Wsrtjtyk

I have been given a problem where I need to find the substring in a char array and have to count how many times that substring occurred.
For eg: "aabbcccddaabbbccc"
O/P:
aa:2
bb:1
ccc:2
dd:1
bbb:1

I tried this code but it is not giving me proper solution, if someone could suggest me what I am doing wrong

public class CountSubString {

     static Set set=new HashSet<>();

     static List list=new ArrayList<>();





    private static Map<char, Integer> count(char charArrayToParse){

        Map<char, Integer> subString = new HashMap<char, Integer>();



        for (int i=0; i<charArrayToParse.length ;)

        {StringBuilder word= new StringBuilder();

            for (int j=i; j<charArrayToParse.length; j++) {



                if(charArrayToParse[i] == charArrayToParse[j]) {

                    word.append(charArrayToParse[j]);

                }

                else {

                    char subStringDone = word.toString().toCharArray();

                    if(subString.isEmpty())

                    subString.put(subStringDone, 1);



                    else if(subString.containsKey(subStringDone)) {

                        subString.put(subStringDone, subString.get(subStringDone)+1);

                    }

                    else {

                        subString.put(subStringDone, 1);

                    }



                //System.out.println("Word value are"+subString.get(key));

                    i=j;

                break;

                }



            }



        }

        Set<char> keyValues=  subString.keySet();

        for(char ch : keyValues) {

            if(subString.get(ch)>1) {

                 System.out.println(ch+"--->"+subString.get(ch));

            }

        }

        return subString;

    }

    public static void main(String args) {

        // TODO Auto-generated method stub

        String str = "aaabbbccddddaaaeebbb"; 

        char charArray = str.toCharArray();

        Map<char, Integer> parsedArray= new HashMap<char, Integer>();

    parsedArray= count(charArray);

    }



}

Since this looks like a homework assignment, I'm not going to post a code solution, but I'll point you in the right direction.

Your double for loop is creating an infinite loop. You're setting i = j and it's not getting set high enough to break out of the loop (never goes past 17). Nested for loops in my book are usually a code smell, you've got something that should be broken into a few more functions (low cohesion).

In this case, your first function should be to break your string down into patterns. Your nested for loops can be reduced down to a single loop, and each time a pattern is found, add it to a Set<String> instead of a map. This will prevent duplicates since a Set can only contain unique values.

You can then move on to the next part which is to figure out your pattern counts. Remember, in that case iterate over the string and check for matches. You'll need to take into account that aa will also match aaa when doing your counts.

Can you use libraries? If so there's StringUtils in the apache.commons.lang3 that can solve the problem in a single line like this:

int count = StringUtils.countMatches("aabbcccddaabbbccc", "aa");

Something to consider.. for bbaaarr are there two sequences of aa or just one? My solution considers that there are two but it can be easily adjusted for the other case.

Comments in my code to help explain it:

/**

 * Returns the number of times the sequence occurs in the string.

 * @param seq - the sequence you are looking for

 * @param str - the string you are searching in

 */

private static int count(String seq, String str) {

    if(seq == null || seq.isEmpty() ||

       str == null || str.isEmpty() ){

        return 0;

    }

    int count = 0;



    // the first character of the sequence you are looking for

    final char seqChar = seq.charAt(0);



    // if there aren't seq.length() chars remaining then

    // it's no longer possible to match your sequence

    // so this is the max index to go to when looking for it

    final int maxIndex = str.length() - seq.length();



    // iterate through the characters in your string

    for (int i = 0; i <= maxIndex; i++) {

        // when you find a character matching the start of your sequence

        // then compare the substring of equal length to your sequence

        // and if it matches then you have a match

        if (seqChar == str.charAt(i) &&

            seq.equals(str.substring(i, i + seq.length()))) {

            count++;

        }

    }

    return count;

}

public static void main(String args){

    String s = "aabbcccddaabbbccc";

    System.out.println(count("aa", s)); // 2

    // [aa]bbcccddaabbbccc

    // aabbcccdd[aa]bbbccc



    System.out.println(count("bb", s)); // 3

    // aa[bb]cccddaabbbccc

    // aabbcccddaa[bb]bccc

    // aabbcccddaab[bbccc



    System.out.println(count("cc", s)); // 4

    // aabb[cc]cddaabbbccc

    // aabbc[cc]ddaabbbccc

    // aabbcccddaabbb[cc]c

    // aabbcccddaabbbc[cc] 

}