Bad solution (the trap):
look at all the possible permutations of wordA and find them in listA.
This is a terrible solution and has a complexity of countOfCharactersInWord(wordA)!
where ! is the factorial symbol :)
take all the characters in wordA and sort them. Then compare it with the each item in listA first sorting the characters of that item as well :)
In python :
This yields for my friend "omair" the following interesting (i made them bold) anagrams:
mario arimo moria omari maori moira
I used the SCOWL wordlist for this. You can get the complete working code (including the wordlist) from here: