Given some assortment of brackets, you must find the largest substring that is a valid matching bracket pattern
- 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 (())
- The valid brackets are ()[]{}
For example for the input {([])()[(])}()]
the answer would be ([])()
as that is the largest substring that has all matches
You must accept the input as a command line argument (entered when your app is ran) and print out the result
(It will be called like node main.js [(]()
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. 3 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
People who completed the challenge:
- @SleveMcDichael@programming.dev -
6.64s
-1203 chars
(rust) - @shape-warrior-t@kbin.social -
0.148s
-1251 chars
(python) - @nieceandtows@programming.dev -
0.008s
-1268 chars
(python) - @brie@beehaw.org -
0.001s
-1516 chars
(c) (fastest) - @Quasari@programming.dev
0.031s
-317 chars
(javascript) (shortest)
submissions open for another day (since the last time I edited the post)
Rust solution:
use std::cmp::min;
use std::ops::Range;
fn completer(bracket: char) -> char {
match bracket {
')' => '(',
'}' => '{',
']' => '[',
_ => unreachable!(),
}
}
pub struct Char {
index: usize,
value: char,
}
fn main() {
let input: String = std::env::args().nth(1).unwrap();
let mut longest = Range::default();
{
let mut current = Range::default();
let mut stack: Vec = Vec::with_capacity(input.len() << 1);
let mut streak = false;
for (i, c) in input.chars().enumerate() {
match c {
']' | '}' | ')' => {
let matched = stack
.last()
.map(|other| completer(c) == other.value)
.unwrap_or_default();
if matched {
current.start = if streak {
min(current.start, stack.pop().unwrap().index)
} else {
stack.pop().unwrap().index
};
current.end = i;
streak = true;
} else {
stack.clear();
if longest.len() < current.len() {
longest = current;
}
current = Range {
start: i + 1,
end: i + 1,
};
streak = false;
}
}
'[' | '{' | '(' => {
stack.push(Char { index: i, value: c });
}
_ => {}
};
}
if streak {
longest = current;
}
}
if longest.start != longest.end {
longest.end += 1;
}
println!("{}", &input[longest]);
}
Also available at: https://pastebin.com/EJsLYPqQ