How fast am I vrooooming?

Question

Introduction

My car speedometer was hacked! Instead of showing me how fast i'm driving, it just shows: "Vroooom!" Please help me know how fast i'm going.

Challenge

Take a string as input, and check if it matches the regex /^[Vv]ro*m!$/m. In English that means any line of the string must start with a capital or lowercase v, then a lowercase r, then any amount (including zero) of the lowercase letter o, then the exact string m!. There may be other lines, but the Vroom string must be on it's own line.

If you find a match, then you must count the amount of o's in the Vroom string and output it. If you don't find a match however, you should output any default value that can't be outputted otherwise (like -1 or an empty string)

Reminders

• I/O is in any reasonable format
• Standard loopholes are banned
• Submission may be a full program or function
• Input is guaranteed to only have 1 Vroom string

Scoring

This is code-golf, so the shortest code in bytes wins. However, I will not mark any answer as accepted.

Test cases

Input

Vrom!

Output 1

Input

vrooooooom!

Output 7

Input

Hello, Vroom!

Output (none)

Input

Foo bar boo baz
Vrooom!
hi

Output 3

Input

Vrm!ooo

Output (none)

Input

PPCG puzzlers pie

Output (none)

Input

hallo
vROOOm!

Output (none)

Answer 1

sed 4.2.2, 20 bytes

-nr options required at the command-line.

s/^[Vv]r(o*)m!$/\1/p

This outputs the speed in unary as the number of os.

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Answer 2

Python 2, 56 53 bytes

lambda x:len(re.search('^[Vv]r(o*)m!$',x,8).group(1))

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Basic regex and grouping, uses re.MULTILINE flag (which has a value of 8) and re.search to ensure it works for multiline inputs. Raises an exception when no match is found. Thanks to @ovs for the -3 bytes from (re.M == 8) tip.

Answer 3

R, 62 60 58 44 bytes

nchar(grep("^[Vv]ro*m!$",readLines(),v=T))-4

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@Giuseppe with 14 bytes golfed.

Original approach with explanation:

function(x)attr(el(regexec("(?m)[Vv]r(o*)m!$",x,,T)),"m")[2]

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R has seven pattern matching functions. The more commonly used ones are grep, grepl, and sub, but here's a nice use for regexec.

regexec gives you a bunch of things, one of which is the length of any captured substring, in this case the (o*) part of the multiline regex.

The attr(el .... "m")[2] stuff is a golfy way to get the desired number.

Returns NA if there is no match.

Answer 4

JavaScript (Node.js), 41 bytes

a=>(l=/[Vv]r(o*)m!/.exec(a))&&l[1].length

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Answer 5

Powershell, 62 58 53 48 bytes bytes

"$($args|sls '(?m-i)^[Vv]ro*m!$'|% M*)".Length-4

returns numbers of o in a first Vroom!, or -4 if Vroom! not found.

Notes:

• sls is alias for Select-String;
• (?m-i) inside regexp means:
  • Use multiline mode. ^ and $ match the beginning and end of a line, instead of the beginning and end of a string.
  • Use case-sensitive matching
• |% M* is shortcut for the property Matches, which gives a first match because we don't use -AllMatches parameter.

Test script:

$f = {

"$($args|sls '(?m-i)^[Vv]ro*m!$'|% M*)".Length-4

}

@(
,('Vrom!',1)
,('vrooooooom!',7)
,('Hello, Vroom!',-4)
,('Foo bar boo baz
Vrooom!
hi',3)
,('Vrm!ooo',-4)
,('PPCG puzzlers pie',-4)
,('hallo
vROOOm!',-4)
,('
Vrooom!
Vrooooom!
',3)        # undefined behavior.
,('vrm!',0) # :)
) | % {
    $n,$expected = $_
    $result = &$f $n
    "$($result-eq$expected): $result"
}

Output:

True: 1
True: 7
True: -4
True: 3
True: -4
True: -4
True: -4
True: 3
True: 0

Answer 6

SNOBOL4 (CSNOBOL4), 99 82 bytes

I	INPUT POS(0) ('V' | 'v') 'r' ARBNO('o') @X 'm!' RPOS(0)	:F(I)
	OUTPUT =X - 2
END

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Pretty direct SNOBOL translation of the spec, reads each line until it finds one that matches ^[Vv]ro*m!$, then outputs the length of the o* bit.

Enters an infinite loop if no Vroom! can be found.

