Trim a possible prefix of a string in Java

Question

I have String str, from which I want to extract the sub-string excluding a possible prefix "abc".

The first solution that comes to mind is:

if (str.startsWith("abc"))
    return str.substring("abc".length());
return str;

My questions are:

1. Is there a "cleaner" way to do it using split and a regular expression for an "abc" prefix?

2. If yes, is it less efficient than the method above (because it searches "throughout" the string)?

3. If yes, is there any better way of doing it (where "better way" = clean and efficient solution)?

Please note that the "abc" prefix may appear elsewhere in the string, and should not be removed.

Thanks

Answer 1

Shorter than above code will be this line:

return str.replaceFirst("^abc", "");

But in terms of performance I guess there wont be any substantial difference between 2 codes. One uses regex and one doesn't use regex but does search and substring.

Answer 2

Using String.replaceFirst with ^abc (to match leading abc)

"abcdef".replaceFirst("^abc", "")     // => "def"
"123456".replaceFirst("^abc", "")     // => "123456"
"123abc456".replaceFirst("^abc", "")  // => "123abc456"

Answer 3

A regex-free solution (I needed this because the string I'm removing is configurable and contains backslashes, which need escaping for literal use in a regex):

Apache Commons Lang StringUtils.removeStart(str, remove) will remove remove from the start of str using String.startsWith and String.substring.

The source code of the method is informative:

public static String removeStart(final String str, final String remove) {
    if (isEmpty(str) || isEmpty(remove)) {
        return str;
    }
    if (str.startsWith(remove)){
        return str.substring(remove.length());
    }
    return str;
}

Answer 4

Try this

str = str.replaceAll("^abc", "");

Answer 5

1. Using String#split can do this, but it's not better solution. Actually it'll be vague and I wouldn't recommend using it for that purpose.
2. Don't waste time about efficiency in this case, it's not significant, focus on logic and clarity. But note that working with regex is usually slower because it involves additional operations so you might want to keep startsWith.
3. Your approach is fine, if you want to check if the String begins with "abc", String#startsWith was designed for that.

---

You can easily measure the time that takes a code to run. Here what you can do:

Create a big loop, inside it you can append the counter of it to some dummy String in order to simulate the Strings you want to check, then try to have startsWith once, and replaceAll after:

for(int i = 0;i<900000;i++) {
    StringBuilder sb = new StringBuilder("abc");
    sb.append(i);
    if(sb.toString().startsWith("abc")) { ... } 
}
long time = System.currentTimeMillis() - start;
System.out.println(time); //Prints ~130

---

for(int i = 0;i<900000;i++){
   StringBuilder sb = new StringBuilder("abc");
   sb.append(i);
   sb.toString().replaceAll("^abc", "");        
}
long time = System.currentTimeMillis() - start;
System.out.println(time);  //Prints ~730