this post was submitted on 20 Aug 2023
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Programming Challenges

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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.

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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:

submissions open for another day (since the last time I edited the post)

all 13 comments
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[–] [email protected] 1 points 1 year ago

This is interesting, though I am 3 weeks late.

Raku code: https://glot.io/snippets/gonclpwl3p

[–] [email protected] 0 points 1 year ago (1 children)

Rust: https://pastebin.com/frYcgdxh

Like last time, a brute force solution: checking if every substring is a set of well formed brackets, and, if so, recording it if its the longest seen so far.

[–] [email protected] 1 points 1 year ago
  • 6/6 Test cases passed
  • Time Taken: 6.64 seconds
  • Characters: 1203
[–] [email protected] 0 points 1 year ago* (last edited 1 year ago) (1 children)

I found this one easier than the medium as well. Maybe because I actually know a strategy for this one. Anywho, solution is in JavaScript. It's super ugly as I went for lowest character count, so no declarations, single char variable names, no semicolons, etc...

Basically use a anchor-runner pointer strategy to find a correct substring, I use a stack based solution for checking if the substring is correct. When the stack is empty, the substring is good, thus I record the start of it and its length. If, I get another good point, I just record the highest. If I hit a point where its no longer good, I jump the start to the end of the most recent good substring. Pretty fun.

Formatting screwed mine up, so heres a pastbin

[–] [email protected] 1 points 1 year ago
  • 6/6 Test cases passed
  • Time taken: 0.031 seconds
  • 317 characters
[–] [email protected] 0 points 1 year ago (1 children)

I actually found this challenge to be easier than this week's medium challenge. (Watch me say that and get this wrong while also getting the medium one correct...) Here's an O(n) solution:

bracket_pairs = {('(', ')'), ('[', ']'), ('{', '}')}

def main(brackets: str) -> str:
    n = len(brackets)
    has_match_at = {i: False for i in range(-1, n + 1)}
    acc = []
    for i, bracket in enumerate(brackets):
        acc.append((i, bracket))
        if len(acc) >= 2:
            opening_idx, opening = acc[-2]
            closing_idx, closing = acc[-1]
            if (opening, closing) in bracket_pairs:
                acc.pop(), acc.pop()
                has_match_at[opening_idx] = has_match_at[closing_idx] = True
    longest_start, longest_end = 0, 0
    most_recent_start = None
    for left_idx, right_idx in zip(range(-1, n), range(0, n + 1)):
        has_match_left = has_match_at[left_idx]
        has_match_right = has_match_at[right_idx]
        if (has_match_left, has_match_right) == (False, True):
            most_recent_start = right_idx
        if (has_match_left, has_match_right) == (True, False):
            most_recent_end = right_idx
            if most_recent_end - most_recent_start > longest_end - longest_start:
                longest_start, longest_end = most_recent_start, most_recent_end
    return brackets[longest_start:longest_end]

if __name__ == '__main__':
    from argparse import ArgumentParser
    parser = ArgumentParser()
    parser.add_argument('brackets')
    print(main(parser.parse_args().brackets))

We start off by doing the same thing as this week's easy challenge, except we keep track of the indices of all of the matched brackets that we remove (opening or closing). We then identify the longest stretch of consecutive removed-bracket indices, and use that information to slice into the input to get the output.

For ease of implementation of the second part, I modelled the removed-bracket indices with a dict simulating a list indexed by [-1 .. n + 1), with the values indicating whether the index corresponds to a matched bracket. The extra elements on both ends are always set to False. For example, {([])()[(])}()] -> FFTTTTTTFFFFFTTFF, and ([{}]) -> FTTTTTTF. To identify stretches of consecutive indices, we can simply watch for when the value switches from False to True (start of a stretch), and from True to False (end of a stretch). We do that by pairwise-looping through the dict-list, looking for 'FT' and 'TF'.

[–] [email protected] 2 points 1 year ago
  • 6/6 Test cases passed
  • 0.148 seconds taken
  • 1251 characters
[–] [email protected] 0 points 1 year ago* (last edited 1 year ago) (1 children)

Please feel free to suggest changes to make the code more efficient.

EDIT: for some reason, < is not showing up properly inside a code block. Kindly replace it with the right symbol when you test.

Python:

import sys

input_string = sys.argv[1]

matches = {
    "}": "{",
    "]": "[",
    ")": "("
}

bracket_matches_strict = []

def get_matching_substring_strict(string):
    substring = ''
    index_start = -1
    index_end = -1
    bracket_counts = {
        "{": 0,
        "[": 0,
        "(": 0
    }
    for index, letter in enumerate(string):
        if letter in matches.values():
            if index_start == -1:
                index_start = index
            substring += letter
            bracket_counts[letter] += 1
        if letter in matches.keys():
            if not substring:
                break
            if substring[-1] == matches[letter]:
                substring = substring[:-1]
                bracket_counts[matches[letter]] -= 1
                if not [cnt for cnt in bracket_counts.values() if cnt]:
                    index_end = index
                if [cnt for cnt in bracket_counts.values() if cnt &lt; 0]:
                    break
            else:
                break

    if index_start != -1 and index_end != -1:
        matching_substring = string[index_start:index_end + 1]
        return len(matching_substring), matching_substring

for i in range(len(input_string)):
    bracket_substring_strict = get_matching_substring_strict(input_string[i:])
    if bracket_substring_strict:
        bracket_matches_strict.append(bracket_substring_strict)

print(sorted(bracket_matches_strict)[-1][1])
[–] [email protected] 2 points 1 year ago (1 children)
  • 6/6 test cases passed
  • 0.008 seconds taken
  • 1268 characters
[–] [email protected] 1 points 1 year ago

I think I could have shaved 50 letters if I named the function and variables differently, 😄