(ns Problem1)
(require '[clojure.set :as cs])
;multiples of 3 and 5

(defn answer [maxrange]
  (apply + (filter (fn [x]
                     (or (zero? (mod x 3))
                         (zero? (mod x 5))))
                   (range maxrange))))
(defn answer2 [maxrange]
  (reduce + (filter #(or (zero? (mod % 3))
                         (zero? (mod % 5)))
                    (range maxrange))))
 ; -> 233168
 ; Answer is correct... but can we do better?
 ; Let's try a different approach and time them side by side.
 ; (I suspect set intersection will be more efficient)


(defn answer3 [maxrange]
  (reduce + (cs/union (set (range 3 maxrange 3))
                     (set (range 5 maxrange 5)))))

(println "================ Strategy 1 ================")
(println "Logic: Apply the addition operator to a filtered subset of range(0 1000)")
(println " (answer 1000):" (answer 1000))
(println "       Correct?" (= (answer 1000) 233168))
(println (time (answer 1000)))
(println "Comments: Logic identical to Strategy 2, except uses '(apply [f coll])' instead of '(reduce [f coll])")
(println "")
(println "================= Strategy 2 ===============")
(println "Logic: Reduce a filtered subset of range(0 1000) using the addition operator")
(println "(answer2 1000):" (answer2 1000))
(println "       Correct?" (= (answer2 1000) 233168))
(println (time (answer2 1000)))
(println "Comments: Faster than Strategy 1 by a small constant factor. Cool!")
(println "")
(println "================= Strategy 3 ===============")
(println "Logic: Set union! Reduce the union of two sets (multiples of 3, multiples of 5) using the addition operator")
(println "(answer3 1000):" (answer3 1000))
(println (time (answer3 1000)))
(println "Comments: I suspect this will be among the fastest solutions to this problem.")
(println "          ***EDIT***: I eat my hat. This is way slower than filtering the entire range! I wonder why??")