May the first be with you

Question

Write a program to replace all occurrences of "force" with "first" and all occurrences of "first" with "force", keeping the original case for all character positions:

"ForcefoRcefOrcE" -> "FirstfiRstfIrsT"
"FirstfiRstfIrsT" -> "ForcefoRcefOrcE"

The rest of the string must stay unchanged, and so running your program twice shall return the original string:

"thirst of forces" -> "thirst of firsts" -> "thirst of forces"

Your program should work on any initial string. So as a hint, you better avoid using magic characters as intermediate representation, because if you try a three pass replacement ("force" -> "zzzzz", "first" -> "force", "zzzzz" -> "first"), it will fail on strings containing "zzzzz".

You should support the full range of characters allowed in a definition of a String by your programming language (in most cases, it's Unicode). Example, using JSON-style representation for non-printable characters (\u + 4 digits):

"\u0000\u0001\u0002\u0003the Force of the firsT"
                     |
                     V
"\u0000\u0001\u0002\u0003the First of the forcE"

Answer 1

Retina, 33 bytes

iT`\OC\E\ocetsiTSI`Ro`first|force

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Edit: Saved 5 bytes thanks to @MartinEnder for pointing out what Ro does.

Answer 2

JavaScript (ES6), 93 88 bytes

f=
s=>s.replace(/force|first/gi,s=>s.replace(/./g,c=>s[s.search(c)^1]||c,s="oicsetOICSET"))

<textarea oninput=o.textContent=f(this.value)></textarea><pre id=o>

Edit: Saved 5 bytes by optimising the unchanged letter case.

Answer 3

APL (Dyalog), 61 bytes

Requires ⎕IO←0 which is default on many systems. Can be four characters shorter using the Unicode symbol ⍠ instead of ⎕OPT .

(t←'force' 'first')⎕R{(m∊⎕A)c¨t⊃⍨~t⍳(c←819⌶)⊂m←⍵.Match}⎕OPT 1

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Answer 4

PHP, 88 Bytes

Online Versions

<?=preg_replace_callback("#first|force#i",function($t){return$t[0]^first^force;},$argn);

PHP, 110 Bytes

<?=preg_replace_callback("#first|force#i",function($t){return strtr($t[0],iIsStToOcCeE,oOcCeEiIsStT);},$argn);

Answer 5

Perl 5, 52 bytes

51 bytes of code + -p flag.

s%first|force%$&=~y/oceOCEistIST/istISToceOCE/r%eig

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Nothing too crazy going on. Find the occurrences of force and first non-case-sensitive (s%force|first%%gi), and then transliterates the characters to convert one to the other.

Answer 6

Java 10, 318 310 284 279 277 bytes

String c(String s){var x=s.toLowerCase();int i=x.indexOf("force")+1,j=x.indexOf("first")+1,t=i>0&j>i|j<1?1:-1;return-j<i?s.substring(0,i=t<0?j:i)+(char)(s.charAt(i++)-t*6)+s.charAt(i++)+(char)(s.charAt(i++)+t*16)+(char)(s.charAt(i++)+t*15)+c(s.length()>i?s.substring(i):""):s;}

-28 bytes thanks to @ceilingcat.

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Explanation:

String c(String s){                     // Recursive method with String as both parameter and return-type
  var x=s.toLowerCase();                //  Temp String as lowercase of the input
  int i=x.indexOf("force")+1,           //  Index of "force" + 1 (becomes 0 if NOT present; >=1 if it is present)
      j=x.indexOf("first")+1,           //  Index of "first" + 1 (becomes 0 if NOT present; >=1 if it is present)
      t=i>0&j>i|j<1?1:-1;               //  Temp integer: -1 if "force" is found first; 1 if "first" is found first
  return-j<i?                           //  If either "force" or "first" is found:
    s.substring(0,i=t<0?j:i)            //   Return the substring before that (if any) + ('f' or 'F')
     +(char)(s.charAt(i++)-t*6)         //   + 'i'↔'o', or 'I'↔'O'
     +s.charAt(i++)                     //   + 'r' or 'R'
     +(char)(s.charAt(i++)+t*16)        //   + 's'↔'c', or 'S'↔'C'
     +(char)(s.charAt(i++)+t*15)        //   + 't'↔'e', or 'T'↔'E'
     +c(s.length()>i?s.substring(i):"") //   + a recursive call for the rest of the input-String (if any)
   :                                    //  Else:
    s;}                                 //   Return the input-String as is

Answer 7

CJam, 66 bytes

qY5m*_"force"{f{_eu}3/:z{~?}f%}:K~\"first"K.{[\]:P~@\/\f/P~@\f*\*}

Goes through every case variation of "first" and "force" and tries to split on it. If it can, it then joins it back with the reverse words.

