# Challenge

I'm sure you read the title, and came in to farm your rep, thinking its kids' stuff, but think again! You have to challenge each other in the shortest code to count the occurences of a string in another. For example, given the following input:

aaaabbbbsssffhd

as a string, and the string

s

should output

3

# Rules

Just before you smile and say, "Hey, I'll use ----," read this:

• No use of external libraries, or your language's API. You have to implement it manually. Which means you can't use your language's built-in function or method for counting occurences
• No file I/O
• No connecting with a server, website, et cetera
• In a case of ababa, where it starts with aba and if you read the last 3 letters it's also aba, you only count one*

Thank you @ProgramFOX for that (the last rule)!
*Hint: When you count occurences, you can remove the ones you counted to avoid disobeying this rule

I think the last 2 rules are just for rule benders!

# Winning Criterion

As previously stated the winner is the code with the less bytes used. The winner will be announced five days later (15 June 2014)

Here's my C++ answer, where it assumes that the li variable holds the string to check occurences in, and l is the string to look for in f:

### Ungolfed

int c = 0;
while (li.find(lf) != string::npos)
{
int p = li.find(lf);
int l = p + lf.length() - 1;
for (p = p; p <= l; p++)
{
li[p] = static_cast<char>(8);
}
++c;
}


Of course, to use std::string, you have to include the string header file!

### Golfed

int c=0;while(li.find(lf)!=string::npos){int p=li.find(lf);int l=p+lf.length()-1;for(p=p;p<=l;p++){li[p]=static_cast<char>(8);}++c;}


### Result

The variable c is going to be the value of how many times the string was found

# Winner

After a long wait @Dennis answer wins with only 3 bytes, written in GolfScript

• Should we count the same character twice? For example, if the input is ababa and aba, should we output 1 or 2? The third a is both the end of the first aba and the begin of the second aba. – ProgramFOX Jun 10 '14 at 15:50
• Does language's API mean libraries that can be imported (that come with the language) or methods of builtin datatypes or commands themselves? – seequ Jun 10 '14 at 15:54
• Does it need to deal with strings longer than s for matching? If not: I have a 5 char J solution: +/a=b – ɐɔıʇǝɥʇuʎs Jun 10 '14 at 15:56
• Does input come through variables or STDIN? – seequ Jun 10 '14 at 15:58
• Function or program? – Kyle Kanos Jun 10 '14 at 18:08

# GolfScript, 3 bytes

/,(


Assumes string and substring are on the stack.

Try it online.

### How it works

/  # Split the string around occurrences of the substring.
,  # Get the length of the split array.
(  # Subtract 1.

• Impressive.. But I have a small question, did you learn GolfScript just for code golf and solving code problems? – user19785 Jun 10 '14 at 17:05
• I did, but I use it occasionally for other tasks by now. Need an ASCII table? golfscript <<< '127,32,-""+' is faster than opening a web browser. – Dennis Jun 10 '14 at 17:09
• @Dennis [win key] jqt [return] a. [return] is even quicker ;) – ɐɔıʇǝɥʇuʎs Jun 10 '14 at 17:12
• @Synthetica: Well, it was just an example. But jqt sounds interesting. What is it? – Dennis Jun 10 '14 at 17:17
• @Dennis It's the standard J console that comes with J (it's one of those graphical consoles, like the iPython QTConsole) – ɐɔıʇǝɥʇuʎs Jun 10 '14 at 17:24

## JavaScript 32

Nothing really interesting here...

(p=prompt)().split(p()).length-1


split main purpose is to create an array from a string using the delimiter in argument.

• split() isn't part of "your language's API"? – svidgen Jun 10 '14 at 16:59
• it is but its purpose is not to count occurrences of a string... so where is the limit of authorized APIs ? – Michael M. Jun 10 '14 at 17:01
• @Michael I apologize of being vague, this is acceptable.. I edited the question to rephrase to "you can't use your language's built-in function or method for counting occurences", so if it is not intended for the purpose of the question, you can use it... – user19785 Jun 10 '14 at 17:04
• @404NotFound Good to know ... this is my answer too then? :) ... (+1) – svidgen Jun 10 '14 at 17:06
• This is the only answer which doesn't assume that the vars have already been initialized, so you've got my upvote. If you assumed the vars were initialized like all the other answers, you would've gotten a.split(b).length-1 19. Maybe @svidgen should clarify initialization in the question. – randunel Jun 11 '14 at 21:52

# J (7)

No use of external libraries Check! , or your language's API. Check...? I don't know what an language API is. You have to implement it manually Check! No file I/O Check! No connecting with a server, website, et cetera Check!

