LRU Caches

Question

Have the function LRUCache(strArr) take the array of characters stored in strArr, which will contain characters ranging from A to Z in some arbitrary order, and determine what elements still remain in a virtual cache that can hold up to 5 elements with an LRU cache algorithm implemented. For example: if strArr is ["A", "B", "C", "D", "A", "E", "D", "Z"], then the following steps are taken:

(1) A does not exist in the cache, so access it and store it in the cache. (2) B does not exist in the cache, so access it and store it in the cache as well. So far the cache contains: ["A", "B"]. (3) Same goes for C, so the cache is now: ["A", "B", "C"]. (4) Same goes for D, so the cache is now: ["A", "B", "C", "D"]. (5) Now A is accessed again, but it exists in the cache already so it is brought to the front: ["B", "C", "D", "A"]. (6) E does not exist in the cache, so access it and store it in the cache: ["B", "C", "D", "A", "E"]. (7) D is accessed again so it is brought to the front: ["B", "C", "A", "E", "D"]. (8) Z does not exist in the cache so add it to the front and remove the least recently used element: ["C", "A", "E", "D", "Z"].

Now the caching steps have been completed and your program should return the order of the cache with the elements joined into a string, separated by a hyphen. Therefore, for the example above your program should return C-A-E-D-Z.

Use the Parameter Testing feature in the box below to test your code with different arguments.

Answer

function LRUCache(strArr) {
  let cache = [];

  for (let char of strArr) {
    // handle exists in cache
    if (cache.includes(char)) {
      cache.splice(cache.indexOf(char), 1);
    } else if (cache.length >= 5) {
      // unshift first el and apped
      cache.shift();
    }

    cache.push(char);
  }

  // code goes here
  return cache.join('-');
}
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