Add solution #121

Author: JoshCrozier

README.md

@@ -35,6 +35,7 @@
 67|[Add Binary](./0067-add-binary.js)|Easy|
 73|[Set Matrix Zeroes](./0073-set-matrix-zeroes.js)|Medium|
 88|[Merge Sorted Array](./0088-merge-sorted-array.js)|Easy|
+121|[Best Time to Buy and Sell Stock](./0121-best-time-to-buy-and-sell-stock.js)|Easy|
 151|[Reverse Words in a String](./0151-reverse-words-in-a-string.js)|Medium|
 152|[Maximum Product Subarray](./0152-maximum-product-subarray.js)|Medium|
 217|[Contains Duplicate](./0217-contains-duplicate.js)|Easy|

solutions/0121-best-time-to-buy-and-sell-stock.js

@@ -0,0 +1,29 @@
+/**
+ * 121. Best Time to Buy and Sell Stock
+ * https://leetcode.com/problems/best-time-to-buy-and-sell-stock/
+ * Difficulty: Easy
+ *
+ * You are given an array prices where prices[i] is the price of a given stock
+ * on the ith day.
+ *
+ * You want to maximize your profit by choosing a single day to buy one stock
+ * and choosing a different day in the future to sell that stock.
+ *
+ * Return the maximum profit you can achieve from this transaction. If you
+ * cannot achieve any profit, return 0.
+ */
+
+/**
+ * @param {number[]} prices
+ * @return {number}
+ */
+var maxProfit = function(prices) {
+  let max = 0;
+
+  for (let i = 0, min = prices[0]; i < prices.length; i++) {
+    min = Math.min(min, prices[i]);
+    max = Math.max(max, prices[i] - min);
+  }
+
+  return max;
+};