Add solution #748

Author: JoshCrozier

README.md

@@ -275,6 +275,7 @@
 739|[Daily Temperatures](./0739-daily-temperatures.js)|Medium|
 746|[Min Cost Climbing Stairs](./0746-min-cost-climbing-stairs.js)|Easy|
 747|[Largest Number At Least Twice of Others](./0747-largest-number-at-least-twice-of-others.js)|Easy|
+748|[Shortest Completing Word](./0748-shortest-completing-word.js)|Easy|
 762|[Prime Number of Set Bits in Binary Representation](./0762-prime-number-of-set-bits-in-binary-representation.js)|Easy|
 784|[Letter Case Permutation](./0784-letter-case-permutation.js)|Medium|
 791|[Custom Sort String](./0791-custom-sort-string.js)|Medium|

solutions/0748-shortest-completing-word.js

@@ -0,0 +1,39 @@
+/**
+ * 748. Shortest Completing Word
+ * https://leetcode.com/problems/shortest-completing-word/
+ * Difficulty: Easy
+ *
+ * Given a string licensePlate and an array of strings words, find the shortest completing
+ * word in words.
+ *
+ * A completing word is a word that contains all the letters in licensePlate. Ignore numbers
+ * and spaces in licensePlate, and treat letters as case insensitive. If a letter appears more
+ * than once in licensePlate, then it must appear in the word the same number of times or more.
+ *
+ * For example, if licensePlate = "aBc 12c", then it contains letters 'a', 'b' (ignoring case),
+ * and 'c' twice. Possible completing words are "abccdef", "caaacab", and "cbca".
+ *
+ * Return the shortest completing word in words. It is guaranteed an answer exists. If there
+ * are multiple shortest completing words, return the first one that occurs in words.
+ */
+
+/**
+ * @param {string} licensePlate
+ * @param {string[]} words
+ * @return {string}
+ */
+var shortestCompletingWord = function(licensePlate, words) {
+  const license = licensePlate.toLowerCase().replace(/[\d\s]+/g, '');
+  const sortedWords = [...words].sort((a, b) => a.length - b.length);
+
+  for (const word of sortedWords) {
+    let updatedLicense = license;
+
+    for (let i = 0; i < word.length; i++) {
+      updatedLicense = updatedLicense.replace(word[i], '');
+      if (!updatedLicense) {
+        return word;
+      }
+    }
+  }
+};