Given a string, compress substrings of repeated characters in to a format similar to aaa -> a3
For example for the string aaabbhhhh33aaa
the expected output would be a3b2h432a3
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 aaaaa
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. 1 point 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 have completed the challenge:
- @[email protected] -
N/A
-84 chars
(javascript) - @[email protected] -
0.011s
-385 chars
(python) - @[email protected]
0.003s
-487 chars
(c) (fastest)0.01s
-60 chars
(javascript) (shortest)
- @[email protected] -
0.005s
-530 chars
(rust) - @[email protected] -
0.191s
-566 chars
(python) - @[email protected] -
0.053s
-359 chars
(factor) If you submitted in factor, ruby, cobol, etc. itll take me a bit longer to check since I havent set up running those yet. Should be soon
Factor:
(command-line) last [ ] group-by [ length 48 + swap ] assoc-map “” concat-as print
Bun, technically prints it to the console:
throw[...Bun.argv[2].matchAll(/(.)+?(?!\1)/g)].map(([a,c])=>c+a.length).join("")
Breaks if there are more than 9 repeats or if you don’t supply an argument.
- 8/8 test cases passed
- no runtime stats since I couldnt use the solution checker to check it (had to do manually)
- Characters: 84
- 8/8 test cases passed
- Time: 0.005s
- Characters: 530
My solution (runs in O(n) time, but so do all the other solutions so far as far as I can tell):
from itertools import pairwise def main(s: str) -> str: characters = [None] + list(s) + [None] transitions = [] for (_, left), (right_idx, right) in pairwise(enumerate(characters)): if left != right: transitions.append((right_idx, right)) repeats = [(stop - start, char) for (start, char), (stop, _) in pairwise(transitions)] return ''.join(f'{char}{length}' for length, char in repeats) if __name__ == '__main__': from argparse import ArgumentParser parser = ArgumentParser() parser.add_argument('s') print(main(parser.parse_args().s))
Runthrough:
'aaabb'
->[None, 'a', 'a', 'a', 'b', 'b', None]
->[(1, 'a'), (4, 'b'), (6, None)]
->[(4 - 1, 'a'), (6 - 4, 'b')]
Golfed (just for fun, not a submission):
import sys from itertools import pairwise as p print(''.join(c+str(b-a)for(a,c),(b,_)in p([(i,r)for i,(l,r)in enumerate(p([None,*sys.argv[1],None]))if l!=r])))
- 8/8 test cases passed
- time taken: 0.191s
- characters: 566
deleted by creator
C:
gcc -O2 easy.c
In case formatting is messed up
#include #include void bail(char const *msg) { fputs(msg, stderr); exit(EXIT_FAILURE); } int main(int argc, char **argv) { int count = 0; char *p, c = 0; if (argc != 2) { bail("Improper invocation.\n"); } for (c = *(p = argv[1]); *p; p++) { if (c == *p) { count++; } else { printf("%c%d", c, count); c = *p; count = 1; } } if (count) { printf("%c%d", c, count); } putchar('\n'); return 0; }
- 8/8 test cases passed
- Time taken: 0.003s
- Characters: 487
Python:
import sys input_string = sys.argv[1] compressed_string = '' last_letter = '' letter_count = 0 for letter in input_string: if letter != last_letter: if last_letter != '': compressed_string += last_letter + str(letter_count) letter_count = 0 last_letter = letter letter_count += 1 compressed_string += last_letter + str(letter_count) print(compressed_string)
- 8/8 test cases passed
- time taken: 0.011s
- characters: 385
JavaScript (
DenoBun):bun easy.js aaabbhhhh33aaa
Golf attempt using RegExp to match homogenous substrings.
Original version
console.log(Deno.args[0].replace(/(.)\1*/g,a=>a[0]+(a.length+'')))
Slightly optimised by removing the frivolous conversion to string. I also made a somewhat silly opitmisation of using
Bun.argv
instead ofDeno.args
to save a character.console.log(Bun.argv[2].replace(/(.)\1*/g,a=>a[0]+a.length))
- 8/8 test cases passed
- Time taken: 0.01s
- Characters: 60
My broken brain just goes “but how would a decompressor know if ‘3101’ was originally ‘30’ or 101 '3’s in a row?”
Factor:
USING: kernel strings command-line namespaces math.parser sequences io ; IN: l : c ( s -- C ) "" swap [ dup empty? not ] [ dup dup dup first '[ _ = not ] find drop dup not [ drop length ] [ nip ] if 2dup tail [ number>string [ first 1string ] dip append append ] dip ] while drop ; MAIN: [ command-line get first c print ]
Benchmark using compiled binary:
$ hyperfine "./compress/compress aaabbhhhh33aaa" Benchmark 1: ./compress/compress aaabbhhhh33aaa Time (mean ± σ): 3.6 ms ± 0.4 ms [User: 1.4 ms, System: 2.2 ms] Range (min … max): 3.0 ms … 6.0 ms 575 runs
- 8/8 test cases passed
- Time taken: 0.163 seconds
- Characters: 359
I ran it using
factor.com -run (path to file)
If theres a better way let me know, first time using factor
Thanks! Yeah to use the optimized compiler to create a faster runnable binary, I’ve put instructions on the first challenge: https://programming.dev/comment/1980496
new time taken: 0.053 seconds