String subtraction

Question

Objective

Create a function to reverse string concatenation

Input

Two strings (alphanumeric + spaces), where one should be subtracted for the other.

• You can assume that the string to be subtracted will never be larger than the other one.

Output

The result from the subtraction

Subtraction

You should remove one string from the start or the end of another string. If the string is present in the start and in the end, you can only remove one, which one will be removed is up to you.
If the string isn't in the start or in the end, or isn't an exact match, it is an invalid subtraction and you should output the original string.

Test Cases

Valid Subtraction

'abcde','ab' -> 'cde'
'abcde','cde' -> 'ab'
'abab','ab' -> 'ab'
'abcab','ab' -> 'abc' or 'cab'
'ababcde','ab' -> 'abcde'
'acdbcd','cd' -> 'acdb'
'abcde','abcde' -> ''
'abcde','' -> 'abcde'
'','' -> ''

Invalid Subtraction (returns original string)

'abcde','ae' -> 'abcde'
'abcde','aa' -> 'abcde'
'abcde','bcd' -> 'abcde'
'abcde','xab' -> 'abcde'
'abcde','yde' -> 'abcde'

Invalid Input (don't need to be handled)

'','a' -> ''

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

Answer 1

Java 8, 46 45 44 40 bytes

-1 byte thanks to TheLethalCoder

-1 byte because I'm dumb (thanks Rod!)

-4 bytes thanks to Kevin Cruijssen

a->b->a.replaceFirst("^"+b+"|"+b+"$","")

Try it online! (includes all test cases)

A Java answer actually beats a few other practical languages. Smiles. (and now it beats JS!)

Answer 2

Brachylog (Try It Online!), 12 bytes

~cpĊh.∧Ċtw|w

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Takes the string to subtract from from standard input, and the string to subtract as a command line argument.

Explanation

~cpĊh.∧Ċtw|w
~c            Split {the input} into pieces
  p           and (possibly) rearrange those pieces
   Ċ          such that there are two pieces
    h         and the first
     .        matches the command line argument
      ∧       then
         w    print
        t     the last
       Ċ      piece.
          |   If all else fails,
           w  print {the input}.

Answer 3

JavaScript (ES6), 41 bytes

s=>t=>s.replace(eval(`/^${t}|${t}$/`),'')

Takes input via currying syntax, i.e. f("abab")("ab").

Answer 4

Retina, 21 bytes

1 byte thanks to Martin Ender.

(.*);(\1|(.*)\1$|)
$3

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Answer 5

JavaScript (ES6), 76 70 45 41 bytes

s=>t=>s.replace(RegExp(`^${t}|${t}$`),"")

---

Try It

f=
s=>t=>s.replace(RegExp(`^${t}|${t}$`),"")
o.innerText=f(i.value="abcde")(j.value="ab")
i.oninput=j.oninput=_=>o.innerText=f(i.value)(j.value)

<input id=i><input id=j><pre id=o>

Answer 6

Perl 6, 21 bytes

->$_,$b {S/^$b|$b$//}

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Expanded:

-> $_, $b {   # pointy block lambda

  S/          # Str replace and return (implicitly against ｢$_｣)

  |   ^ $b    # starting with the second argument
  |     $b $  # or ending with the second argument

  //          # replace with nothing.

}

Answer 7

Japt, 11 9 bytes

k"^|$"¬qV

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Answer 8

TI-Basic (TI-84 Plus CE), 63 bytes

Prompt Str0,Str2
inString(Str0,Str2
If Ans
sub(Str0,1,Ans-1)+sub(Str0,Ans+length(Str2),length(Str0)-Ans+1-length(Str2→Str0
Str0

Answer 9

Mathematica, 162 bytes

(c=Characters;a=c@#;b=c@#2;l=Length;t={};If[l@Intersection[a,b]==l@b,If[MemberQ[Partition[a,l@b,1],b],t=a;Table[t=DeleteCases[t,b[[i]],1,1],{i,l@b}]],t=a];""<>t)&

test input style ["abcde","ab"]

Answer 10

Python, 69 68 64 57 51 45 bytes

This ended up being a completely different solution with Regex.

