this post was submitted on 17 Aug 2023
14 points (100.0% liked)

Programming Challenges

234 readers
1 users here now

Welcome to the programming.dev challenge community!

Three challenges will be posted every week to complete

Easy challenges will give 1 point, medium will give 2, and hard will give 3. If you have the fastest time or use the least amount of characters you will get a bonus point (in ties everyone gets the bonus point)

Exact duplicate solutions are not allowed and will not give you any points. Submissions on a challenge will be open for a week.

A leaderboard will be posted every month showing the top people for that month

founded 1 year ago
MODERATORS
[email protected]
14
[Med] Bracket Match Generation (programming.dev)
submitted 1 year ago* (last edited 1 year ago) by [email protected] to c/[email protected]
16 comments fedilink hide all child comments
 

Bracket Inc. wants to ship out new products using their excess brackets. They have tasked you with generating every possible assortment of brackets for some n brackets where the brackets will match

  • A bracket match is an opening and closing version of the same kind of bracket beside each other ()
  • If a bracket matches then outer brackets can also match (())
  • n will be an even number
  • The valid brackets are ()[]{}

For example for n = 4 the options are

  • ()()
  • (())
  • [][]
  • [[]]
  • {}{}
  • {{}}
  • []()
  • ()[]
  • (){}
  • {}()
  • []{}
  • {}[]
  • ({})
  • {()}
  • ([])
  • [()]
  • {[]}
  • [{}]

You must accept n as a command line argument (entered when your app is ran) and print out all of the matches, one per line

(It will be called like node main.js 4 or however else to run apps in your language)

You can use the solution tester in this post to test you followed the correct format https://programming.dev/post/1805174

Any programming language may be used. 2 points will be given if you pass all the test cases with 1 bonus point going to whoevers performs the quickest and 1 for whoever can get the least amount of characters

To submit put the code and the language you used below

you are viewing a single comment's thread
view the rest of the comments
[–] [email protected] 4 points 1 year ago* (last edited 1 year ago)

Rust solution:

fn completer(bracket: char) -> char {
    match bracket {
        '[' => ']',
        '{' => '}',
        '(' => ')',
        ')' => '(',
        '}' => '{',
        ']' => '[',
        _ => unreachable!(),
    }
}

fn rec(v: &mut Vec, stack: &mut Vec, depth: usize) {
    // If we have more characters in the stack to complete
    // than the space avaiable, bail!
    if stack.len() > depth {
        return;
    }

    if depth == 0 {
        let output: String = v.iter().collect();
        println!("{}", output);
        return;
    }

    for b in &['[', '{', '('] {
        v.push(*b);
        stack.push(*b);
        rec(v, stack, depth - 1);
        stack.pop();
        v.pop();
    }

    if let Some(c) = stack.pop() {
        v.push(completer(c));
        rec(v, stack, depth - 1);
        stack.push(c);
        v.pop();
    }
}

fn main() {
    let depth: usize = std::env::args().nth(1).unwrap().parse().unwrap();
    let mut v = Vec::with_capacity(depth);
    let mut s = Vec::with_capacity(depth / 2);
    rec(&mut v, &mut s, depth);
}

Hope you enjoy this!

Edit: Lemmy makes the usize in angle brackets disappear after parse::. Idk what to do.

Edit: I kinda fixed it by assigning the type as a part of the variable definition.

Note to self: the angle bracket problem persists nonetheless.