Hackerrank step-perms

Challenge

The base cases for steps are 0, 1, 2, 4, and I add I recurse within step-perms’. The solution is correct for the test cases where the time limit does not exceed.

(defn step-perms [n]
  "TOC: <20m
   Complexity: "
  (letfn [(step-perms' [n]
            (cond
              (= n 0) 0
              (= n 1) 1
              (= n 2) 2
              (= n 3) 4
              :else (+ (step-perms' (dec n))
                       (step-perms' (- n 2))
                       (step-perms' (- n 3)))))]
    (mod (step-perms' n) 10000000007)))

(step-perms 5)

I memoize to avoid computing same results by storing values already computed in an atom map. All tests pass:

(defn step-perms
  "TOC: <1h
   Complexity: TBD"
  [n]
  (let [memoized-step-perms (atom {})]
    (letfn [(memo-step-perms [k]
              (if (get @memoized-step-perms k)
                (get @memoized-step-perms k)
                (let [result (step-perms' k)]
                  (swap! memoized-step-perms assoc k result) ;; looked up swap! online
                  result)))
            (step-perms' [n]
              (cond
                (= n 0) 0
                (= n 1) 1
                (= n 2) 2
                (= n 3) 4
                :else (+ (memo-step-perms (dec n))
                         (memo-step-perms (- n 2))
                         (memo-step-perms (- n 3)))))]
      (mod (step-perms' n) 10000000007))))

(step-perms 5)