Thanks to Value Ink for -2 bytes!
and Felipe Nardi Batista for the massive -6 bytes!

import re
lambda s,c:re.sub(c+'$|^'+c,'',s,1)

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Answer 11

Bash, 66 61 49 bytes

case $1 in *$2)echo ${1%$2};;*)echo ${1#$2};;esac

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less golfed:

a=$1;
case $1 in 
    *$2)  c=${a%$2};;       
    $2*)  c=${a#$2};;
      *)  c=$1;;
esac;
echo $c

Uses case to test begining or end, and array prefix/suffix (% / #) substraction

Answer 12

APL (Dyalog), 31 30 bytes

-1 thanks to Zacharý.

This actually uses reverse (i.e. the inverse of) concatenation! Takes original string as left argument, and what to subtract as right argument.

{0::⍺{0::⍺⋄,∘⍵⍣¯1⊢⍺}⍵⋄⍵,⍣¯1⊢⍺}

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Ungolfed:

{
    0::⍺{         ⍝ if an error happens, apply the following function on the arguments
        0::⍺          ⍝ if an error happens, return the left argument unmodified
        ,∘⍵⍣¯1⊢⍺      ⍝ inverse-append right argument on left argument
        }⍵
    ⍵,⍣¯1⊢⍺      ⍝ inverse-prepend the right argument on the left argument
}

Legend:

{…} anonymous function

⍺ left argument of the current function

⍵ right argument of the current function

0::… if any error happens, execute this, else…

⍣¯1⊢ inverse

,∘⍵ concatenate ⍵ on the right

⍵, concatenate ⍵ on the left

Answer 13

PHP, 54 Bytes

[,$x,$y]=$argv;echo preg_replace("#^$y|$y$#","",$x,1);

Testcases

Answer 14

Python 2, 68 bytes

def f(s,c):v=len(c);print[s[v:],s[:-v],s][[s[:v],s[-v:],c].index(c)]

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Answer 15

Haskell, 49 bytes

f s a b|s==b=a|a/=b,h:t<-a=f(s++[h])t b|1<3=s
f""

Try it online! Usage: f"" "abcdef" "ab". Alternatively, define (-)=f""and use like "abcdef" - "ab".

This regex-free solution works by recursively splitting the string in all its pre- and postfixes and checking whether the string to be substracted matches one of them.

Answer 16

Python 2, 72 65 bytes

a,b=input()
l=len(b)
print[[a,a[:-l]][b==a[-l:]],a[l:]][b==a[:l]]

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-7 bytes thanks to @FelipeNardiBatista

Answer 17

C#, 88 bytes

s=>r=>s.StartsWith(r)?s.Substring(r.Length):s.EndsWith(r)?s.Substring(0,s.IndexOf(r)):s;

Compiles to a Func<string, Func<string, string>>.

Answer 18

Ruby (lambda expression), 29 bytes

->a,b{a.sub /^#{b}|#{b}$/,""}

Yay for regex interpolation! Requires regex-safe subtrahends, but that's okay as per the challenge.

Answer 19

Tcl, 37 bytes

proc s {a b} {regsub "^$b|$b$" $a {}}

Try it online! (now running all tests)

Tcl is straightforward. proc s {a b} defines a function named s which takes parameters a and b. regsub substitutes {}, which is an empty string, for the value of b when it's at the start or end of a. The return is implicit.

Answer 20

C, 96 bytes

It's common knowledge that string manipulation in C is cumbersome, as an extension golfing would be borderline masochistic. Sounds alright to me.

f(a,b,t,l)char**a,*b,*t;{t=*a;l=strlen(b);bcmp(t,b,l)?bcmp(t+=strlen(t)-l,b,l)||(*t=0):(*a+=l);}

One of the less readable programs I've written. Takes two inputs (despite how the function looks), a char** pointing to the string to deconcatenate and a char* which is the string to remove. The input pointer is edited in place and becomes the output (who cases about memory leaks anyway).

Example usage:

char *a = malloc(6);
strcpy(a, "abcde");
char *b = malloc(4);
strcpy(b, "abc");
f(&a,b);
printf("%s\n", a); // "de"

Answer 21

AWK, 21 32 bytes

{sub("^"$2"|"$2"$",z,$1);$0=$1}1

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Original submission naively replaced text within first string, not just at beginning or end.

{sub($2,z,$1);$0=$1}1

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Originally tried without the braces, but it required tricks to print out empty lines and or no-matches which ended up adding more bytes than this version.

Answer 22

R, 20 42 41 bytes

pryr::f(sub(sprintf('^%s|%s$',b,b),'',a))

-1 byte thanks to MickyT!

