Add solution #2000

Author: JoshCrozier

README.md

@@ -94,6 +94,7 @@
 1472|[Design Browser History](./1472-design-browser-history.js)|Medium|
 1598|[Crawler Log Folder](./1598-crawler-log-folder.js)|Easy|
 1880|[Check if Word Equals Summation of Two Words](./1880-check-if-word-equals-summation-of-two-words.js)|Easy|
+2000|[Reverse Prefix of Word](./2000-reverse-prefix-of-word.js)|Easy|
 
 ## License
 

solutions/2000-reverse-prefix-of-word.js

@@ -0,0 +1,22 @@
+/**
+ * 2000. Reverse Prefix of Word
+ * https://leetcode.com/problems/reverse-prefix-of-word/
+ * Difficulty: Easy
+ *
+ * Given a 0-indexed string word and a character ch, reverse the segment of word that
+ * starts at index 0 and ends at the index of the first occurrence of ch (inclusive).
+ * If the character ch does not exist in word, do nothing.
+ *
+ * For example, if word = "abcdefd" and ch = "d", then you should reverse the segment
+ * that starts at 0 and ends at 3 (inclusive). The resulting string will be "dcbaefd".
+ */
+
+/**
+ * @param {string} word
+ * @param {character} ch
+ * @return {string}
+ */
+var reversePrefix = function(word, ch) {
+  const index = word.indexOf(ch) + 1;
+  return word.slice(0, index).split('').reverse().join('') + word.slice(index);
+};