Answer 7

PowerShell, 83 bytes

($args-split"`n"|%{if(($x=[regex]::Match($_,"^[Vv]ro*m!$")).success){$x}}).length-4

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-splits the input $args on `newlines, pipes those into a for loop. Each iteration, we check whether our [regex]::Match is a .success or not. If so, we leave $x (the regex results object) on the pipeline. Outside the loop, we take the .length property -- if it's the regex results object, this is the length of the match (e.g., "Vroom!" would be 6); if it's not a regex results object, the length is zero. We then subtract 4 to remove the counts for the Vrm! and leave that on the pipeline. Output is implicit. Outputs a -4 if no match is found.

Answer 8

Retina, 21 bytes

L$m`^[Vv]r(o*)m!$
$.1

Try it online! Explanation: L lists matches, so if the regex fails to match then output is empty. $ causes the result to be the substitution rather than the match. m makes it a multiline match (the equivalent to the trailing m in the question). The . in the substitution makes it output the length of the capture in decimal.

Answer 9

Perl 6, 26 bytes

{-!/^^[V|v]r(o)*m\!$$/+$0}

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Answer 10

C (gcc), 188 183 bytes

Why use regexes when you can use a state machine instead? :-)

a,b;f(char*s){for(a=b=0;a<5;s++){!a&*s==86|*s=='v'?a++:a==1&*s=='r'?a++:a==2?*s-'o'?*s-'m'?0:a++:b++:a==3&*s==33?a++:!*s&a==4?a++:*s-10?(a=-1):a-4?(a=0):a++;if(!*s)break;}s=a<5?-1:b;}

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Answer 11

Stax, 16 bytes

∞╠mQ╛3mQ->n▀÷↕┐ò

Run and debug it

Answer 12

Japt, 18 bytes

fè`^[Vv]*m!$` ®èo

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Saved a byte by taking input as an array of lines.

Includes an unprintable character between ] and *.

Explanation:

fè                   Get the line(s) that match
  `^[Vv]*m!$`          The provided RegEx with a little compression
              ®èo    Count the number of "o" in that line if it exists

Answer 13

Haskell, 75 71 69 bytes

f s=[length n-2|r<-lines s,n<-scanr(:)"m!"$'o'<$r,v<-"Vv",r==v:'r':n]

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No regex. Instead builds all valid Vrooom!-strings up to a sufficient length and compares the lines of the input against them, collecting the number of os in a list. Thus for invalid inputs an empty list is returned.

Answer 14

C (gcc), 138 124 bytes

Here's the boring regex way.

#include<regex.h>
f(char*s){regmatch_t m[9];regcomp(m+2,"^[Vv]r(o*)m!$",5);s=regexec(m+2,s,2,m,0)?-1:m[1].rm_eo-m[1].rm_so;}

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Answer 15

C (gcc), 104 100 bytes

s;main(c,n){for(;gets(&s);sscanf(&s,"v%*[o]%nm%c%c",&n,&c,&c)-1||c-33?:printf("%d",n-2))s=s-768|32;}

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Output the n for each valid line, exactly in the requirement(nothing if no valid line, the n if exactly one)

int s; // Use as a char[]
main(c){
  while(gets(&s)) {
    s=s-768|32; // byte 0: 'V'=>'v'; byte 1: 'r'=>'o', 'o'=>'l'
    if (sscanf(&s,"v%[o]m%c%c",&s,&c,&c)==2 && c=='!') {
    // The last '%c' get nothing if it's EndOfLine
      printf("%d",strlen(&s)-1))
    }
  }
}

Answer 16

Pyth, 28 bytes

IJjb.nm:rd0"^vro*m!$"1.z/J\o

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Answer 17

Pyth, 20 bytes

/R\o:#"^Vro*m!$"1cQb

Outputs as a list containing only the number of 'o's, or an empty list if there's no Vroom.
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Explanation

/R\o:#"^Vro*m!$"1cQb
                 cQb  Split on newlines.
    :#"^Vro*m!$"1     Filter the ones that match the regex.
/R\o                  Count the `o`s in each remaining element.

Answer 18

C# (.NET Core), 134 122 bytes

for(var a="";a!=null;a=Console.ReadLine())if(new Regex(@"^[Vv]ro*m!$").Match(a).Success)Console.Write(a.Count(x=>x=='o'));

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-12 bytes: moved null check into for loop and removed brackets

Ungolfed:

for(var a = ""; a != null; a = Console.ReadLine())  // initialize a, and while a isn't null, set to new line from console
    if(new Regex(@"^[Vv]ro*m!$")                        // set regex
                        .Match(a).Success)              // check if the line from the console matches
        Console.Write(a.Count(x => x == 'o'));              // write the number of 'o's to the console

Answer 19

Pip, 21 bytes

a~,`^[Vv]r(o*)m!$`#$1

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Match regex ^[Vv]r(o*)m!$ in multiline mode; output length of capture group.

