Question: I am new to LINUX. This question sounds simple and stupid, but I suppose this has a lot of meaning behind it. "HOW COMMANDS ARE PROCESSED IN LINUX?". Which means suppose if I give ls command,what makes it to display list of all files inside the directory?. I have been searching for the answer, and I could not find any clear explanation for the same. Please help me to solve the same.

Answer: 
If you want the length of trailing zeroes, you could do this regex :

```
public static void main(String[] args) {
    String s = "8320987112741390144276341183223364380754172606361245952449277696409600000000000000";
    Pattern p = Pattern.compile("0+$");
    Matcher m = p.matcher(s);
    m.find();
    String val = m.group();
    System.out.println(val);
    System.out.println(val.length());

}

```

O/P :

```
00000000000000
14

```
Does this answer fulfills your requirement?

```
String str = "8320987112741390144276341183223364380754172606361245952449277696409600000000000000";
    String withoutTrailingZeroes = "";
    for (int i = str.length() - 1; i >= 0; i--) {
        String charFromString = Character.toString(str.charAt(i));
        if (charFromString.equals("0")) {
            withoutTrailingZeroes += str.charAt(i);
        } else {
            break;
        }
    }
    System.out.println(str);
    System.out.println(str.replace(withoutTrailingZeroes, "")); // string without trailing zeros
    System.out.println(withoutTrailingZeroes); // trailing zeroes
    System.out.println(withoutTrailingZeroes.length()); // trailing zeroes length

```