4Clojure #53 Longest Increasing Sub-Seq

Longest Increasing Sub-Seq

Difficulty: Hard Topics: seqs

Given a vector of integers, find the longest consecutive sub-sequence of increasing numbers. If two sub-sequences have the same length, use the one that occurs first. An increasing sub-sequence must have a length of 2 or greater to qualify. test not run

• (= (__ [1 0 1 2 3 0 4 5]) [0 1 2 3])
• (= (__ [5 6 1 3 2 7]) [5 6])
• (= (__ [2 3 3 4 5]) [3 4 5])
• (= (__ [7 6 5 4]) [])

;; Misty
;; TODO 有部分不知足題意
(defn sub-seq [in]
  {:pre [(vector? in)]}
  (letfn [(compare [prev next]
            (if (and (> (count next) 1) (> (count next) (count prev)))
              next
              prev))]
    (loop [ret [] sub [(first in)] r (rest in)]
      (if (seq sub)
        (let [a (last sub)
              b (first r)]
          (if (= (inc a) b)
            (recur ret (conj sub b) (rest r))
            (recur (compare ret sub) (when b [b]) (rest r))))
        ret))))

;; Lo
(fn [v]
  (let [subseqs (for [i (range (count v)) 
                      j (range (+ 2 i) (inc (count v)))]
                 (subvec v i j))]
    (->> subseqs
         (filter #(apply < %))
         reverse ;have to reverse here as max-key returns the last element if count is the same
         (apply max-key count [])))) ;have to add in a blank vector here, in case there is no increasing subvec