Given a text txt[0..n-1] and a pattern pat[0..m-1], write a function search(char pat[], char txt[]) that prints all occurrences of pat[] and its permutations (or anagrams) in txt[]. You may assume that n > m.E.g. : I/p : txt[] = “BACDGABCDA” pat[] = “ABCD”o/p :0,5,6

import java.util.ArrayList;
import java.util.HashMap;
import java.util.Map;

/**
 * Created by Arun on 8/9/14.
 */
public class PatternMatching {
    public static boolean isPrime(int n) {
        if (n <= 1) {
            return false;
        }
        for (int i = 2; i < Math.sqrt(n); i++) {
            if (n % i == 0) {
                return false;
            }
        }
        return true;
    }

    public static ArrayList<Integer> getPrimes() {
        ArrayList<Integer> arr = new ArrayList<Integer>();
        for(int i=0, j=0;j<26;i++){
            if(isPrime(i)){
                arr.add(i);
                j++;
            }
        }
        return arr;
    }

    private static ArrayList<Integer> matchPattern(String str, String pattern) {
        ArrayList<Integer> a = getPrimes();
        String alphabets = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
        Map<String, Integer> valMap = new HashMap<String, Integer>();
        int patternProduct = 1;
        for(int i=0;i<alphabets.length();i++) {
            valMap.put(String.valueOf(alphabets.charAt(i)), a.get(i));
        }
        for(int i=0;i<pattern.length();i++) {
            patternProduct = patternProduct * valMap.get(String.valueOf(pattern.charAt(i)));
        }
        for(int i=0;i<=str.length()-pattern.length();i++) {
            int tempVal=1;
            for(int j=i, k=0;k<pattern.length();k++,j++){
                tempVal = tempVal * valMap.get(String.valueOf(str.charAt(j)));
            }
            if(tempVal==patternProduct) {
                System.out.println(i);
            }
        }
        return null;
    }
    public static void main(String args[]) {
        matchPattern("BACDGABCDABCD", "ABCD");
    }
}