Add solution #565

Author: JoshCrozier

README.md

@@ -21,6 +21,7 @@
 451|[Sort Characters By Frequency](./0451-sort-characters-by-frequency.js)|Medium|
 459|[Repeated Substring Pattern](./0459-repeated-substring-pattern.js)|Easy|
 541|[Reverse String II](./0541-reverse-string-ii.js)|Easy|
+565|[Array Nesting](./0565-array-nesting.js)|Medium|
 606|[Construct String from Binary Tree](./0606-construct-string-from-binary-tree.js)|Easy|
 617|[Merge Two Binary Trees](./0617-merge-two-binary-trees.js)|Easy|
 648|[Replace Words](./0648-replace-words.js)|Medium|

solutions/0565-array-nesting.js

@@ -0,0 +1,34 @@
+/**
+ * 565. Array Nesting
+ * https://leetcode.com/problems/array-nesting/
+ * Difficulty: Medium
+ *
+ * A zero-indexed array A of length N contains all integers from 0 to N-1.
+ * Find and return the longest length of set S, where S[i] = {A[i],
+ * A[A[i]], A[A[A[i]]], ... } subjected to the rule below.
+ *
+ * Suppose the first element in S starts with the selection of element
+ * A[i] of index = i, the next element in S should be A[A[i]], and then
+ * A[A[A[i]]]… By that analogy, we stop adding right before a duplicate
+ * element occurs in S.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var arrayNesting = function(nums) {
+  const history = new Set();
+  let max = 0;
+  for (let i = 0; i < nums.length; i++) {
+    let count = 0;
+    let j = i;
+    while (!history.has(nums[j])) {
+      j = nums[j];
+      history.add(j);
+      count++;
+    }
+    max = Math.max(max, count);
+  }
+  return max;
+};