+/a E.b


How it works:

E. is WindowedMatch: the J Refsheet gives 're' E. 'reread' as example. This gives 1 0 1 0 0 0. Then, the only thing left to do is simply adding this with +/ (basically sum).

I don't think this counts as using your language's built-in function or method for counting occurences, but that's disputable.

EDIT: Just to be clear:

   +/'aba'E.'ababa'
2

• I thought ababa -> aba was to return 1? – Kyle Kanos Jun 10 '14 at 18:08
• @KyleKanos It was, but the author specifically said @Synthetica Absolutley go for it! when I asked him, so I suppose it's okay. – ɐɔıʇǝɥʇuʎs Jun 10 '14 at 18:09

# C# - 73

//a = "aba";
//b = "ababa";

Console.Write(b.Split(new string[]{a},StringSplitOptions.None).Length-1);

// output = "1"

• It's wrong! aba appears twice in ababa: aba ba ab aba. – user23125 Jun 11 '14 at 1:23
• @Runemoro "In a case of ababa, where it starts with aba and if you read the last 3 letters it's also aba, you only count one*" << In the rules.. – mnsr Jun 11 '14 at 1:25
• Oh, didn't read... – user23125 Jun 11 '14 at 1:26

# Python 2.x - 49 23 22 bytes

This is assuming variable input is okay. Both strings can be of any length.

Shortened @avall.

a='s'
b='aaaabbbbsssffhd'
print~-len(b.split(a))


49 bytes version, counts every instance of the substring ('aba' is in 'ababa' twice).

a='s'
b='aaaabbbbsssffhd'
print sum(a==b[i:i+len(a)]for i in range(len(b)))

• Worked with every test I put it in... – user19785 Jun 10 '14 at 16:43
• And language's API means you have to use your own function or method, not use the language's built-in way to do it (if it has one, I'm a C++ programmer;( ) – user19785 Jun 10 '14 at 16:44
• Well the second one absolutely is language's own method for it. Also, the first one outputs 3 if a='aa' and b is the same. Is that okay? – seequ Jun 10 '14 at 16:48
• Yes, absolutely – user19785 Jun 10 '14 at 16:48
• Also, it outputs 2 for the a='aba'; b='ababa' case. – seequ Jun 10 '14 at 16:50

# Powershell 32

($args[0]-split$args[1]).count-1


Works like this:

PS C:\MyFolder> .\ocurrences.ps1 ababa aba
1


Explanation: Uses -split to separate the first argument by the second one, returns the size of the array resulting from the split (minus 1)

# Applescript, 106 bytes

Applescript is a fun, but silly language to golf in.

on run a
set AppleScript's text item delimiters to (a's item 1)
(count of (a's item 2)'s text items)-1
end


Run with osascript:

$osascript instr.scpt s aaaabbbbsssffhd 3$


# C# - 66 Bytes

//s = "aba"
//t = "ababa"

Console.Write(t.Split(new[]{s},StringSplitOptions.None).Length-1);

//Output: 1

• lol. That's basically my answer. Usually on this site, if you cant come up with something different than an existing answer, you can give hints to the existing answer on how to save a few extra bytes. – mnsr Jun 12 '14 at 6:49
• @malik Listen, I wasn't able to comment at that time. Otherwise I'd commented on your answer. I'm willing to take my answer down, if you want to. – tsavinho Jun 12 '14 at 6:53
• Fair enough. I don't mind either way. I was just letting you know. – mnsr Jun 12 '14 at 7:18

# C 130 120

Note: will probably crash if called with incorrect arguments.

r;main(int c,char**a){char*p=*++a,*q,*t;while(*p){for(q=a[1],t=p;*q&&*q==*t;q++)t++;*q?p++:(p=t,r++);}printf("%d\n",r);}


