Add solution #893

Author: JoshCrozier

README.md

@@ -703,6 +703,7 @@
 890|[Find and Replace Pattern](./0890-find-and-replace-pattern.js)|Medium|
 891|[Sum of Subsequence Widths](./0891-sum-of-subsequence-widths.js)|Hard|
 892|[Surface Area of 3D Shapes](./0892-surface-area-of-3d-shapes.js)|Easy|
+893|[Groups of Special-Equivalent Strings](./0893-groups-of-special-equivalent-strings.js)|Medium|
 901|[Online Stock Span](./0901-online-stock-span.js)|Medium|
 905|[Sort Array By Parity](./0905-sort-array-by-parity.js)|Easy|
 909|[Snakes and Ladders](./0909-snakes-and-ladders.js)|Medium|

solutions/0893-groups-of-special-equivalent-strings.js

@@ -0,0 +1,40 @@
+/**
+ * 893. Groups of Special-Equivalent Strings
+ * https://leetcode.com/problems/groups-of-special-equivalent-strings/
+ * Difficulty: Medium
+ *
+ * You are given an array of strings of the same length words.
+ *
+ * In one move, you can swap any two even indexed characters or any two odd indexed characters of
+ * a string words[i].
+ *
+ * Two strings words[i] and words[j] are special-equivalent if after any number of moves,
+ * words[i] == words[j].
+ * - For example, words[i] = "zzxy" and words[j] = "xyzz" are special-equivalent because we may
+ *   make the moves "zzxy" -> "xzzy" -> "xyzz".
+ *
+ * A group of special-equivalent strings from words is a non-empty subset of words such that:
+ * - Every pair of strings in the group are special equivalent, and
+ * - The group is the largest size possible (i.e., there is not a string words[i] not in the
+ *   group such that words[i] is special-equivalent to every string in the group).
+ *
+ * Return the number of groups of special-equivalent strings from words.
+ */
+
+/**
+ * @param {string[]} words
+ * @return {number}
+ */
+var numSpecialEquivGroups = function(words) {
+  return new Set(words.map(normalizeWord)).size;
+
+  function normalizeWord(word) {
+    const evenChars = [];
+    const oddChars = [];
+    for (let i = 0; i < word.length; i++) {
+      if (i % 2 === 0) evenChars.push(word[i]);
+      else oddChars.push(word[i]);
+    }
+    return `${evenChars.sort().join('')}-${oddChars.sort().join('')}`;
+  }
+};