Add solution #238

Author: JoshCrozier

README.md

@@ -149,6 +149,7 @@
 234|[Palindrome Linked List](./0234-palindrome-linked-list.js)|Easy|
 235|[Lowest Common Ancestor of a Binary Search Tree](./0235-lowest-common-ancestor-of-a-binary-search-tree.js)|Easy|
 237|[Delete Node in a Linked List](./0237-delete-node-in-a-linked-list.js)|Easy|
+238|[Product of Array Except Self](./0238-product-of-array-except-self.js)|Medium|
 242|[Valid Anagram](./0242-valid-anagram.js)|Easy|
 263|[Ugly Number](./0263-ugly-number.js)|Easy|
 264|[Ugly Number II](./0264-ugly-number-ii.js)|Medium|

solutions/0238-product-of-array-except-self.js

@@ -0,0 +1,30 @@
+/**
+ * 238. Product of Array Except Self
+ * https://leetcode.com/problems/product-of-array-except-self/
+ * Difficulty: Medium
+ *
+ * Given an integer array nums, return an array answer such that answer[i] is equal to the product
+ * of all the elements of nums except nums[i].
+ *
+ * The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.
+ *
+ * You must write an algorithm that runs in O(n) time and without using the division operation.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var productExceptSelf = function(nums) {
+  const emptyResult = new Array(nums.length).fill(0);
+  const zeroCount = nums.filter(n => n === 0).length;
+  if (zeroCount > 1) {
+    return emptyResult;
+  }
+  const product = nums.reduce((product, n) => product * (n === 0 ? 1 : n), 1);
+  if (zeroCount === 1) {
+    emptyResult[nums.indexOf(0)] = product;
+    return emptyResult;
+  }
+  return nums.map(n => product / n);
+};