Programming Challenges

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Three challenges will be posted every week to complete

Tuesday (Easy)
Thursday (Medium)
Saturday (Hard)

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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[Easy] String Compression (self.challenges)

submitted 1 year ago* (last edited 1 year ago) by Ategon to c/challenges

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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

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[–] Quasari 2 points 1 year ago* (last edited 1 year ago)

Ruby, just because I wanted a bit more practice. Again, I went to lowering character count, so it is ugly. I'm just using a anchor-runner pointer strategy to do the work. I tried to think of a regex to do the work for me, but couldn't think of one that would work.

i=ARGV[0]
o=''
a=r=0
l=i.length
while l>a
  r+=1
  if r>l||i[a]!=i[r] then
    o+=i[a]+(r-a).to_s
    a=r
  end
end
puts o