Add solution #820

Author: JoshCrozier

README.md

@@ -628,6 +628,7 @@
 817|[Linked List Components](./0817-linked-list-components.js)|Medium|
 818|[Race Car](./0818-race-car.js)|Hard|
 819|[Most Common Word](./0819-most-common-word.js)|Easy|
+820|[Short Encoding of Words](./0820-short-encoding-of-words.js)|Medium|
 821|[Shortest Distance to a Character](./0821-shortest-distance-to-a-character.js)|Easy|
 824|[Goat Latin](./0824-goat-latin.js)|Easy|
 827|[Making A Large Island](./0827-making-a-large-island.js)|Hard|

solutions/0820-short-encoding-of-words.js

@@ -0,0 +1,38 @@
+/**
+ * 820. Short Encoding of Words
+ * https://leetcode.com/problems/short-encoding-of-words/
+ * Difficulty: Medium
+ *
+ * A valid encoding of an array of words is any reference string s and array of indices
+ * such that:
+ * - words.length == indices.length
+ * - The reference string s ends with the '#' character.
+ * - For each index indices[i], the substring of s starting from indices[i] and up to
+ *   (but not including) the next '#' character is equal to words[i].
+ *
+ * Given an array of words, return the length of the shortest reference string s possible
+ * of any valid encoding of words.
+ */
+
+/**
+ * @param {string[]} words
+ * @return {number}
+ */
+var minimumLengthEncoding = function(words) {
+  const uniqueWords = [...new Set(words)];
+  const wordSet = new Set(uniqueWords);
+
+  for (const word of uniqueWords) {
+    for (let i = 1; i < word.length; i++) {
+      const suffix = word.slice(i);
+      wordSet.delete(suffix);
+    }
+  }
+
+  let result = 0;
+  for (const word of wordSet) {
+    result += word.length + 1;
+  }
+
+  return result;
+};