Occurrences of substring in a string

How about using StringUtils.countMatches from Apache Commons Lang?

String str = "helloslkhellodjladfjhello";
String findStr = "hello";

System.out.println(StringUtils.countMatches(str, findStr));

That outputs:

3

Your lastIndex += findStr.length(); was placed outside the brackets, causing an infinite loop (when no occurence was found, lastIndex was always to findStr.length()).

Here is the fixed version :

String str = "helloslkhellodjladfjhello";
String findStr = "hello";
int lastIndex = 0;
int count = 0;

while (lastIndex != -1) {

    lastIndex = str.indexOf(findStr, lastIndex);

    if (lastIndex != -1) {
        count++;
        lastIndex += findStr.length();
    }
}
System.out.println(count);

A shorter version. ;)

String str = "helloslkhellodjladfjhello";
String findStr = "hello";
System.out.println(str.split(findStr, -1).length-1);

The last line was creating a problem. lastIndex would never be at -1, so there would be an infinite loop. This can be fixed by moving the last line of code into the if block.

String str = "helloslkhellodjladfjhello";
String findStr = "hello";
int lastIndex = 0;
int count = 0;

while(lastIndex != -1){

    lastIndex = str.indexOf(findStr,lastIndex);

    if(lastIndex != -1){
        count ++;
        lastIndex += findStr.length();
    }
}
System.out.println(count);

Do you really have to handle the matching yourself ? Especially if all you need is the number of occurences, regular expressions are tidier :

String str = "helloslkhellodjladfjhello";
Pattern p = Pattern.compile("hello");
Matcher m = p.matcher(str);
int count = 0;
while (m.find()){
    count +=1;
}
System.out.println(count);     

String str = "helloslkhellodjladfjhello";
String findStr = "hello";
int lastIndex = 0;
int count = 0;

while((lastIndex = str.indexOf(findStr, lastIndex)) != -1) {
     count++;
     lastIndex += findStr.length() - 1;
}
System.out.println(count);

at the end of the loop count is 3; hope it helps

Increment lastIndex whenever you look for next occurrence.

Otherwise it's always finding the first substring (at position 0).

public int indexOf(int ch,
                   int fromIndex)

Returns the index within this string of the first occurrence of the specified character, starting the search at the specified index.

So your lastindex value is always 0 and it always finds hello in the string.

try adding lastIndex+=findStr.length() to the end of your loop, otherwise you will end up in an endless loop because once you found the substring, you are trying to find it again and again from the same last position.

This solution prints the total number of occurrence of a given substring throughout the string, also includes the cases where overlapping matches do exist.

class SubstringMatch{
    public static void main(String []args){
        //String str = "aaaaabaabdcaa";
        //String sub = "aa";
        //String str = "caaab";
        //String sub = "aa";
        String str="abababababaabb";
        String sub = "bab";

        int n = str.length();
        int m = sub.length();

        // index=-1 in case of no match, otherwise >=0(first match position)
        int index=str.indexOf(sub), i=index+1, count=(index>=0)?1:0;
        System.out.println(i+" "+index+" "+count);

        // i will traverse up to only (m-n) position
        while(index!=-1 && i<=(n-m)){   
            index=str.substring(i, n).indexOf(sub);
            count=(index>=0)?count+1:count;
            i=i+index+1;  
            System.out.println(i+" "+index);
        }
        System.out.println("count: "+count);
    }
}

Just a little more peachy answer

    public int countOccurrences(String str, String sub) {
        if (str == null || str.length() == 0 || sub == null || sub.length() == 0) return 0;
        int count = 0;
        int i = 0;
        while ((i = str.indexOf(sub, i)) != -1) {
            count++;
            i += sub.length();
        }
        return count;
    }

I was asked this question in an interview just now and I went completely blank. (Like always, I told myself that the moment the interview ends ill get the solution) which I did, 5 mins after the call ended :(

    int subCounter=0;
    int count =0;
    for(int i=0; i<str.length(); i++) {
        if((subCounter==0 && "a".equals(str.substring(i,i+1))) 
                || (subCounter==1 && "b".equals(str.substring(i,i+1)))
                || (subCounter==2 && "c".equals(str.substring(i,i+1)))) {
            ++subCounter;
        }
        if(subCounter==3) {
            count = count+1;
            subCounter=0;
        }
    }
    System.out.println(count);