tail recursion - Return a sequence with the elements not in common to two original sequences by using clojure -

i have 2 sequences, can vector or list. want return sequence elements not in common 2 sequences.

here example:

(removedupl [1 2 3 4] [2 4 5 6]) = [1 3 5 6] (removeddpl [] [1 2 3 4]) = [1 2 3 4] 

i pretty puzzled now. code:

(defn remove-dupl [seq1 seq2]     (loop [a seq1 b seq2]         (if (not= (first a) (first b))             (recur (rest b))))) 

but don't know next.

i encourage think problem in terms of set operations

(defn extrasection [& ss]   (clojure.set/difference     (apply clojure.set/union ss)     (apply clojure.set/intersection ss))) 

such formulation assumes inputs sets.

(extrasection #{1 2 3 4} #{2 4 5 6}) => #{1 6 3 5} 

which achieved calling (set ...) function on lists, sequences, or vectors.

even if prefer stick sequence oriented solution, keep in mind searching both sequences o(n*n) task if scan both sequences [unless sorted]. sets can constructed in 1 pass, , lookup fast. checking duplicates o(nlogn) task using set.