Finding all 3 character length substrings in a string

Question

I am trying to find all three letter substrings from a string in Java.

For example from the string "example string" I should get "exa", "xam", "amp", "mpl", "ple", "str", "tri", "rin", "ing".

I tried using the Java Regular expression "([a-zA-Z]){3}" but I only got "exa", "mpl", "str", "ing".

Can someone tell me a regex or method to correct this.

Solution 1

Implementing Juvanis' idea somewhat, iterate to get your substrings, then use a regular expression to make sure the substring is all letters:

String s = "example string";
for (int i = 0; i <= s.length() - 3; i++) {
    String substr = s.substring(i, i + 3);
    if (substr.matches("[a-zA-Z]+")) { System.out.println(substr); }
}

Solution 2

try this

    Matcher m = Pattern.compile("([a-zA-Z]){3}").matcher("example string");
    for (int i = 0; m.find(i); i = m.start() + 1) {
        System.out.print(m.group() + " ");
    }

output

exa xam amp mpl ple str tri rin ing 

Solution 3

When a character is consumed in one regex, it cannot be used in other regexes. In your example, a is consumed in exa so amp will not be listed as output. You should try traditional iterative approach. It is easier to implement.

Solution 4

This can be done using regex as follows:

1. Find the position of all matches for the string using the regex \w(?=\w\w). This will give you the start index of the first character of each required sub-string.

   In this case, you would get: 0, 1, 2, 3, 4, 8, 9, 10 and 11.

2. Get what you need by taking the sub-strings starting from each position going upto that plus 2.

   In this case, that would mean, my_string.substring(0,3), my_string.substring(1,4) and so on, as the begin index parameter is inclusive while the end index parameter is exclusive.

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Comments

[Jim Garrison]

This is hammer/nail syndrome. You have a brand new hammer (regex) and everything looks like a nail. This is a case where regex is the wrong tool to use. Just iterate from position 0 to length-3, taking the substring at each index. If you need to ignore spaces, build a temp string with spaces removed first.

[jpmc26]

@JimGarrison Just removing spaces won't work. You'll get the invalid results les and est in the example.

[MLD_Saturn]

Apart from minor parentheses error in the if and print statement this is in my opinion the simplest correct solution thus far.

[jpmc26]

@MLD_Saturn Thanks. The idea is mainly Juvanis'; I just implemented it and added verification that it contains letters. Please at least give her/him an upvote.