Ungolfed (kinda):

int main(int argc, char *argv[]) {
int result = 0;
char *ptr = argv[1];
while (*ptr) {
char *tmp, *tmp2 = ptr;
// str(n)cmp
for (tmp = argv[2]; *tmp; tmp++, tmp2++)
if (*tmp != *tmp2)
break;
if (*tmp) {
ptr++;
} else {
result++;
ptr += tmp;
}
}
printf("%d\n", result);
}


Old version with strstr and strlen: 103

l;main(int c,char**a){char*p=a[1];l=strlen(a[2]);while(c++,p>l)p=strstr(p,a[2])+l;printf("%d\n",c-5);}


# Delphi XE3 (113)

Takes 2 strings, removes substring from string and substracts new length from old length followed by a division of the substring length.

function c(a,b:string):integer;begin c:=(Length(a)-Length(StringReplace(a,b,'',[rfReplaceAll])))div Length(b)end;


Testing:

c('aaaabbbbsssffhd','s') = 3
c('aaaabbbbsssffhd','a') = 4
c('ababa','aba') = 1
c('ababa','c') = 0

# Lua (48)

So i thought I could just submit another answer, this time in lua. Its very possible this could be improved a lot, im very new to this.

print((a.len(a)-a.len(a.gsub(a,b,"")))/b.len(b))


# Fortran 90: 101

Standard abuse of implicit typing, works for any length arrays a and b, though one should expect that len(a) < len(b).

function i();i=0;k=len(trim(b))-1;do j=1,len(trim(a))-k;if(a(j:j+k)==b(1:1+k))i=i+1;enddo;endfunction


This function must be contained within a full program to work. a and b are received from stdin and can be entered either on the same line (either comma or space separated) or on different lines. Compile via gfortran -o main main.f90 and execute as you would any other compiled program.

program main
character(len=256)::a,b
print*,i()
contains
function i()
i=0
k=len(trim(b))-1
do j=1,len(trim(a))-k
if(a(j:j+k)==b(1:1+k))i=i+1
end do
end function
end program main


Tests:

>ababa aba
2


I could make the above return 1 if I add 4 characters (,k+1) for the do loop

> aaaabbbbbsssffhd s
3


# Mathematica 26 23

Works like Dennis' algorithm, but wordier:

Length@StringCases[a,b]


Three chars shaved off by Szabolics.

• Length@StringCases[a,b] is shorter. This problem is not well specified because if we're not allowed to use StringCount, it's not clear whether any of the other builtin String* functions should be allowed either ... they're all based on the same underlying code. – Szabolcs Jun 11 '14 at 17:53
• Thanks. I forgot about StringCases. It was unclear to me what the OP meant by the language's API. – DavidC Jun 11 '14 at 18:25
• Using the new composition operator in Mathematica 10, we could even just write a function as Length@*StringCases. This is shorter than Length@StringCases[##]& if we were to go for just writing a function but not actually apply it to a and b. – Szabolcs Jun 11 '14 at 18:43

## C++ 225

int n,k,m;
int main()
{
string s1,s2;
cin>>s1;
cin>>s2;
int x=s1.size(),y=s2.size();
if(x>=y)
{
for(int i=0;i<x;i++)
{
k=0,m=0;
for(int j=0;j<y;j++)
{
if(s2[j]==s1[i+m])
{
k++,m++;
}
else break;
}
if(k==y)
{
n++;
i+=(y-1);
}
}
}
cout<<n<<endl;
return 0;
}


## Java (38)

System.out.print(a.split(b).length-1);


(The question did not require a complete program or function.)

# Cobra - 25

print a.split(b).length-1


# K/Kona 6

+/y~'x


where x is the string and y the substring. ~ is the negate operator, with ', it is applied to every element in x; it will return 0 if it does not match and 1 if it does match. Since it is applied element-wise, the result of y~'x is a vector, the +/ then sums the result giving the total number of occurrences.

Unfortunately, this method requires that y is only one character, otherwise we will be comparing a multi-character string to a single character string, resulting in a length error.