Answer 20

C++, MSVC, 164 159 bytes

-5 bytes thanks to Zacharý

It compiles even with the regex header only

#include<regex>
using namespace std;int f(vector<string>i){smatch m;for(auto&e:i)if(regex_match(e,m,regex("^[Vv]ro*m!$")))return m[0].str().size()-4;return-1;}

Tests :

std::cout << "Vrom!" << " -> " << f({ "Vrom!" }) << '\n';
std::cout << "vrooooooom!" << " -> " << f({ "vrooooooom!" }) << '\n';
std::cout << "Hello, Vroom!" << " -> " << f({ "Hello, Vroom!" }) << '\n';
std::cout << "Foo bar boo baz \\n Vrooom! \\n hi" << " -> " << f({ "Foo bar boo baz", "Vrooom!", "hi" }) << '\n';
std::cout << "Vrm!ooo" << " -> " << f({ "Vrm!ooo" }) << '\n';
std::cout << "PPCG puzzlers pie" << " -> " << f({ "PPCG puzzlers pie" }) << '\n';
std::cout << "hallo \\n vROOOm!" << " -> " << f({ "hallo", "vROOOm!" }) << '\n';

Answer 21

sfk, 94 bytes

xex -i -case "_[lstart][char of Vv]r[chars of o]m![lend]_[part 4]o_" +linelen +calc "#text-1" 

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Gives -1 when you're not vrooming.

Answer 22

JavaScript, 90 73 61 bytes

_=>_.replace(/^[Vv]r(o*)m!$|[^\1]/mg,(m,a)=>a||'').length||-1

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Replace characters that are not captured at (o*) with empty string, return length of string containing only "o" or -1 if resulting string is empty.

Answer 23

Ruby, 28 29 bytes

p$_[/^[vV]r(o*)m!$/].count ?o

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Multi-line strings require three more bytes. I'm not sure if that's a hard requirement. If so, I'll update this.

->l{l[/^[Vv]r(o*)m!$/].count ?o}

Answer 24

Red, 104 bytes

func[s][n:""if parse/case s[opt[thru"^/"]["V"|"v"]"r"copy n any"o""m!"opt["^/"to end]][print length? n]]

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A straightforward solution. Red's parse is cool and readable, but too long compared to regex

Red []
f: func [ s ] [
    n: ""
    if parse/case s [
             opt [ thru newline ]
             [ "V" | "v" ]
             "r"
             copy n any "o"
             "m!"
             opt [ newline to end ]
    ] [ print length? n ]
]

Answer 25

J, 35 bytes

(]{~0<{.)(1{'^[Vv]r(o*)m!'rxmatch])

Returns negative 1 if the pattern doesn't match.

Answer 26

Ruby, 32 bytes

->n{n=~/^[Vv]r(o*)m!$/m;$1.size}

Matches the string to the regex, then uses Ruby's magic regex group variables to get the first group's size.

Call it like so:

x=->n{n=~/^[Vv]r(o*)m!$/m;$1.size}
x["Vrooooooooooooooooooooom!"] # returns 21

Answer 27

Clojure, 90 bytes

#(do(def a(clojure.string/replace % #"(?ms).*^[Vv]r(o*)m!$.*""$1"))(if(= a %)-1(count a)))

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This anonymous function returns the number of "o"s in the vroom string, or -1 if there is no valid vroom string.

Readable version

(fn [s]
  (def a (clojure.string/replace s #"(?ms).*^[Vv]r(o*)m!$.*" "$1"))
  (if (= a s) -1 (count a)))

Explanation

#"(?ms).*^[Vv]r(o*)m!$.*" ; This regex matches any string that contains a valid vroom string. The first capturing group contains only the "o"s in the vroom string
(clojure.string/replace s #"(?ms).*^[Vv]r(o*)m!$.*" "$1") ; Replaces a match of the above regex with its first capturing group. The resulting string is stored in the variable a
(if (= a s) -1 (count a))) ; a equals s if and only if there is no valid vroom string, so if a equal s we return -1. If there is a valid vroom string, a contains only the "o"s from the vroom string, so we return the length of a

Answer 28

perl -nE, 35 bytes

$s=length$1if/^[Vv]r(o*)m!$/}{say$s

This uses the Eskimo greeting (}{) which abuses the a quick on how the -n option is dealt with by perl.

