Okay, so today I wanted to mess around with this “longest barbell” problem. I’d seen it floating around and it seemed like a fun little challenge. The basic idea is you’re given a list of numbers, and you need to find the longest “barbell” – that’s a sequence where you have a number, then some other stuff in the middle, and then the same number at the end. The length of the barbell is the total count of numbers, including the two matching ends.
First thing I did was just try to understand the problem. I jotted down some example lists on a piece of paper and manually worked out the longest barbells. That helped me get a feel for what I was looking for.
My initial thought was, “Okay, I’ll just loop through the list, and for each number, I’ll search the rest of the list for a match.” Sounds simple, right? So I started coding that up:
- Outer loop: Goes through each number in the list, one by one. Let’s call this number the ‘start’.
- Inner loop: Starts from the position after the ‘start’ and goes to the end of the list, looking for a match.
- If a match is found: Calculate the length (the difference in positions + 1) and keep track of the longest one so far.
- If no match is found for outer loop, move the loop and start over.
I banged out this code, and it…sort of worked. On small lists, it was fine. But then I threw a bigger list at it, and it started to get really slow. I mean, really slow. I realized I’d created something with nested loops, and that’s not usually a good sign for performance.
So, back to the drawing board. I thought, “There’s gotta be a better way than searching the entire rest of the list for every single number!” And then it hit me: What if I used a dictionary (or a “hash map,” whatever you want to call it)?
The idea is this:
- Go through the list once.
- For each number, store its position in the dictionary. if find it, then that’s another barbell, update it.
I rewrote my code using this dictionary approach. Instead of nested loops, I now had a single loop. And guess what? It was way faster. Like, night and day difference. The big list that was taking forever before? Now it was done in a blink of an eye.
Here’s the gist of how the code works, I think:
- Create an empty dictionary.
- Loop the list.
- If a number is found: calculate the position gap and compared with current barbell and update the longest length.
- If a number isn’t found: add it to the dictionary, storing its position.
- Finish and return. I should got the right answer.
It felt pretty good to take a problem that was initially kicking my butt and then find a much more elegant and efficient solution. That’s one of the things I love about coding – there’s always something new to learn, and there’s usually more than one way to get the job done.
