String Concatenation

Question

The Challenge

Given two strings containing only lowercase letters and no spaces, the result should be the shorter string, followed by an underscore, followed by the longer string with the first instance of a character removed for each character it contains that is in the shorter string.

You can assume the strings will always be different lengths.

Test Cases:

sale salinewater Result: sale_inwater (only the first 'a' in salinewater is removed) jostling juggle Result: juggle_ostin (juggle is the shorter string) juggle juggler Result: juggle_r (a 'g' is removed for every 'g' in the shorter string)

Rules

This is code-golf, so shortest answer in bytes wins!

What does "remove the first occurrence" mean when the shorter string has duplicate characters? — Peter Taylor
– Peter Taylor, Commented Sep 21, 2016 at 7:21
Four days is very short to pick a winner, I'd suggest at least a couple of week at minimum. Also, in the future I'd suggest first posting to the Sandbox where challenges can be clarified and improved before going live. — xnor
– xnor, Commented Sep 21, 2016 at 7:21
smaller string is shorter string......sorry for my bad english — Ajay
– Ajay, Commented Sep 21, 2016 at 7:30
What should the output for juggle juggler be? juggle_r (remove for each character instance) or juggle_gr (remove for each distinct character)? — PurkkaKoodari
– PurkkaKoodari, Commented Sep 21, 2016 at 8:06
@Pietu1998 That's a good point. We should put this on hold until clarified — Luis Mendo
– Luis Mendo, Commented Sep 21, 2016 at 8:09

Community · Accepted Answer · 2020-06-17 09:04:33Z

5

Pyth, 13 bytes

++hAlDQ\_.-HG

Try it online.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 21, 2016 at 8:03

PurkkaKoodari

17.9k2 gold badges37 silver badges92 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

JavaScript (ES6), 78 75 69 Bytes

const g=(x,y)=>x[y.length]?g(y,x):[...x].map(c=>y=y.replace(c,''))&&x+'_'+y ; console.log(g.toString().length + 2); // 69 console.log(g('sale', 'salinewater')) // sale_inwater console.log(g('juggle', 'juggler')) // juggle_r console.log(g('jostling','juggle')) // juggle_ostin

Breakdown

x[y.length]?g(y,x): \\ Make sure that x is the shorter string [...x] \\ Spread string in array of characters .map(c=>y=y.replace(c,'')) \\ For each character remove its first occurence in y &&x+'_'+y \\ Concat x and changed y

dianne · Accepted Answer · 2016-09-23 16:22:39Z

2

Haskell, 56 55 bytes

import Data.List x%y|(0<$y)<(0<$x)=y%x|z<-y\\x=x++'_':z

-1 byte thanks to @xnor

edited Sep 23, 2016 at 16:22

answered Sep 23, 2016 at 5:37

dianne

1,1337 silver badges7 bronze badges

\$\begingroup\$ You can cut a byte by binding y\\x for the 0<1 otherwise guard. \$\endgroup\$

xnor
– xnor

2016-09-23 09:10:07 +00:00
Commented Sep 23, 2016 at 9:10
\$\begingroup\$ filter(`notElem`x)y is shorter than y\\x with import \$\endgroup\$

Damien
– Damien

2016-09-23 16:10:04 +00:00
Commented Sep 23, 2016 at 16:10
\$\begingroup\$ @Damien I think that would remove all occurrences of elements of x, not just the first ones. \$\endgroup\$

dianne
– dianne

2016-09-23 16:23:57 +00:00
Commented Sep 23, 2016 at 16:23
\$\begingroup\$ Oh, yes you are right. \$\endgroup\$

Damien
– Damien

2016-09-23 16:26:16 +00:00
Commented Sep 23, 2016 at 16:26

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Community · Accepted Answer · 2020-06-17 09:04:33Z

Java 7, 262 bytes

import java.util.*;String c(String z,String y){int i=0,l=y.length();if(z.length()>l)return c(y,z);List x=new ArrayList();for(;i<l;x.add(y.toCharArray()[i++]));for(Object q:z.toCharArray())x.remove(q);String r="";for(i=0;i<x.size();r+=x.get(i++));return z+"_"+r;}

Can probably be golfed some more by just using arrays without the lists..

Ungolfed & test cases:

Try it here.

import java.util.*; class M{ static String c(String z, String y){ int i = 0, l = y.length(); if(z.length() > l){ return c(y, z); } List x = new ArrayList(); for(; i < l; x.add(y.toCharArray()[i++])); for(Object q : z.toCharArray()){ x.remove(q); } String r = ""; for(i = 0; i < (x.size()); r += x.get(i++)); return z+"_"+r; } public static void main(String[] a){ System.out.println(c("sale", "salinewater")); System.out.println(c("jostling", "juggle")); System.out.println(c("juggle", "juggler")); } }

Output:

sale_inwater juggle_ostin juggle_r

Community · Accepted Answer · 2020-06-17 09:04:33Z

Java 8, 156 Bytes

String a(String x,String y){int l=x.length(),m=y.length();String b=l>m?x:y,s=m<l?y:x;for(char c:s.toCharArray()){b=b.replaceFirst(""+c,"");}return s+"_"+b;}

This can probably be golfed some more.

