Check for occurences of string in other string

Question

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)

My Small Answer

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

---

                                                                          Enjoy!

Winner

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

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

C# - 73

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

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

// output = "1"

Answer 5

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)))

Answer 6

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)

Answer 7

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
$ 

Answer 8

C# - 66 Bytes

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

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

//Output: 1

Answer 9

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);}

Answer 10

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

Answer 11

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))

Answer 12

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
   read*,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

Answer 13

Mathematica 26 23

Works like Dennis' algorithm, but wordier:

Length@StringCases[a,b]

Three chars shaved off by Szabolics.

Answer 14

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;
}

Answer 15

Java (38)

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

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

Answer 16

Cobra - 25

print a.split(b).length-1

Answer 17

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.