Add solution #937

Author: JoshCrozier

README.md

@@ -315,6 +315,7 @@
 926|[Flip String to Monotone Increasing](./0926-flip-string-to-monotone-increasing.js)|Medium|
 929|[Unique Email Addresses](./0929-unique-email-addresses.js)|Easy|
 933|[Number of Recent Calls](./0933-number-of-recent-calls.js)|Easy|
+937|[Reorder Data in Log Files](./0937-reorder-data-in-log-files.js)|Medium|
 966|[Vowel Spellchecker](./0966-vowel-spellchecker.js)|Medium|
 970|[Powerful Integers](./0970-powerful-integers.js)|Easy|
 976|[Largest Perimeter Triangle](./0976-largest-perimeter-triangle.js)|Easy|

solutions/0937-reorder-data-in-log-files.js

@@ -0,0 +1,40 @@
+/**
+ * 937. Reorder Data in Log Files
+ * https://leetcode.com/problems/reorder-data-in-log-files/
+ * Difficulty: Medium
+ *
+ * You are given an array of logs. Each log is a space-delimited string of words,
+ * where the first word is the identifier.
+ *
+ * There are two types of logs:
+ * - Letter-logs: All words (except the identifier) consist of lowercase English
+ *   letters.
+ * - Digit-logs: All words (except the identifier) consist of digits.
+ *
+ * Reorder these logs so that:
+ * 1. The letter-logs come before all digit-logs.
+ * 2. The letter-logs are sorted lexicographically by their contents. If their
+ *   contents are the same, then sort them lexicographically by their identifiers.
+ * 3. The digit-logs maintain their relative ordering.
+ *
+ * Return the final order of the logs.
+ */
+
+/**
+ * @param {string[]} logs
+ * @return {string[]}
+ */
+var reorderLogFiles = function(logs) {
+  const split = logs.map(l => l.split(' '));
+  const sortedA = split.filter(l => isNaN(parseInt(l[1], 10))).sort((a, b) => {
+    return a[1] !== b[1]
+      ? a[1].localeCompare(b[1]) : a[2] !== b[2]
+        ? a[2].localeCompare(b[2]) : a.length === b.length
+          ? a[0].localeCompare(b[0]) : a.length - b.length;
+  }).map(l => l.join(' '));
+  const sortedB = split.filter(l => !isNaN(parseInt(l[1], 10)))
+    .sort((a, b) => b - a)
+    .map(l => l.join(' '));
+
+  return [...sortedA, ...sortedB];
+};