Add solution #880

Author: JoshCrozier

README.md

@@ -690,6 +690,7 @@
 877|[Stone Game](./0877-stone-game.js)|Medium|
 878|[Nth Magical Number](./0878-nth-magical-number.js)|Hard|
 879|[Profitable Schemes](./0879-profitable-schemes.js)|Hard|
+880|[Decoded String at Index](./0880-decoded-string-at-index.js)|Medium|
 884|[Uncommon Words from Two Sentences](./0884-uncommon-words-from-two-sentences.js)|Easy|
 889|[Construct Binary Tree from Preorder and Postorder Traversal](./0889-construct-binary-tree-from-preorder-and-postorder-traversal.js)|Medium|
 890|[Find and Replace Pattern](./0890-find-and-replace-pattern.js)|Medium|

solutions/0880-decoded-string-at-index.js

@@ -0,0 +1,46 @@
+/**
+ * 880. Decoded String at Index
+ * https://leetcode.com/problems/decoded-string-at-index/
+ * Difficulty: Medium
+ *
+ * You are given an encoded string s. To decode the string to a tape, the encoded string is read
+ * one character at a time and the following steps are taken:
+ * - If the character read is a letter, that letter is written onto the tape.
+ * - If the character read is a digit d, the entire current tape is repeatedly written d - 1
+ *   more times in total.
+ *
+ * Given an integer k, return the kth letter (1-indexed) in the decoded string.
+ */
+
+/**
+ * @param {string} s
+ * @param {number} k
+ * @return {string}
+ */
+var decodeAtIndex = function(s, k) {
+  let tapeLength = 0;
+  let i = 0;
+
+  while (tapeLength < k) {
+    if (isLetter(s[i])) {
+      tapeLength++;
+    } else {
+      tapeLength *= parseInt(s[i]);
+    }
+    i++;
+  }
+
+  while (i--) {
+    if (isLetter(s[i])) {
+      if (tapeLength === k) return s[i];
+      tapeLength--;
+    } else {
+      tapeLength = Math.floor(tapeLength / parseInt(s[i]));
+      k = k % tapeLength || tapeLength;
+    }
+  }
+};
+
+function isLetter(char) {
+  return /[a-z]/.test(char);
+}