Best time to Buy and Sell Stock 🤑 Solution Added

02. Algorithms/01. Arrays/27.Best time to Buy and Sell Stock/README.md

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+# Best Time To Buy and Sell Stocks🤑
+
+**Problem Link : https://leetcode.com/problems/best-time-to-buy-and-sell-stock/**
+
+![image](https://user-images.githubusercontent.com/93209316/184504891-d6b768b4-8f52-45f5-b888-5cfad513c861.png)
+
+
+**Lets Analyse this test case once**
+
+Input: prices = [7,1,5,3,6,4]
+Output: 5
+
+* Satring day the price is 7, which is greater so this wont be best time to buy and sell stocks
+* as we move further if we buy the stock on 2nd day which is *leastSoFar* and sell on 3rd day. the *profitsofar* will be 5-1 = 4
+* but this isnt *bestOverallProfit*
+* the next option is to buy the stocks on 2nd day and sell it on 5th day. in this case the profit will be 6-1 = 5 which is greater than *profiteSofar* so this will be the *bestOverallProfit*
+
+
+If we see carefully the approach will be :
+1. We will maintain 3 variables leastSofar , profitSofar, bestOverallProfit
+2. While traversing from index 0 till the end of an array, will update *leastSofar* if prices[i] < leastSoFar
+3. After that we will calculate *profitSoFar* by doing pricaes[i] - leastSofar if this value is greater than *bestOverallProfit** then will update variable *bestOverallProfit*
+4. In the end we will return Max Profit which is *(beastOverallProfit)*
+
+**Solution :-**
+
+```
+class Solution {
+ public int maxProfit(int[] prices) {
+ int leastSoFar = Integer.MAX_VALUE;
+ int profitIfSoldToday = 0;
+ int overallProfit = 0;
+
+ for(int i=0; i < prices.length;i++){
+ if(prices[i] < leastSoFar){
+ leastSoFar = prices[i]; 
+ }
+ profitIfSoldToday = prices[i] - leastSoFar;
+ if(profitIfSoldToday > overallProfit){
+ overallProfit = profitIfSoldToday;
+ }
+ }
+ return overallProfit;
+ }
+}
+```