Return the number of times the string "hello/HELLO/hellO ..." appears anywhere in the given string

Question

Return the number of times the string "hello/Hello/...etc" appears anywhere in the given string.

The different in the problem is that

The string hello can be in any case i.e either upper case or lower case.

Sample Input #1
count("abc hello def")

Sample Output #1
1

Sample Input #2
count("Hi. Hello. Hello. Ok")

Sample Output #2
2

Sample Input #3
count("hi")

Sample Output #3
0

MyApproach

public int count(String str)
    {
      String str1="Hello";

        int l=str.length();
        int l1=str1.length();
        if(l<l1)
        {
         return 0;
        }
        else
        {
           int count=0;
           int p=0;
           int j=0;
           while(j<l)
           {
            char c=str.charAt(j);
            char c1=str1.charAt(p);
            if(c==c1)
             {
               p++;

               if(p==l1)
               {
                count++;
                p=0;
               }  

             }
             else
             {
                 p=0;
             }
             j++;

           } 

     return count;
    }

}

Output       TestcaseParameters Testcase Actual Answer  Expected

No output    'HELLO how are you'               0             1

I am getting the following output.

Can anyone tell me what I am doing wrong?

Possible duplicate of [Occurrences of substring in a string](http://stackoverflow.com/questions/767759/occurrences-of-substring-in-a-string) — Marvin, Nov 21 '15 at 15:45
Is the `+ 32` and `- 32` logic intended to handle upper case vs. lower case? You could look at http://stackoverflow.com/questions/10320200/turn-a-user-input-string-to-upper-case-java to get a better way to do that. — Erick G. Hagstrom, Nov 21 '15 at 15:55
I have updated my code.The only difference is the pattern of string which I need to find can be in any case. — Jason arora, Nov 21 '15 at 16:55
@Marvin I think it isn't because of case sensitive nature of the pattern to be found. — Jason arora, Nov 21 '15 at 16:57
@Jasonarora: Wasn't clear to me in the first version. But couldn't you convert all to lowercase and use one of the methods from the other question? — Marvin, Nov 21 '15 at 17:33
@Marvin Yes Marwin that would be much better.Thanku for the hint. — Jason arora, Nov 21 '15 at 17:35

score 6 · Answer 1 · 2015-11-21T17:10:34.937

I'm wondering how your code compiles. c1 is not even declared. I would probably use such logic:

str1.toUpperCase().split(str2.toUpperCase).length()-1

The code above does not return correct answer when str1 beginsWith or endsWith str2. Here is a better version:

public static int count(String str1, String str2) {
    int count = 0;
    int len1 = str1.length();
    int len2 = str2.length();
    for (int i=0;i<=len1-len2;++i){
        if ((str1.substring(i,i+len2)).equalsIgnoreCase(str2)) {
            ++count;
        }
    }
    return count;
}

The code moves str2 rightwards once a step, and compares if str2 is the same as the substring of str1. Please note in the situation str1 = "AAA", and str2 = "AA", the result is 2, since both the first and the second occurrences of AA will have a match.
Step 1:

str1: *****
str2: **

Step 2:

str1: *****
str2:  **

Step 3:

str1: *****
str2:   **

Step 4:

str1: *****
str2:    **

If you wish in the above situation, AA only counts once, here is the code for you:

public static int count(String str1, String str2) {
    int count = 0;
    int len1 = str1.length();
    int len2 = str2.length();
    for (int i=0;i<=len1-len2;){
        if ((str1.substring(i,i+len2)).equalsIgnoreCase(str2)) {
            ++count;
            i+=len2;
        }else{
            ++i;
        }

    }
    return count;
}

This version above works this way:

Step 1:

str1: *****
str2: **

Step 2:

str1: *****
str2:   **

It will not compare any character already compared in the previous steps.

I tested the code against your sample input data, and they all return the expected results.

I upvoted you but this is not correct at all actually and I think it should be edited. The result could differ if hello is at the beginning or at the end or in the middle only. — Yassin Hajaj, Nov 21 '15 at 16:05
Thanks @YassinHajaj. This is some pseudo code off the top of my head. I will edit and refine it. — , Nov 21 '15 at 16:20
@ElgsQianChen I have updated my code.Sir,Can you tell me what error I did.As I will keep your code for reference. — Jason arora, Nov 21 '15 at 16:58
Thanks @YassinHajaj. I edited my answer and made a more serious version. — , Nov 21 '15 at 17:09

score 0 · Answer 2 · edited May 23 '17 at 10:27

0

If this is a school assignment then pls say so. Otherwise You will need to ask a question on google like: 'count occurrences in string java' or something. This will give you - tada - stackoverflow:

Occurrences of substring in a string

Hope to be of help.

edited May 23 '17 at 10:27

Community

1
1

answered Nov 21 '15 at 15:42

user2692263

475
4
8

score 0 · Answer 3 · answered Nov 21 '15 at 16:06

Here is a pretty intuitive solution :

public static void main(String[] args) {
    System.out.println(count("hellohellohellohihihihihihihellohihhihello")); // Prints : 5
}

public static int count(String str){
    String str1 = "hello";
    String str2 = str.toLowerCase();
    int index = 0;
    int count = 0;

    while (true){
        if (str2.contains(str1)){
            count++;
            str2 = str2.substring(str2.indexOf(str1) + str1.length(), str2.length());
        } else {
            return count;
        }
    }
}

score -1 · Answer 4 · answered Dec 09 '15 at 05:26

//in this we have changed every upper case character in lowercase for comparing it with s2 public int count(String str){ String s1="";String s2="hello"; int l2=s2.length(); int count=0;

    int l1=str.length();
    for(int i=0;i<=l1-l2;i++){
        for(int j=i;j<l2+i;j++){
            char ch=str.charAt(j);
            if(ch>=65&&ch<=90){
                ch=(char)(ch+32);
                s1=s1+ch;
            }
            else
            s1=s1+ch;
        }
        if(s1.equals(s2)){
            count++;
        }
//in that line i have removed the data of string s1 to form new string again
           s1="";
    }return count;

}

Return the number of times the string "hello/HELLO/hellO ..." appears anywhere in the given string

4 Answers4