Pseudocode:

input_chars = list(read_all_input()) # CJam: q
power = cartesian_power(2, 5) # CJam: Y4m*_
def case_variations(s): # CJam: {...}:K
    temp0 = [[i, j, upper(j)] for i, j in zip(power, s)] # CJam: f{_eu}3/
    temp1 = map(transpose, temp0) # CJam: :z
    ret = []
    for i in ret:
        for j in i: # CJam: {...}f%
            ret.append(j[1] if j[0] else j[2]) # CJam: ~?
    return ret
force_var = K("force") # CJam: "force"{...}:K~
first_var = K("first") # CJam: \"first"K
for force, first in zip(force_var, first_var): # CJam: .{...}
    current = [force, first] # CJam: [\]:P~
    input_chars = list_split(input_chars, force) # CJam: @\/
    input_chars = [list_split(i, first) for i in input_chars] # CJam: \f/
    input_chars = [list_join(i, force) for i in input_chars] # CJam: P~@\f*
    input_chars = list_split(input_chars, first) # CJam: \*

Answer 8

Jelly, 37 36 bytes

Is there is a way to use a reduce across slices of length 5 instead?

®‘©ị“Ɓu“¡Ḣƭ»
Œlœṣ¢œṣ€¢j€¢j¢Œu⁸=ŒuT¤¦

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How?

®‘©ị“Ɓu“¡Ḣƭ» - Link 1 helper that fetches the next word to use: no arguments
®            - recall value from register (initially zero)
 ‘           - increment
  ©          - place the result into the register
    “Ɓu“¡Ḣƭ» - literal dictionary compressed string list ["first","force"]
   ị         - index into (1-indexed and modular)
             - so this link first yields "first", then "force", then "first" and so on.

Œlœṣ¢œṣ€¢j€¢j¢Œu⁸=ŒuT¤¦ - Main link: list of characters, S
Œl                      - convert S to lower case
  œṣ                    - split on sublists equal to:
    ¢                   -   call the last link (1) as a nilad ("first")
     œṣ€                - split €ach on sublists equal to:
        ¢               -   call the last link (1) as a nilad ("force")
         j€             - join €ach with:
           ¢            -   call the last link (1) as a nilad ("first")
            j           - join with:
             ¢          -   call the last link (1) as a nilad ("force")
                      ¦ - apply a link to sparse indices:
              Œu        -   convert to upper case
                     ¤  -   nilad followed by link(s) as a nilad:
                ⁸       -     chain's left argument, S
                  Œu    -     convert to upper case
                 =      -     equal to S? (vectorises)
                    T   -     truthy indexes (indexes at which input is upper case)

Answer 9

MATL, 47 bytes

5W:qB!"o['first';'force']@!32*-cZ}_Zt5M_6MoZt|c

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This uses negative values as the intermediate step, and after the two passes it takes the absolute value.

Answer 10

Pyth, 36 35 bytes

K"first"srVjJ"force"mjKcdJcr0QKqVr1

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Pyth is not especially good at string manipulations.

Answer 11

Flex (lexer), 72 bytes

%%
 #define x(a) yytext[a]^=
(?i:first|force) x(1)6;x(3)16;x(4)17;ECHO;

To compile and run:

flex first.l
gcc lex.yy.c -lfl # -ll on Macs, apparently
./a.out

Answer 12

Python 2, 171 bytes

I wanted to try to do this using built-ins, but it can't beat the messy method with all the splitting and zipping.

import re,string as g
def f(s):f="istISTECOeco";l=re.split("(first|force)",s,0,re.IGNORECASE);l[1::2]=[t.translate(g.maketrans(f,f[::-1]))for t in l[1::2]];print"".join(l)

I think it's pretty clear what I'm doing here. Split the string on instances of first and force (case-insensitive), replace those instances with versions translated using str.translate, and join it back into a string again.

