From 6bceb9ce2835942c6f7dfcaa420ea18d5c9935a6 Mon Sep 17 00:00:00 2001
From: Navneet Jain <njain4u@gmail.com>
Date: Wed, 31 Oct 2018 10:23:49 +0530
Subject: [PATCH] added code for best time to buy sell stock (#136)

---
 .../buy_sell_stocks/C++/buy_sell_stock_dp.cpp | 29 +++++++++++++++++++
 Dynamic Programming/buy_sell_stocks/README.md | 14 +++++++++
 2 files changed, 43 insertions(+)
 create mode 100644 Dynamic Programming/buy_sell_stocks/C++/buy_sell_stock_dp.cpp
 create mode 100644 Dynamic Programming/buy_sell_stocks/README.md

diff --git a/Dynamic Programming/buy_sell_stocks/C++/buy_sell_stock_dp.cpp b/Dynamic Programming/buy_sell_stocks/C++/buy_sell_stock_dp.cpp
new file mode 100644
index 00000000..14c7d2f3
--- /dev/null
+++ b/Dynamic Programming/buy_sell_stocks/C++/buy_sell_stock_dp.cpp	
@@ -0,0 +1,29 @@
+/*@author Navneet Jain
+ * Say you have an array for which the ith element is the price of a given stock on day i.
+ * If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.
+
+ */
+#include<iostream>
+#include<vector>
+#include<climits>
+using namespace std;
+
+int maxProfit(const vector<int> &A) {
+	int buy = A[0], flag = 0, i = 1, max_sell = INT_MIN;
+	int sol = 0;
+	while(i < A.size()){
+		int diff = A[i] - A[i-1];
+		//It's a greedy appoach. If current diff is greater than 0, then add that to the solution.
+		if(diff > 0){
+			sol = sol + diff;
+		}
+		i++;
+	}
+
+	return sol;
+}
+
+int main(){
+	vector<int> A = {1,2};
+	cout << maxProfit(A);
+}
diff --git a/Dynamic Programming/buy_sell_stocks/README.md b/Dynamic Programming/buy_sell_stocks/README.md
new file mode 100644
index 00000000..75291b0d
--- /dev/null
+++ b/Dynamic Programming/buy_sell_stocks/README.md	
@@ -0,0 +1,14 @@
+Problem Statement:
+Say you have an array for which the ith element is the price of a given stock on day i.
+If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.
+
+Solution Approach:
+If you buy your stock on day i, you’d obviously want to sell it on the day its price is maximum after that day. 
+So essentially at every index i, you need to find the maximum in the array in the suffix. 
+Now this part can be done in 2 ways : 
+1) Have another array which stores that information. 
+max[i] = max(max[i+1], A[i])
+
+2) Start processing entries from the end maintaining a maximum till now. Constant additional space requirement.
+
+