Adding Efficient Solution - Problem - 121 - Best Time Buy Sell Stock

Author: Vanditg

Best_Time_Buy_Sell_Stock/EfficientSolution.py

@@ -0,0 +1,37 @@
+##==================================
+## Leetcode
+## Student: Vandit Jyotindra Gajjar
+## Year: 2020
+## Problem: 121
+## Problem Name: Best Time to Buy and Sell Stock          
+##===================================
+#
+#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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.
+#
+#Note that you cannot sell a stock before you buy one.
+#
+#Example 1:
+#
+#Input: [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.
+#             Not 7-1 = 6, as selling price needs to be larger than buying price.
+#Example 2:
+#
+#Input: [7,6,4,3,1]
+#Output: 0
+#Explanation: In this case, no transaction is done, i.e. max profit = 0.
+class Solution:
+    def maxProfits(self, prices):
+        if prices == []:	#Condition-check: If prices is empty list 
+            return 0	#Return 0 
+        tmp, min = 0, prices[0]	#Initialize tmp and min which is 0 and first element of the list respectively
+        for i in range(1, len(prices)):	#Loop through prices  
+            if prices[i] < min:	#Condition-check: If prices is less thatn min  
+                min = prices[i]	#We update min by set to prices[i]
+            else:	#Condition-check: Else 
+                tmp = max(tmp, prices[i] - min)	#Update tmp by taking max of tmp and difference of prices[i] and min 
+        return tmp	#We return tmp				
+