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Answer 13

Python 2.7, 173 165 bytes

8 bytes saved by quintopia

This one got gross:

lambda S:`[(t[0],t[0].upper())[t[1]]for t in zip("".join("first".join(s.replace("first","force")for s in S.lower().split("force"))),[l.isupper() for l in S])]`[2::5]

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Breaking it down step by step:

1. S.lower().split("force"): take the string, unify to lowercase, split into substrings separated by "force"
2. s.replace("first","force")for s in <STEP 1>: Replace all "first"'s with "force"
3. _`.join("first".join(<STEP 2>)`[2::5]`_: replace all "force"'s with "first" by recombining the "force" delineated substrings with "first" and rejoin into single string (underscores added to get tick marks correct)
4. zip(<STEP 3>,[(2,1)[l.isupper()]for l in S]): zip each character of replaced phrase with case encoding of original string (2 for lowercase, 1 for uppercase)
5. _`[(t[0],t[0].upper())[t[1]==1]for t in <STEP 4>]`[2::5]`_: Restore original casing, converts list to string (underscores added to get tick marks correct)

Answer 14

C (clang), 201 183 226 214 bytes

Had some bugs... Still needs to be golfed down quite a lot

(saved 12 thanks to ceilingcat)

char*s,*p,*q;main(i,v)char**v;{puts(s=v[1]);do{p=strcasestr(s,"first");q=strcasestr(s,"force");if(p&&(!q|p<q))p[1]+=6,p[3]-=16,p[4]-=15;else if(q)q[1]-=6,q[3]+=16,q[4]+=15;s=p&&(!q|p<q)?p:q;}while(s++);puts(v[1]);}

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Answer 15

C# 273 bytes

string c(string s){var x=s.ToLower();int i=x.IndexOf("force")+1,j=x.IndexOf("first")+1,t=i>0&j>i?0:j>0?1:0;return i>0|j>0?s.Substring(0,t>0?(i=j):i)+(char)(s[i++]-(t>0?-6:6))+s[i++]+(char)(s[i++]+(t>0?-16:16))+(char)(s[i++]+(t>0?-15:15))+c(s.Length>i?s.Substring(i):""):s;}

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Direct port of Kevin Cruijssen's Java answer, turns out when it comes to getting the char in a string at a given index, C# is much golfier than java (s[i++] instead of s.charAt(i++))

Answer 16

Japt, 41 bytes

r"first|force"_d"i1o s1c t1e"¸m²®+ZuÃq}'i

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This would be considerably shorter if Japt had a sane transliterate function...

Alternate version:

r"first|force"_d"io sc te"¸®¬¸²Ã®+ZuÃq}'i

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Answer 17

C#, 235 chars

string a(string s){var l=s.ToLower();int f=l.IndexOf("first"),F=l.IndexOf("force"),m=f<F&f>-1?f:F>-1?F:f;return ++m>0?s.Substring(0,m)+(char)(s[m]^6)+s[m+1]+(char)(s[m+2]^16)+(char)(s[m+3]^17)+(s.Length-m>5?c(s.Substring(m+4)):""):s;}

Answer 18

Ruby, 55 bytes

gsub(/first|force/i){$&.tr(s="iIsStTEeCcOo",s.reverse)}

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Answer 19

Java, 382 bytes non-comptent

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String f(String t){String s="";for(String w:t.split(" "))if(w.equalsIgnoreCase("force")|w.equalsIgnoreCase("first"))s+=" "+w.charAt(0)+(char)(w.charAt(1)+(w.charAt(1)=='o'|w.charAt(1)=='O'?-6:6))+w.charAt(2)+(char)(w.charAt(3)+(w.charAt(3)=='c'|w.charAt(3)=='C'?16:-16))+(char)(w.charAt(4)+(w.charAt(4)=='e'|w.charAt(4)=='E'?15:-15));else s+=" "+w;return s.substring(1,s.length());}

Answer 20

C# (269 Bytes)

string s(string z){var u=z.ToUpper();var a=new[]{"FIRST","FORCE"};return String.Join("",u.Split(a,StringSplitOptions.None).Aggregate((c,n)=>c+(u.Substring(c.Length,5)==a[0]?a[1]:a[0])+n).Select((c,i)=>Char.IsLower(z[i])?Char.ToLower(c):c));}

yet another c# solution, only the second-smallest because I declared two variables and so can't use lambda syntax. oh well, I had fun. :)

explanation:

• upshift the original string, then split on "FORCE" and "FIRST".

• aggregate the results and on every split, find the five-character substring that was used to split the original string using the length so far of the string being aggregated. if it was "force" make it "first" and vice versa

• select all the characters of the newly created all caps string and check if the original string was lowercase at the same index. if yes, return lowercased character at that index in the new string, otherwise return the uppercase character