检查字符串是否在Octave中包含三个相同的字母

Question

I'm trying to make a function in Octave to check whether a string contains three consecutive same characters. That is, if my string is "asdf" it should return 0 and if it's like "asdfffg" it should return 1. What I did so far is this

if(length(findstr(word,"aaa",0)) > 1 || length(findstr(word,"bbb",0)) > 1 ||   ..

It's costly and I think not that really inefficient. Any suggestions?

Answer 1

Use a regular expression :

match = regexp(word, '(.)\1{2}', 'once');

This means: match any character ( (.) ), followed by that same character ( \\1 ) twice ( {2} ). It will return the starting index of the first match, or an empty array if there isn't any match. So your desired result would be

result = ~isempty(match);

---

Another possibility is to use convolution :

result = any(conv([1 1], +~diff(word))==2);

This works as follows: diff will give 0 when two consecutive characters are the same. So you want to detect if the output of diff contains two consecutive zeros. This is done by negating ( ~ ), converting to double ( + ), convolving with the sequence [1 1] ( conv([1 1], ...) ), and seeing if 2 is present in the output.