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"))
  1. 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.
  2. If the characters are in order, we return nil.
  3. If we’ve found our missing letter, we return it.
  4. keep returns a sequence of all non-nil values, so the value of keep will be the sequence of missing letters.
  5. Since the problem states there will only be one missing letter, we return the first one (keep is lazy; when we find our first value, cljs will stop processing the rest of the list).

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