Same Name, Lame!

Question

Write a function or program that, when given a list of names, outputs or returns a list where duplicates of given names have a unique shortened version of their surname.

Input:

A list of names, where a name is defined by a given name and a last name separated by a space. Names are non-empty strings containing only uppercase and lowercase letters. The list can be an array of strings, or the names separated by a constant non-alpha, non-space character, but the output must be in the same format as the input.

Output:

A list of the inputted names in the same order and format as the input that have been modified by these rules:

• For unique given names, output just the given name
• For names that share the same given name:
  • As well as their given name, add the shortest unique version of their surname that that is not shared by another name, followed by a period. For instance: John Clancy, John Smith becomes John C.,John S. and James Brown, James Bratte becomes James Bro.,James Bra.
  • If one surname is a subset of another, such as Julian King,Julian Kingsley, return the full surname of the smaller one without a period. The example would become Julian King,Julian King.
• Basically a period represents the regex .+, where only one name should match it.
• You may assume that no-one will share both the same given name and the same surname
• Names are case-sensitive

Test Cases:

• John Clancy,Julie Walker,John Walker,Julie Clancy -> John C.,Julie W.,John W.,Julie C.
• Julian King,Jack Johnson,Julian Kingsley> Julian King,Jack,Julian King.
• Jack Brown,Jack Black,Jack Blue> Jack Br.,Jack Bla.,Jack Blu.
• John Storm,Jon Snow,Johnny Storm> John,Jon,Johnny
• Jill DeSoma,Jill Desmond> Jill DeS.,Jill Des.
• XxXnO sCOppeXxX,XxXNO MERCYXxX> XxXnO,XxXNO

This is code-golf, so the lowest byte count for each language wins.

• Jo King: Related, related (Thanks @Laikoni). Sandbox Post
• caird coinheringaahing: Where does the Angela C. come from the in the test case?
• Jo King: Meant to be a Julie oops, thanks. Fixed

Answer 1

Jelly, (削除) 34 33 32 (削除ここまで) 30 bytes

;\ċÐf6t6ドル;€JṖḊ$$¦".μ€ċ@ÐṂ€Ẏ$Ḣ€

A monadic link taking a list of lists of characters (i.e. a list of "strings") and returning the abbreviations in the same format and relative order.

Try it online! (a full program test suite)

How?

;\ċÐf6t6ドル;€JṖḊ$$¦".μ€ċ@ÐṂ€Ẏ$Ḣ€ - Link: list of lists e.g. ["Sam Ng","Sam Li","Sue Ng"]
 μ€ - monadically for €ach: e.g. "Sue Ng"
 \ - cumulative reduce with:
; - concatenation ["S","Su","Sue","Sue ","Sue N","Sue Ng"]
 6 - literal space character ' '
 Ðf - filter keep if:
 ċ - count (spaces) ["Sue ","Sue N","Sue Ng"]
 6 - literal space character ' '
 t€ - trim from €ach ["Sue","Sue N","Sue Ng"]
 ". - literal period character '.'
 ¦ - sparse application...
 ;€ - ...of: concatenate €ach (with a period)
 - ...only for these indexes:
 $ - last two links as a monad:
 J - range of length [1,2,3]
 $ - last two links as a monad:
 Ṗ - pop [1,2]
 Ḋ - dequeue [2] (i.e. 2,3,...,length-1)
 - ...i.e.: ["Sue","Sue N.","Sue Ng"]
 - yielding: [["Sam","Sam N.","Sam Ng"],["Sam","Sam L.","Sam Li"],["Sue","Sue N.","Sue Ng"]]
 $ - last two links as a monad:
 Ẏ - tighten ["Sam","Sam N.","Sam Ng","Sam","Sam L.","Sam Li","Sue","Sue N.","Sue Ng"]
 ÐṂ€ - filter keep minimals for €ach: 
 ċ@ - count (sw@ping args) [["Sam N.","Sam Ng"],["Sam L.","Sam Li"],["Sue","Sue N.","Sue Ng"]]
 Ḣ€ - head €ach ["Sam N.","Sam L.","Sue"]

Answer 2

Python 2, 130 bytes

def f(a):n=[[x[:i]+'.'*(' 'in x[:i]<x)for i in range(x.find(' '),len(x)+1)]for x in a];print[min(x,key=sum(n,[]).count)for x in n]

Try it online!

First generates all the nicknames, as follows:

n == [
 ['John', 'John .', 'John C.', 'John Cl.', 'John Cla.', 'John Clan.', 'John Clanc.', 'John Clancy'],
 ['Julie', 'Julie .', 'Julie W.', 'Julie Wa.', 'Julie Wal.', 'Julie Walk.', 'Julie Walke.', 'Julie Walker'],
 ['John', 'John .', 'John W.', 'John Wa.', 'John Wal.', 'John Walk.', 'John Walke.', 'John Walker'],
 ['Julie', 'Julie .', 'Julie C.', 'Julie Cl.', 'Julie Cla.', 'Julie Clan.', 'Julie Clanc.', 'Julie Clancy'],
 ['Jill', 'Jill .', 'Jill D.', 'Jill De.', 'Jill Des.', 'Jill Desm.', 'Jill Desmo.', 'Jill Desmon.', 'Jill Desmond']
]

Then picks the first* one from each list that is least frequent in sum(n,[]). This will always be the first unique nickname.

Note that n includes the erroneous nicknames 'John .' etc., but they will never be picked.

(*CPython 2.7’s min does so, anyway. This code may not be portable!)

Answer 3

Ruby (削除) 165 162 161 (削除ここまで) 160 bytes

Includes 1 trailing space if only the given name is returned, eg. "John "

->a,n=0,s=0{a.group_by{|i|i[n]}.values.flat_map{|i|j=i[0];k=j.index' ';i.size<2?j.size>n ?j[0,[n+1,k].max-(s>1?0:1)]+(n>k ??.:''):j:f[i,n+=1,i.count{|l|l[n]}]}}

Try it online!

163 bytes if you don't want the trailing space

->a,n=0,s=0{a.group_by{|i|i[n]}.values.flat_map{|i|j=i[0];k=j.index' ';i.size<2?j.size>n ?j[0..[n,k-1].max-(s>1?0:1)]+(n>k ??.: ''):j:f[i,n+1,i.count{|l|l[n+1]}]}}

Try it online!