How to calculate how many times a string appears in another string?

Question

I need to find out for example how many times test appears in ttest and answer for that would be 2, or for example world in w1o1r1l1d and answer would be one. I have already written a code which finds all possibilities, and then checks if it is the string I am searching for, but that is too slow.

Did you check any of those? http://en.wikipedia.org/wiki/String_searching_algorithm — Ashalynd, Apr 12 '14 at 20:14
Could you please share your code and how did you measure that it is too slow.. — Mantosh Kumar, Apr 12 '14 at 20:15
Without seeing your code, no one can tell how to improve it ... — πάντα ῥεῖ, Apr 12 '14 at 20:15
http://stackoverflow.com/questions/23032730/can-you-help-me-make-this-program-faster here is my complete code in pascal, but no one helped me :\ — Luka, Apr 12 '14 at 20:17
Just because no one answered your question the first time doesn't mean you should post it again. — Luigi, Apr 12 '14 at 20:18
Search for index of first occurrence of the string, then resume at the index + string length until another occurrence, and so on until the end, while counting. — dtech, Apr 12 '14 at 20:38

pmg · Answer 1 · 2014-04-12T21:11:25.900

I'd try a recursive solution.

The number of times a one-letter string appears in another string is the number of times that characters appears there.

the number of time "r" appears in "program" is 2

The number of times a n-letter string appears in another string is:
the number of times the (n-1)- string appears after the first match for the first letter plus the number of times the n-letter string appears after the first match

the number of times "test" appears in "ttest" is
      the number of times "est" appears in "test"
    + the number of times "test" appears in "test"

#include <stdio.h>
#include <string.h>

int count(const char *needle, const char *stack) {
  int n = 0;
  const char *p;
  if (*stack == 0) return 0;
  if (*needle == 0) return 0;
  p = strchr(stack, *needle);
  if (needle[1] == 0) n += !!p;
  if (p) {
    n += count(needle + 1, p + 1);
    n += count(needle, p + 1);
  }
  return n;
}

int main(void) {
  const char *needle, *stack;

  needle = "a"; stack = "";
  printf("[%s] exists %d times in [%s]\n", needle, count(needle, stack), stack);

  needle = ""; stack = "a";
  printf("[%s] exists %d times in [%s]\n", needle, count(needle, stack), stack);

  needle = "a"; stack = "abracadabra";
  printf("[%s] exists %d times in [%s]\n", needle, count(needle, stack), stack);

  needle = "br"; stack = "abracadabra";
  printf("[%s] exists %d times in [%s]\n", needle, count(needle, stack), stack);

  needle = "test"; stack = "ttest";
  printf("[%s] exists %d times in [%s]\n", needle, count(needle, stack), stack);

  needle = "world"; stack = "w1o1r1l1d";
  printf("[%s] exists %d times in [%s]\n", needle, count(needle, stack), stack);

  return 0;
}

You can see the code "running" at codepad.org: http://codepad.org/Jktgdbp7 — pmg, Apr 12 '14 at 20:53

How to calculate how many times a string appears in another string?

1 Answers1