added code for best time to buy sell stock (#136)

Author: find3rskeeper

Diff for: Dynamic Programming/buy_sell_stocks/C++/buy_sell_stock_dp.cpp

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+/*@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 for: Dynamic Programming/buy_sell_stocks/README.md

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+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.
+
+