Add solution #1286

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,209 LeetCode solutions in JavaScript
+# 1,210 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -970,6 +970,7 @@
 1277|[Count Square Submatrices with All Ones](./solutions/1277-count-square-submatrices-with-all-ones.js)|Medium|
 1278|[Palindrome Partitioning III](./solutions/1278-palindrome-partitioning-iii.js)|Hard|
 1284|[Minimum Number of Flips to Convert Binary Matrix to Zero Matrix](./solutions/1284-minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix.js)|Hard|
+1286|[Iterator for Combination](./solutions/1286-iterator-for-combination.js)|Medium|
 1287|[Element Appearing More Than 25% In Sorted Array](./solutions/1287-element-appearing-more-than-25-in-sorted-array.js)|Easy|
 1290|[Convert Binary Number in a Linked List to Integer](./solutions/1290-convert-binary-number-in-a-linked-list-to-integer.js)|Easy|
 1291|[Sequential Digits](./solutions/1291-sequential-digits.js)|Medium|

solutions/1286-iterator-for-combination.js

@@ -0,0 +1,51 @@
+/**
+ * 1286. Iterator for Combination
+ * https://leetcode.com/problems/iterator-for-combination/
+ * Difficulty: Medium
+ *
+ * Design the CombinationIterator class:
+ * - CombinationIterator(string characters, int combinationLength) Initializes the object with a
+ *   string characters of sorted distinct lowercase English letters and a number combinationLength
+ *   as arguments.
+ * - next() Returns the next combination of length combinationLength in lexicographical order.
+ * - hasNext() Returns true if and only if there exists a next combination.
+ */
+
+/**
+ * @param {string} characters
+ * @param {number} combinationLength
+ */
+var CombinationIterator = function(characters, combinationLength) {
+  this.chars = characters;
+  this.length = combinationLength;
+  this.indices = new Array(combinationLength).fill(0).map((_, i) => i);
+  this.hasMore = true;
+};
+
+/**
+ * @return {string}
+ */
+CombinationIterator.prototype.next = function() {
+  const result = this.indices.map(i => this.chars[i]).join('');
+
+  this.hasMore = false;
+  for (let i = this.length - 1; i >= 0; i--) {
+    if (this.indices[i] < this.chars.length - (this.length - i)) {
+      this.indices[i]++;
+      for (let j = i + 1; j < this.length; j++) {
+        this.indices[j] = this.indices[j - 1] + 1;
+      }
+      this.hasMore = true;
+      break;
+    }
+  }
+
+  return result;
+};
+
+/**
+ * @return {boolean}
+ */
+CombinationIterator.prototype.hasNext = function() {
+  return this.hasMore;
+};