Answer 29

Java 8, 109 bytes

s->{int r=-1;for(var l:s.split("\n"))r=l.matches("[Vv]ro*m\\!")?l.replaceAll("[^o]","").length():r;return r;}

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Explanation:

s->{                             // Method with String parameter and integer return-type
  int r=-1;                      //  Result-integer, starting at -1
  for(var l:s.split("\n"))       //  Loop over the lines:
    r=l.matches("[Vv]ro*m\\!")?  //   If the current line matches the regex:
       l.replaceAll("[^o]","").length()
                                 //    Change `r` to the amount of "o"'s in it
      :                          //   Else:
       r;                        //    Leave the result `r` unchanged
  return r;}                     //  Return the result

Answer 30

05AB1E, 39 37 bytes

|ʒć„VvsåsÁÁD…m!rÅ?s¦¦¦Ù'oså)P}Dgi`'o¢

Although 05AB1E is a golfing languages, regex-based challenges are definitely not its strong suite, since it has no regex-builtins.

Outputs [] if no match was found.

Try it online or verify all test cases.

Explanation:

|              # Get the input split by newlines
 ʒ             # Filter it by:
  ć            #  Head extracted: Pop and push the remainder and head-character
               #   i.e. "vrm!" → "rm!" and "v"
               #   i.e. "Vroaom!" → "roaom!" and "V"
   „Vvså       #  Is this head character in the string "Vv"?
               #   i.e. "v" → 1 (truthy)
               #   i.e. "V" → 1 (truthy)
  s            #  Swap so the remainder is at the top of the stack again
   ÁÁ          #  Rotate it twice to the right
               #   i.e. "rm!" → "m!r"
               #   i.e. "roaom!" → "m!roao"
     D         #  Duplicate it
      …m!rÅ?   #  Does the rotated remainder start with "m!r"?
               #   i.e. "m!r" → 1 (truthy)
               #   i.e. "m!roao" → 1 (truthy)
  s¦¦¦         #  Remove the first three characters from the duplicated rotated remainder
               #   i.e. "m!r" → ""
               #   i.e. "m!roao" → "oao"
      Ù        #  Uniquify, leaving only distinct characters
               #   i.e. "" → ""
               #   i.e. "oao" → "oa"
       'oså   '#  Is this uniquified string in the string "o"?
               #   i.e. "" → 1 (truthy)
               #   i.e. "oa" → 0 (falsey)
  )P           #  Check if all three checks above are truthy
               #   i.e. [1,1,1] → 1 (truthy)
               #   i.e. [1,1,0] → 0 (falsey)
 }             # Close the filter
  D            # After the filter, duplicate the list
   gi          # If its length is 1:
               #   i.e. ["vrm!"] → 1 (truthy)
               #   i.e. [] → 0 (falsey)
     `         #  Push the value in this list to the stack
               #   i.e. ["vrm!"] → "vrm!"
      'o¢     '#  And count the amount of "o" in it (and output implicitly)
               #   i.e. "vrm!" → 0
               # (Implicit else:)
               #  (Implicitly output the duplicated empty list)
               #   i.e. []

Answer 31

16-bit x86 assembly code, 35 bytes

l1: lodsb          ;fetch a byte
    cbw            ;zero AH to allow returning 0
    cwd            ;zero counter
l2: and   al, 0dfh ;lower- to upper-case
    je    l4       ;exit if end of string
    cmp   al, 'V'
    jne   l1
    lodsb          ;fetch a byte
    cmp   al, 'r'
    jne   l2       ;if not 'r', branch to try 'v' or 'V' instead
l3: lodsb          ;fetch a byte
    test  al, al
    je    l4       ;exit if end of string
    inc   dx       ;speculatively count 'o's
    cmp   al, 'o'
    je    l3       ;loop while 'o'
    lodsw          ;fetch two bytes
    dec   si       ;align pointer as though only one byte fetched
                   ;in case of mismatch at the end
    cmp   ax, '!m' ;Intel backwards ordering
    jne   l2       ;loop while not matching complete string
    xchg  dx, ax   ;return count only if match
l4: ret

Accepts a pointer to the string in si, returns the count (or 0) in ax.

Answer 32

Kotlin, 55 bytes

{Regex("^[Vv]ro*m!\$").find(it)?.value?.count{it=='o'}}

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