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Daily Challenge #66- Friend List

dev.to staff on September 14, 2019

John has invited some friends. His list is: s = "Fred:Corwill;Wilfred:Corwill;Barney:Tornbull;Betty:Tornbull;Bjon:Tornbull;Raphael:Corwill;Alfred:...

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La blatte • Sep 14 '19

Lets do it:

s
.split(';')
.map(n=>n.split(':').reverse().join(':'))
.sort()
.map(n=>`(${n.split(':')[0]}, ${n.split(':')[1]})`.toUpperCase())
.join``

It's pretty straightforward:

split(';') gets us all Name:Surname
.map(n=>n.split(':').reverse().join(':')) reverses them to be Surname:Name
.sort() sorts them alphabetically
.map(n=>(${n.split(':')[0]}, ${n.split(':')[1]}).toUpperCase()) reformats every name (uppercase it, place parentheses and commas)
The last step is equivalent to .join(''), which transforms the name array to an output string

Here is the output:

"Fred:Corwill;Wilfred:Corwill;Barney:Tornbull;Betty:Tornbull;Bjon:Tornbull;Raphael:Corwill;Alfred:Corwill"
.split(';')
.map(n=>n.split(':').reverse().join(':'))
.sort()
.map(n=>`(${n.split(':')[0]}, ${n.split(':')[1]})`.toUpperCase())
.join``;


// "(CORWILL, ALFRED)(CORWILL, FRED)(CORWILL, RAPHAEL)(CORWILL, WILFRED)(TORNBULL, BARNEY)(TORNBULL, BETTY)(TORNBULL, BJON)"

Ben Calder • Sep 15 '19 • Edited

Same but different:

  const s = "Fred:Corwill;Wilfred:Corwill;Barney:Tornbull;Betty:Tornbull;Bjon:Tornbull;Raphael:Corwill;Alfred:Corwill";
  return s.toUpperCase().split(';')
          .map(name => name.split(':').reverse())
          .sort()
          .reduce((accumulator, name) => ( accumulator += `(${name[0]}, ${name[1]})`), "");

You don't need the arbitrary join(':') after reverse() since sort() coerces the nested arrays to strings for you. That means you can avoid having to split again later.

But Amin is right:

  const s = "Fred:Corwill;Wilfred:Corwill;Barney:Tornbull;Betty:Tornbull;Bjon:Tornbull;Raphael:Corwill;Alfred:Corwill";
  return s.toUpperCase().split(';')
             .map(name=> `(${name.split(':')[1]}, ${name.split(':')[0]}`)
             .sort()
             .join('');

La blatte • Sep 16 '19

That's way shorter! Thanks a lot for the input

Amin • Sep 15 '19

I see we got a similar solution you and I!

I think you can even shorten (if I'm not mistaken) your solution by reducing one .map down and doing like me the surrounding in your first call to the .map method!

Should such a .surround method exist in String.prototype? I have came accross several cases where I would need it like in a WYSIWYG editor. But I don't want to risk proposing such a method to the community haha!

Good take btw!

Idan Arye • Sep 15 '19

s="Fred:Corwill;Wilfred:Corwill;Barney:Tornbull;Betty:Tornbull;Bjon:Tornbull;Raphael:Corwill;Alfred:Corwill"

echo $s \
    | tr ';' '\n' \
    | awk -F: '{print toupper("("$2":"$1")")}' \
    | sort \
    | paste -sd '' -

Oleksii Filonenko • Sep 19 '19

Rust:

fn meeting(list: &str) -> String {
    let list = list.to_uppercase();
    let list: Vec<_> = list.split(|c| [';', ':'].contains(&c)).collect();
    let mut list: Vec<_> = list
        .chunks_exact(2)
        .map(|chunk| format!("({}, {})", chunk[1], chunk[0]))
        .collect();
    list.sort();
    list.join("")
}

Amin • Sep 15 '19 • Edited

My take at the challenge written in Elm

Source-Code

$ touch src/FriendList.elm

module FriendList exposing (friendList)


surround : String -> String -> String -> String
surround left right string =
    left ++ string ++ right


toUpperTuples : String -> String
toUpperTuples =
    String.split ":"
        >> List.reverse
        >> String.join ", "
        >> surround "(" ")"
        >> String.toUpper


friendList : String -> String
friendList =
    String.split ";"
        >> List.map toUpperTuples
        >> List.sort
        >> String.join ""

Unit test source-code

$ touch tests/FriendListTest.elm

module FriendListTest exposing (suite)

import Expect
import FriendList exposing (friendList)
import Test exposing (Test)


suite : Test
suite =
    Test.describe "Friend list"
        [ Test.test "It should return the correct friend list" <|
            \_ ->
                let
                    expectations =
                        "(CORWILL, ALFRED)(CORWILL, FRED)(CORWILL, RAPHAEL)(CORWILL, WILFRED)(TORNBULL, BARNEY)(TORNBULL, BETTY)(TORNBULL, BJON)"

                    friends =
                        "Fred:Corwill;Wilfred:Corwill;Barney:Tornbull;Betty:Tornbull;Bjon:Tornbull;Raphael:Corwill;Alfred:Corwill"

                    result =
                        friendList friends
                in
                Expect.equal expectations result
        ]

Zohar Peled • Sep 16 '19

Not much of a challenge, I'm afraid. Pretty straight forward split, split, join, sort and join again. A one liner in most modern languages, I suspect.
In c# it would look like this:

string meeting(string s)
{
    return "(" + string.Join(")(", 
        s.
        ToUpper().
        Split(';').
        Select(n => n.Split(':')).
        Select(n => string.Join(":", n.Reverse())).
        OrderBy(n => n)
    ) + ")"; 
}

E. Choroba • Sep 17 '19

Perl solution:

#!/usr/bin/perl
use warnings;
use strict;
use feature qw{ say };

my $in = 'Fred:Corwill;Wilfred:Corwill;Barney:Tornbull;Betty:Tornbull;Bjon:Tornbull;Raphael:Corwill;Alfred:Corwill';

say join "",
    map "($_->[1], $_->[0])",
    sort { $a->[1] cmp $b->[1] || $a->[0] cmp $b->[0] }
    map [split /:/],
    split /;/,
    uc $in;

Kevin Ard • Sep 16 '19 • Edited

Anyone who says rex is too clever clearly hasn't bumped into bit-shifting code-golf lol

Rex is a very appropriate amount of cleverness 😉

Amin • Sep 15 '19

Clever use of the regex split. I like it!