How to match string (with regular expression) that begins with a string

Question

In a bash script I have to match strings that begin with exactly 3 times with the string lo; so lololoba is good, loloba is bad, lololololoba is good, balololo is bad.

I tried with this pattern: "^$str1/{$n,}" but it doesn't work, how can I do it?

EDIT:

According to OPs comment, lololololoba is bad now.

Answer 1

This should work:

pat="^(lo){3}"
s="lolololoba"
[[ $s =~ $pat ]] && echo good || echo bad

EDIT (As per OPs comment):

If you want to match exactly 3 times (i.e lolololoba and such should be unmatched):

change the pat="^(lo){3}" to:

pat="^(lo){3}(l[^o]|[^l].)"

Answer 2

You can use following regex :

^(lo){3}.*$

Instead of lo you can put your variable.

See demo https://regex101.com/r/sI8zQ6/1