Adding Efficient Solution - Problem - 122 - Best Time Buy Sell Stock II

Author: Vanditg

Best_Time_Buy_Sell_Stock_II/EfficientSolution.py

@@ -0,0 +1,42 @@
+##==================================
+## Leetcode
+## Student: Vandit Jyotindra Gajjar
+## Year: 2020
+## Problem: 122
+## Problem Name: Best Time to Buy and Sell Stock II         
+##===================================
+#
+#Say you have an array prices for which the ith element is the price of a given stock on day i.
+#
+#Design an algorithm to find the maximum profit. 
+#You may complete as many transactions as you like 
+#(i.e., buy one and sell one share of the stock multiple times).
+#
+#Note: You may not engage in multiple transactions at the same time 
+#(i.e., you must sell the stock before you buy again).
+#
+#Example 1:
+#
+#Input: [7,1,5,3,6,4]
+#Output: 7
+#Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
+#             Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
+#Example 2:
+#
+#Input: [1,2,3,4,5]
+#Output: 4
+#Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
+#             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
+#             engaging multiple transactions at the same time. You must sell before buying again.
+#Example 3:
+#
+#Input: [7,6,4,3,1]
+#Output: 0
+#Explanation: In this case, no transaction is done, i.e. max profit = 0.
+class Solution:
+    def maxProfit(self, prices):	
+        tmp = 0	#Initialze tmp
+        for i in range(1, len(prices)):	#Loop through prices from 1 to prices
+            if prices[i - 1] < prices[i]:	#Condition-check:	If price[i-1] is less than price[i]
+                tmp += (prices[i] - prices[i-1])	#We update tmp by adding tmp and difference of price from i to i-1 untill we reach the end of array
+        return tmp	#We return tmp which is max Profit