Valid Parentheses Leetcode Solution

Working Through the Valid Parentheses Problem

The Valid Parentheses Leetcode Solution is one of those problems that looks stupid simple until you actually sit down to code it and realize how many edge cases exist. I've watched people breeze through the description and then fail on test cases like "][" or "}{" because they didn't think through the logic properly. Here's how to actually do it right. You use a stack. That's it. Push opening brackets onto the stack, and when you hit a closing bracket, pop the stack and check if it matches. If the stack is empty when you need to pop, or if the popped value doesn't match the current closing bracket, return false immediately. If you finish iterating and the stack still has items in it, also return false because you have unclosed brackets. I remember running into a weird case a while back where the test suite included strings with nested mixed brackets like "{[()]}" and people were messing up the mapping between opening and closing types. The fix is straightforward — use a hash map or dictionary that maps closing brackets to their corresponding opening brackets. Check the map when you encounter a closing bracket instead of writing a bunch of if/else chains.

Implementation Details

Here's the clean version in Python: The key line that trips people up is stack.pop() if stack else '#'. Without that conditional, you get an index out of bounds error when the string starts with a closing bracket. That '#' sentinel value will never match any opening bracket in your mapping, so the function correctly returns False. It's a minor detail that separates people who've actually coded this before from people who copy-pasted a solution they didn't understand. The JavaScript || '#' trick works because empty arrays are truthy, but pop() on an empty array returns undefined, which is falsy. So when the stack is empty, you fall back to the sentinel value.

An empty string should return true. This one is deceptively simple — an empty string is technically valid because there are no unmatched brackets. Don't overthink it, just let the loop finish and return true since the stack will be empty. Strings with odd lengths can never be valid. You could add an early return here to skip processing entirely, but honestly it's not worth the extra lines of code unless you're optimizing for competitive programming. The algorithm handles odd-length strings fine on its own — it'll just hit a mismatch somewhere. I once spent twenty minutes debugging a solution that failed on a string like "()" when it should have passed. The bug was that I had the mapping backwards — I was checking if the closing bracket matched itself instead of checking if the popped opening bracket matched the mapped value. Pretty embarrassing in an interview setting, but it's a mistake people make when they're tired and rushing.

Performance Notes

Time complexity is O(n) where n is the length of the string. You iterate through each character exactly once. Space complexity is also O(n) in the worst case where the string is all opening brackets like "(((((" — every character goes onto the stack. There's no real way to beat O(n) here since you have to examine every character at least once. Some people try to optimize by pre-checking string length parity, but that doesn't change the asymptotic complexity and makes the code less readable. Don't do it.

When This Approach Fails

This stack-based approach assumes all characters in the string are brackets. If the input can contain arbitrary characters and you need to ignore non-bracket characters, the logic changes slightly — you'd skip any character that isn't in your mapping keys or values. The original LeetCode problem guarantees only bracket characters, so this isn't an issue there, but it comes up in real-world validation scenarios where you might be parsing expressions with numbers and operators mixed in. If you're dealing with deeply nested structures (thousands of levels), you might hit stack overflow issues in some languages. Python's recursion limit won't affect you here since we're using an iterative approach with a list as a stack, but languages with smaller default stack sizes could be a concern. In practice this is rare for this problem, but it's worth knowing about if you scale this pattern to more complex parsing tasks.

Why You Should Care About This Problem

The valid parentheses pattern shows up everywhere beyond LeetCode. Database query validation, HTML/XML tag matching, calculator expression evaluation — they all rely on the same stack-based logic. Understanding this problem thoroughly means you can spot similar patterns faster during technical interviews and in actual work. Don't memorize the code. Understand why the stack is necessary. The stack enforces the LIFO (last-in-first-out) property that matching brackets require — the most recently opened bracket must be the first one closed. This intuition transfers to harder problems involving bracket sequences, expression parsing, and syntax validation.