Returns an anonymous function (which has arguments in the order b,a). Computes the string difference a-b. sub is a simple substitution that swaps the first occurrence of the pattern with, in this case, the empty string ''. Constructs the regex with sprintf to match only at the beginning and end of string. Requires the pryr package to be installed.

On the TIO link, uses the more verbose function(a,b) definition for the function for four more bytes.

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Answer 23

Common Lisp, 121 bytes

(lambda(x y)(cond((equal(#1=subseq x 0 #3=(length y))y)(#1#x #3#))((equal(#1#x #2=(-(length x)#3#))y)(#1#x 0 #2#))(t x)))

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The usual wordy Common Lisp!

Ungolfed version:

(defun f(x y)
  (cond ((equal (subseq x 0 (length y)) y)               ; if x starts with y
         (subseq x (length y)))                          ; return rest of x
        ((equal (subseq x (- (length x) (length y))) y)  ; if x ends with x
         (subseq x 0 (- (length x) (length y))))         ; return first part of x
        (t x)))                                          ; else return x

Answer 24

Kotlin, 91 bytes

{a,b->val v=b.length
if(a.startsWith(b))a.drop(v)else if(a.endsWith(b))a.dropLast(v)else a}

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Answer 25

Powershell, 34 40 bytes

+6 bytes when Invalid Subtraction test cases added

param($s,$t)$s-replace"^$t(?!.*$t$)|$t$"

Comment:

The regexp expression ^$t|$t$ does not work as expected: it replaces both matches instead one (flag g always on). So, we are forced to use the negative lookahead group.

Test script:

$f = {
    param($s,$t)$s-replace"^$t(?!.*$t$)|$t$"
}

@(
    ,('abcde','ab', 'cde')
    ,('abcde','cde', 'ab')
    ,('abab','ab', 'ab')
    ,('abcab','ab', 'abc', 'cab')
    ,('ababcde','ab', 'abcde')
    ,('acdbcd','cd', 'acdb')
    ,('abcde','abcde', '')
    ,('abcde','', 'abcde')
    ,('','', '')

    ,('abcde','ae', 'abcde')
    ,('abcde','aa', 'abcde')
    ,('abcde','bcd', 'abcde')
    ,('abcde','xab', 'abcde')
    ,('abcde','yde', 'abcde')

    ,('','a', '')
) | % {
    $s,$t,$e = $_
    $r = &$f $s $t
    "$($r-in$e): $r"
}

Output:

True: cde
True: ab
True: ab
True: abc
True: abcde
True: acdb
True:
True: abcde
True:

Answer 26

QBIC, 57 bytes

Whegh, this is a mess in QBIC/QBasic...

B=@ `+B┘x=instr(;,;)~x|?_t_sB,x-1|+_sB,x+_lC|,_lB|||\?B

B=@ `+B          Prepend a string to B$. Thisis a hack to avoid errors with 
                 removing substrings stating at index 1
┘                Line-break in QBasic output
       (;,;)     Read the string (B$) and the to-be-removed substring (C$)
x=instr          And make x to be the starting index of the first C$ in B$
~x|              IF X <> 0 (ie C$ is present in B$)
?                PRINT
 _t                trimmed version (drops the prepended space)
  _sB,x-1|+        of a substring from 1 to x (the start of C$) -1
  _sB,x+_lC|,_lB   and the rest of the string, starting after C$
                     _l takes the length of a string
  |||              End TRIM, end Substring, end Length
\?B              When missing C$, just print B$

Answer 27

Lua, 71 65 bytes

Accepting suggestions

a,b=...p='(.*)'print(a:match('^'..b..p)or a:match(p..b..'$')or a)

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Answer 28

I initially misread the instructions. Thanks, Ørjan Johansen for pointing out my mistake!

PowerShell, 46 51 bytes

Function F($o,$a){([regex]"^$a").replace($o,'',1);}

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Answer 29

Excel, 129 bytes

=IFERROR(IF(FIND(B1,A1)=1,SUBSTITUTE(A1,B1,"",1),IF(FIND(B1,A1,LEN(A1)-LEN(B1))>LEN(A1)-LEN(B1),LEFT(A1,LEN(A1)-LEN(B1)),A1)),A1)

Answer 30

sed, 56 53 bytes

s/^(.*),(.*)\1$/\2/
s/^(.*),\1(.*)$/\2/
s/.*,(.*)/\1/

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