JS to CLJS # 4
I’ll regularly take a JavaScript problem and work it out in ClojureScript. This will hopefully take away some of the Magic™ of functional programming and show how to solve real problems. I promise you won’t have to learn what a monad is.
Problem
Write a function that takes an array of consecutive (increasing) letters as input and returns the missing letter.
- You will always get an valid array.
- The array will be always exactly one letter be missing.
- The length of the array will always be at least 2.
- The array will always contain letters in only one case.
findMissingLetter(['a','b','c','d','f']) === 'e';
findMissingLetter(['O','Q','R','S']) === 'P';Solution
;;#cljs
(defn missing-letter [letters]
(first
(keep
(fn [[a b]]
(let [next-char-code (inc (.charCodeAt a 0))]
(if (not= next-char-code (.charCodeAt b 0))
(js/String.fromCharCode next-char-code))))
(partition 2 1 letters))))Thoughts
The main breakthrough is realizing that the fundamental problem is comparing one character to the next character. A nice cljs trick is to use partition to get the data structure that better represents our problem:
cljs.user=> (partition 2 1 ["a" "b" "c" "d" "f"])
=> (("a" "b") ("b" "c") ("c" "d") ("d" "f"))- With our problem-aligned data structure, we work our way through the list, comparing the character code of the first letter to that of the second.
- If the characters are in order, we return
nil. - If we’ve found our missing letter, we return it.
keepreturns a sequence of all non-nil values, so the value ofkeepwill be the sequence of missing letters.- Since the problem states there will only be one missing letter, we return the
firstone (keepis lazy; when we find our first value, cljs will stop processing the rest of the list).