正则表达式-左右字符数相等

Question

So I have this regular expression

[a+][a-z-[a]]{1}[a+]

which will match string "aadaa"

but it will also match string "aaaaaaaadaa"

Is there any way to force it to match only those strings in which left side a's and right side a's occurrence count should be same?

so that it will match only "aadaa" and not this "aaaaaaaadaa"

Edit

With the help of Peter's answer I could make it working, this is the working version for my requirement

(a+)[a-z-[a]]{1}\1

Answer 1

You can use a back reference , as follows:

 console.log(check("ada")); console.log(check("aadaa")); console.log(check("aaaaaaaadaa")); console.log(check("aaadaaaaaaa")); function check(str) { var re = /^(.*).\\1$/; return re.test(str); } 

Or to only match a 's and d 's:

 console.log(check("aca")); console.log(check("aadaa")); console.log(check("aaaaaaaadaa")); console.log(check("aaadaaaaaaa")); function check(str) { var re = /^(a*)d\\1$/; return re.test(str); } 

Or to only match a 's that surround not-an- a :

 console.log(check("aca")); console.log(check("aadaa")); console.log(check("aaaaaaaadaa")); console.log(check("aaadaaaaaaa")); function check(str) { var re = /^(a*)[bz]\\1$/; return re.test(str); } 

---

I realize all the above is javascript , which was easy for quick demoing within the context of SO.

I made a working DotNetFiddle with the following C# code that is similar to all the above:

public static Regex re = new Regex(@"^(a+)[b-z]\1$");

public static void Main()
{
    check("aca");
    check("ada");
    check("aadaa");
    check("aaddaa");
    check("aadcaa");
    check("aaaaaaaadaa");
    check("aadaaaaaaaa");
}

public static void check(string str)
{
    Console.WriteLine(str + " -> " + re.IsMatch(str));
}

Answer 2

You can also use the following regex for the same although I would prefer the one suggested by @PeterB

 console.log(check("aca")); console.log(check("aadaa")); console.log(check("aaaaaaaadaa")); console.log(check("aaadaaaaaaa")); function check(str) { var re = /^(\\w+)[A-Za-z]\\1$/; return re.test(str); } 

The code is similar to the one in Peter B's answer, but the regex is the one changed by me.