Permutation of a string so that the patterns match - c

Question

The question is to count how many permutations of a string B have an equivalent pattern into a bigger string A. For example, if A="aabbccd" and B="xx", then it should print 3, since "aa", "bb", "cc" are all substrings of A which share the same pattern as B.
I have tried to pass the substrings as numbers, such as xx becomes "11" and do the same for string A, but I still can't get it to work. Any ideas? Length can be up to 10^7.
Here's the code for changing pattern:
void transform(int* dest, char* original, int len) {
int j=1;
Al[original[0]-'a']=j;
dest[0]=j;
j++;
for (int i=1;i<len;i++) {
if (Al[original[i]-'a']==0)
Al[original[i]-'a']=j++;
dest[i]=Al[original[i]-'a'];
}
}

Answer 1

Concept: Use Regular Expressions
You would need the following regular expression (\\w)\\1{(REPETITIONS-1)}
I don't know about C but Java provides a library to compile RegEx patterns. Here's a class that implements just what you want:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class StringPatternPermutation {
public static void main(String[] args) {
int REPETITIONS = 3;
String REGEX = "(\\w)\\1{" + (REPETITIONS-1) + "}";
String INPUT = "abbbbbbccddeffff";
Pattern p = Pattern.compile(REGEX);
Matcher m = p.matcher(INPUT);
int count = 0;
while(m.find()){
String match = m.group();
System.out.println(match);
count++;
}
System.out.println(count);
}
}
Here's a test of the code above: https://ideone.com/5nztaa
Here's a useful website to test any RegEx: https://regexr.com/

Answer 2

Without Regular Expressions
public class StringPatternPermutation {
public static void main(String[] args) {
String a = "abjjjiixsssppw";
String b = "qwwwee";
String patternA = detectPattern(a);
String patternB = detectPattern(b);
System.out.println("-String A: " + a);
System.out.println("-Pattern A: " + patternA);
System.out.println("-String B: " + b);
System.out.println("-Pattern B: " + patternB);
System.out.println("-A contains B? " + patternA.contains(patternB));
int count = 0;
int index = 0;
while((index = patternA.indexOf(patternB)) != -1){
count++;
patternA = patternA.substring(index+1, patternA.length());
}
System.out.println("-Number of occurances: " + count);
}
private static String detectPattern(String a){
StringBuilder sb = new StringBuilder();
char prev = a.charAt(0);
int count = 1;
for(int i = 1; i < a.length(); i++){
char curr = a.charAt(i);
if(curr == prev)
count++;
else {
sb.append(count + ", ");
prev = curr;
count = 1;
}
if(i == a.length() - 1){
sb.append(count);
}
}
return sb.toString();
}
}
Test it on ideOne: https://ideone.com/w422Du