Conquering Valid Parentheses with Stack Magic

🔥 Today's Challenge Problem: Given a string containing just (, ), {, }, [ and ], determine if the input string is valid.

Example:

Input: "()[]{}"
Output: true
🧠 The Hunter's Strategy Weapon of Choice: Stack Data Structure When dealing with nested structures, stacks are your best friend. Here's why:

Last-In-First-Out (LIFO) principle perfectly matches bracket validation

O(n) time complexity - we process each character exactly once

Early termination possible when mismatches occur

Approach Breakdown: Initialize an empty stack

Create a bracket pairing map (] → [, etc.)

Iterate through the string:

Push opening brackets onto stack

For closing brackets: check if top of stack matches

Final stack check ensures all brackets were closed

💻 TypeScript Implementation

function isValid(s: string): boolean {
const stack: string[] = [];
const bracketPairs: Record = {
')': '(',
'}': '{',
']': '['
};

for (const char of s) {
    if (!bracketPairs[char]) {
        stack.push(char);
    } else if (stack.pop() !== bracketPairs[char]) {
        return false;
    }
}

return stack.length === 0;

}
Key Optimizations:

Used a hashmap for O(1) bracket lookups

Early return on mismatch

Single pass through the string

🎯 Edge Cases That Made Me Wiser Empty string: Technically valid (handled by final stack check)

Nested valid brackets: ({[]}) → true

Mismatched brackets: ([)] → false

Single bracket: [ → false

🚀 Performance Analysis Metric Value Time Complexity O(n) Space Complexity O(n) LeetCode Runtime 72ms (faster than 92.5%) 💡 Pro Tips Visualize the stack like a tower - you can only remove the top block!

Test these cases:

"(([]){})" (complex but valid)

"]" (immediate fail)

Try solving it without stack (hint: recursive approach exists)

What's your favorite stack-based algorithm? Let's discuss in the comments! 👇

Tomorrow's Preview: We'll tackle "Merge Two Sorted Lists" using pointer manipulation!