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!

Answer 1

Pyth, 13 bytes

++hAlDQ\_.-HG

Try it online.

Answer 2

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 

Answer 3

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

Answer 4

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

Answer 5

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"));
    }
}

Answer 6

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!

Answer 7

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);

Answer 8

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
}

Answer 9

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

Answer 10

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
}