Ungolfed & test cases

interface A { static String a(String x,String y){ int l=x.length(),m=y.length(); String b=l>m?x:y,s=m<l?y:x; for(char c:s.toCharArray()){ b=b.replaceFirst(""+c,""); } return s+"_"+b; } static void main(String[]a) { System.out.println(a("sale","salinewater")); System.out.println(a("jostling","juggle")); System.out.println(a("juggle","juggler")); } }

Community · Accepted Answer · 2017-04-13 12:39:03Z

1

Ruby, 65 bytes

->a,b{a,b=b,a if a.size>b.size;a.chars.map{|e|b.sub!e,""};a+?_+b}

ungolfed

->a, b{ a, b = b, a if a.size > b.size a.chars.map { |e| b.sub! e, "" } a + ?_ + b }

61 bytes(in case the argument is an array of strings)

->a{a.sort_by!(&:size);a[0].chars.map{|c|a[1].sub!c,""};a*?_‌}

Thank you, m-chrzan!

edited Apr 13, 2017 at 12:39

CommunityBot

1

answered Sep 23, 2016 at 10:45

cia_rana

4812 silver badges5 bronze badges

1

\$\begingroup\$ gsub! doesn't work here - you're supposed to remove the first occurance of each letter. Fortunately, sub!, which does exactly that, is a byte shorter. \$\endgroup\$

m-chrzan
– m-chrzan

2016-09-23 14:23:11 +00:00
Commented Sep 23, 2016 at 14:23
1

\$\begingroup\$ Also, I think you need to name the lambda in this case, since you do end up calling it. However, doing a,b=b,a if a.size>b.size to exchange the strings is non-recursive, and saves you another byte. \$\endgroup\$

m-chrzan
– m-chrzan

2016-09-23 14:39:05 +00:00
Commented Sep 23, 2016 at 14:39
\$\begingroup\$ @m-chrzan Oh, I didn't observe carefully the question. Thanks! \$\endgroup\$

cia_rana
– cia_rana

2016-09-23 15:00:14 +00:00
Commented Sep 23, 2016 at 15:00
\$\begingroup\$ 61 bytes, input as an array of strings: ->a{a.sort_by!(&:size);a[0].chars.map{|c|a[1].sub!c,""};a*?_} \$\endgroup\$

m-chrzan
– m-chrzan

2016-09-23 15:01:20 +00:00
Commented Sep 23, 2016 at 15:01
\$\begingroup\$ @m-chrzan Great! I added your answer. \$\endgroup\$

cia_rana
– cia_rana

2016-09-23 15:14:18 +00:00
Commented Sep 23, 2016 at 15:14

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Jörg Hülsermann · Accepted Answer · 2016-09-23 09:04:42Z

PHP, 154 Bytes

list($f,$s)=strlen($b=$argv[2])<strlen($a=$argv[1])?[$b,$a]:[$a,$b];foreach(str_split($f)as$x)$s=preg_replace("#(.*?)".$x."(.*)#","$1$2",$s);echo$f."_$s";

Instead of $s=preg_replace("#(.*?)".$x."(.*)#","$1$2",$s); you can also use if($z=strstr($s,$x))$s=strstr($s,$x,1).substr($z,1);

Billywob · Accepted Answer · 2016-09-23 11:51:12Z

R, 161 bytes

This turned out to be much longer than I expected, albeit, string manipulation is usually tedious in R. I feel that this should easily be golfable by just using another approach.

function(x,y){s=strsplit;if(nchar(x)>nchar(y)){v=y;w=x}else{v=x;w=y};v=s(v,"")[[1]];w=s(w,"")[[1]];j=0;for(i in v){j=j+1;if(i==w[j])w[j]=""};cat(v,"_",w,sep="")}

Ungofled

function(x,y){ s=strsplit # alias for stringsplit if(nchar(x)>nchar(y)){v=y;w=x} # assign v/w for the short/long strings else{v=x;w=y} v=s(v,"")[[1]] # split short string into vector w=s(w,"")[[1]] # split long string into vector j=0 for(i in v){ # for each char in short string, check j=j+1 # if is equal to corresponding char in if(i==w[j])w[j]="" # long, replace long with "" if true } cat(v,"_",w,sep="") # insert _ and print }

Community · Accepted Answer · 2020-06-17 09:04:33Z

0

Python 2, 81 72 bytes

a,b=sorted(input(),key=len) for c in a:b=b.replace(c,"",1) print a+"_"+b

Try it online

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 22, 2016 at 20:48

mbomb007

23.6k7 gold badges66 silver badges143 bronze badges

2

\$\begingroup\$ I think you can save 9 bytes by replacing the first two lines with a,b=sorted(input(),key=len) \$\endgroup\$

dianne
– dianne

2016-09-23 08:07:38 +00:00
Commented Sep 23, 2016 at 8:07

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corvus_192 · Accepted Answer · 2016-09-23 15:43:31Z

Scala, 78 bytes

def f(a:String,b:String):String={if(a.size>b.size)f(b,a)else a+"_"+(b diff a)}

Explanation:

def f(a:String,b:String):String={ //define a method f which has two String parameters and returns a String //(because it's recursive, scala can't figure out the return type) if (a.size > b.size) //make sure that a is the shorter string f(b, a) else a+"_"+(b diff a) //`b diff a` removes all elements/chars of a from b }

Stack Exchange Network

String Concatenation

10 Answers 10

Pyth, 13 bytes

JavaScript (ES6), 78 75 69 Bytes

Breakdown

Haskell, 56 55 bytes

Java 7, 262 bytes

Java 8, 156 Bytes

Ruby, 65 bytes

ungolfed

61 bytes(in case the argument is an array of strings)

PHP, 154 Bytes

R, 161 bytes

Ungofled

Python 2, 81 72 bytes

Scala, 78 bytes

Hot Network Questions

String Concatenation

10 Answers 10

Pyth, 13 bytes

JavaScript (ES6), 78 75 69 Bytes

Breakdown

Haskell, 56 55 bytes

Java 7, 262 bytes

Java 8, 156 Bytes

Ruby, 65 bytes

ungolfed

61 bytes(in case the argument is an array of strings)

PHP, 154 Bytes

R, 161 bytes

Ungofled

Python 2, 81 72 bytes

Scala, 78 bytes

Related

Hot Network Questions