', ')', '{', '}', '[', and ']'. 🎯 The task at hand is to determine if the input string is valid. But what makes a string valid, you ask? Well, let's break it down:

1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.
3. Every closing bracket must have a corresponding open bracket of the same type. 🔄

Example Cases:
Let's illuminate this problem with some examples to help you grasp it better:

1. Input: s = "()" Output: true ✅
2. Input: s = "()[]{}" Output: true ✅
3. Input: s = "(]" Output: false ❌

The Solution:
Ah, the moment you've been waiting for! Fear not, for we shall conquer this challenge with elegance and efficiency. Our weapon of choice? The mighty stack data structure! 🛡️ Here's the battle plan:

• Iterate through the string, pushing opening brackets onto the stack.
• When encountering a closing bracket, check if it matches the top element of the stack.
• If it does, pop the corresponding opening bracket from the stack.
• If not, or if the stack is empty, the string is invalid.
• Finally, if the stack is empty after processing all characters, the string is valid. 🎉

Implementation:

def isValid(s: str) -> bool:
    # Mapping closing brackets to their corresponding opening brackets
    Map = {")": "(", "]": "[", "}": "{"}
    stack = []

    for char in s:
        if char not in Map:  # If it's an opening bracket
            stack.append(char)
            continue

        if (not stack) or (stack[-1] != Map[char]):  # If it's a closing bracket
            return False

        stack.pop()  # Matching opening bracket found, pop from stack

    return not stack  # Return True if stack is empty, False otherwise

Time and Space Complexity Analysis:

• Time Complexity: O(n) - where n is the length of the input string. We traverse the string once, performing constant-time operations for each character.
• Space Complexity: O(n) - In the worst case, the stack could contain all opening brackets from the input string, resulting in linear space usage. 📏

Some Notable Applications of Parentheses Validation:

1. Compiler Design and Parsing: Crucial for ensuring syntactic correctness in compilers, aiding in efficient parsing of programming language constructs.

2. Text Editors and IDEs: Enables real-time error detection and code readability enhancement in text editors and IDEs.

3. Mathematical Expressions Evaluation: Essential for accurate evaluation of mathematical expressions, supporting algorithms like Shunting Yard or recursive descent parsers.

These areas demonstrate the wide-ranging significance of parentheses validation in computer science and software engineering.

Conclusion:
Congratulations! You've mastered the art of validating parentheses in strings like a true programming ninja! 🥷 Armed with this knowledge, you're ready to tackle even more challenging problems in the vast realm of computer science and programming. Stay tuned for more thrilling adventures and mind-bending solutions right here on our blog! 🚀

In conclusion, our journey through the realm of validating parentheses has equipped us with invaluable insights and skills. With LeetCode as our guiding star, we've ventured into the heart of programming challenges, emerging victorious and ready to tackle new frontiers. Stay tuned for more adventures and solutions as we continue to explore the boundless possibilities of computer science! 🌟

For further exploration and practice, you can visit the LeetCode problem "Valid Parentheses" at this link.