Add solution #914

Author: JoshCrozier

README.md

@@ -238,6 +238,7 @@
 884|[Uncommon Words from Two Sentences](./0884-uncommon-words-from-two-sentences.js)|Easy|
 890|[Find and Replace Pattern](./0890-find-and-replace-pattern.js)|Medium|
 905|[Sort Array By Parity](./0905-sort-array-by-parity.js)|Easy|
+914|[X of a Kind in a Deck of Cards](./0914-x-of-a-kind-in-a-deck-of-cards.js)|Medium|
 916|[Word Subsets](./0916-word-subsets.js)|Medium|
 922|[Sort Array By Parity II](./0922-sort-array-by-parity-ii.js)|Easy|
 925|[Long Pressed Name](./0925-long-pressed-name.js)|Easy|

solutions/0914-x-of-a-kind-in-a-deck-of-cards.js

@@ -0,0 +1,24 @@
+/**
+ * 914. X of a Kind in a Deck of Cards
+ * https://leetcode.com/problems/x-of-a-kind-in-a-deck-of-cards/
+ * Difficulty: Medium
+ *
+ * You are given an integer array deck where deck[i] represents the number written on the ith card.
+ *
+ * Partition the cards into one or more groups such that:
+ * - Each group has exactly x cards where x > 1, and
+ * - All the cards in one group have the same integer written on them.
+ *
+ * Return true if such partition is possible, or false otherwise.
+ */
+
+/**
+ * @param {number[]} deck
+ * @return {boolean}
+ */
+var hasGroupsSizeX = function(deck) {
+  const gcd = (a, b) => !b ? a : gcd(b, a % b);
+  const map = new Map();
+  deck.forEach(n => map.set(n, (map.get(n) || 0) + 1));
+  return [...map.values()].reduce(gcd) > 1;
+};