Gee, what a newb. Here's how to do it in O(1): return userNamesStr.indexOf(curFo...

Gee, what a newb. Here's how to do it in O(1):

    return userNamesStr.indexOf(curForwardUserName) >= 0

;)

(I'd love to not have to explain sarcasm, but people have an incredible difficult time understanding it here.)

lol, I was thinking that too -- "but indexOf is faster..."