From e24aeb3a67d29fbd71ca345778c537c986f6d2e7 Mon Sep 17 00:00:00 2001
From: Neha Gupta <78951376+Neha611@users.noreply.github.com>
Date: Tue, 5 Mar 2024 20:36:04 +0530
Subject: [PATCH] Create BestTimetoBuyandSellStock.js

This problem is one of the most important concepts of Dynamic Programming which is more often asked in interviews.
I have provided problem statement , test example , code and explanation of code here.
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 .../BestTimetoBuyandSellStock.js              | 49 +++++++++++++++++++
 1 file changed, 49 insertions(+)
 create mode 100644 Dynamic-Programming/BestTimetoBuyandSellStock.js

diff --git a/Dynamic-Programming/BestTimetoBuyandSellStock.js b/Dynamic-Programming/BestTimetoBuyandSellStock.js
new file mode 100644
index 0000000000..bb47bb2e9f
--- /dev/null
+++ b/Dynamic-Programming/BestTimetoBuyandSellStock.js
@@ -0,0 +1,49 @@
+/*
+  Author : Neha611 
+  github Link:https://github.com/Neha611
+  
+  Problem Description:
+  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.
+
+  Example:
+  Input: prices = [7,1,5,3,6,4]
+  Output: 5
+  Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
+  Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.
+
+  Problem Link:
+  https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/
+*/
+
+/**
+ * @param {number[]} prices
+ * @return {number}
+ */
+var maxProfit = function(prices) {
+    let buy_day=-1;
+    let buy=Number.MAX_VALUE;
+    let sell_day=-1;
+    let profit=0;
+    for(let i=0;i<prices.length;i++)
+    {
+        if(prices[i]<buy)
+        {
+            /*we will buy the stock with lower price*/
+            buy=prices[i];
+            buy_day=i;
+        }
+        else if(prices[i]>prices[buy_day])
+        {
+            /*if the current price of stock is higher than the price it is bought with we can or cannot sell on this day*/
+            sell_day=i;
+        }
+        if(buy_day<sell_day)
+        {
+            /*get profit*/
+            profit=Math.max(profit,prices[sell_day]-prices[buy_day]);
+        }
+    }
+    return profit;
+};