regular expression - match word only once in line

Question

Case:

1. ehello goodbye hellot hello goodbye
2. ehello goodbye hello hello goodbye

I want to match line 1 (only has 'hello' once!) DO NOT want to match line 2 (contains 'hello' more than once)

Tried using negative look ahead look behind and what not... without any real success..

Answer 1

A simple option is this (using the multiline flag and not dot-all):

^(?!.*\bhello\b.*\bhello\b).*\bhello\b.*$

First, check you don't have 'hello' twice, and then check you have it at least once.
There are other ways to check for the same thing, but I think this one is pretty simple.

Of course, you can simple match for \bhello\b and count the number of matches...

Answer 2

A generic regex would be:

^(?:\b(\w+)\b\W*(?!.*?\b\1\b))*\z

Altho it could be cleaner to invert the result of this match:

\b(\w+)\b(?=.*?\b\1\b)

This works by matching a word and capturing it, then making sure with a lookahead and a backreference that it does/not follow anywhere in the string.

Answer 3

Since you're only worried about words (ie tokens separated by whitespace), you can just split on spaces and see how often "hello" appears. Since you didn't mention a language, here's an implementation in Perl:

use strict;
use warnings;

my $a1="ehello goodbye hellot hello goodbye";
my $a2="ehello goodbye hello hello goodbye";

my @arr1=split(/\s+/,$a1);
my @arr2=split(/\s+/,$a2);

#grab the number of times that "hello" appears

my $num_hello1=scalar(grep{$_ eq "hello"}@arr1);
my $num_hello2=scalar(grep{$_ eq "hello"}@arr2);

print "$num_hello1, $num_hello2\n";

The output is

1, 2