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)
Hackerrank step-perms
Hackerrank step-perms
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)