A simple solution would be checking if the first and last character are the same and can have any word in b/w. ^a+[\w]+a$

Then we can repeat it for each of the vowels (added spaces for sake for readability. `/ ^a+[\w]+a$|

^e+[\w]+e$|

^i+[\w]+i$|

^o+[\w]+o$|

^u+[\w]+u$ /`

An advance solution would be to check for the matched value at the starting of the string to the value at end of string. ^([aeiou]).*\1$

let re = /^([aeiou]).*\1$/i

this is for my case.

my regex: const re = /^([aeiou])\w*\1$/;

passes all the sample test provided because the sample test are all 3 letter strings but it should fail the constraint: **'' The length of string s is >= 3" **

better regex: const re = /^([aeiou]).+\1$/;

the above regex checks for above constrain.

quickest possible way ik

function regexVar() {
    return /^([aeiou]).*\1$/;
}