I have a sting and need to find first character without adjacent duplicates. I wonder how to do it with regexp. Here is what i supposed should it be.
For example for string=aabbbcddg answer is c.
const paragraph='aabbbcddg';
console.log(paragraph.match(/(.)\1{1}/));
or
const paragraph='aabbbcddg';
console.log(paragraph.match(/[.]{1}/));
Can you explain why my solutions don't work?
Your pattern (.)\1{1} (you can omit {1}) matches 2 consecutive characters in the string and does not take the following character into account.
You could match adjacent characters from the start of the string ^ and then repeat that 1+ times. Then capture in group 2 the first character:
^(?:(.)\1+)+(.)
^ Start of string(?: Non capturing group
(.)\1+ Capture any character except a newline in group 1 and repeat the backreference \1 1+ times)+ Close non capturing group and repeat 1+ times
(.) Match any character except a newline in group 2If you want to get multiple matches and your string does not start with consecutive characters, you can omit the ^ anchor:
const regex = /(?:(.)\1+)+(.)/;
const str = `aabbbcddg`;
console.log(str.match(regex)[2]);
a in the beginning, instead it matches the second occurrence of s in the string, ignoring a, s, a, and d which are all valid matches for "single characters"
(.)\1{1} matches two of the same character.
You could replace all the (.)\1+ to get all the characters which are not followed by itself
let str = 'aabbbcddg'
let unmatched = str.replace(/(.)\1+/g, '')[0]
console.log(unmatched)You could also use a for loop if you want to break when a match is found
let str = 'aabbbcddg',
found;
for (let i = 0; i < str.length; i++) {
if (str[i - 1] !== str[i] && str[i] !== str[i + 1]) {
found = str[i];
break;
}
}
console.log(found)
bbb because the first two characters are followed by another b but is not matching aa because there is not a third one and I'm fairly sure that's not what you meant.