121. Best Time to Buy and Sell Stock (#100)

Author: ignacio-chiazzo

LeetcodeProblems/Algorithms/easy/Best_Time_to_buy_and_sell_stock.js

@@ -0,0 +1,95 @@
+/*
+121. Best Time to Buy and Sell Stock
+https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/
+
+Problem:
+You are given an array prices where prices[i] is the price of a given stock on the ith day.
+
+You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.
+
+Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.
+
+ 
+
+Example 1:
+
+Input: prices = [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.
+Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.
+Example 2:
+
+Input: prices = [7,6,4,3,1]
+Output: 0
+Explanation: In this case, no transactions are done and the max profit = 0.
+
+Constraints:
+
+1 <= prices.length <= 10^5
+0 <= prices[i] <= 10^4
+*/
+/*
+Approach:
+We will use a Two pointers strategy (Left and Right pointers). 
+The first will start pointing to the first element, and the right to the second position of array.
+The Left is to buy stock and Right is to sell stock
+
+We initialize our maxProfitValue as 0.
+
+Now we will start our while loop, and we will run till our
+Right pointer less than the array's length.
+
+For Example:
+prices=[7,1,5,3,6,4]
+Note:
+prices[left] --> buy stock
+prices[right] --> sell stock
+We will check the price at the right and left pointer
+
+step 1:
+
+price[left]=7 price[right]=1 profit=-6
+here, price[left] is greater than price[right], so we will move the left pointer to the right position
+and increment our right pointer by 1. We always want our left point to be the minimum.
+
+step 2:
+
+price[left]=1 price[right]=5 profit=4
+here, price[left] is less than price[right], which means we will get profit, 
+so we will update our maxProfitValue and move our right pointer alone
+
+step 3:
+
+price[left]=1 price[right]=3 profit=2
+here, price[left] is less than price[right], we will get profit, so we will compare the maxProfitValue with the current profit. 
+We will update our maxProfitValue and move our right pointer alone
+
+step 4:
+
+price[left]=1 price[right]=6 profit=5
+same logic as above
+
+step 5:
+
+price[left]=1 price[right]=4 profit=3
+same logic as above
+*/
+
+const maxProfit = (prices) => {
+  let left = 0; // Buy
+  let right = 1; // sell
+  let maxProfitValue = 0;
+  while (right < prices.length) {
+    if (prices[left] < prices[right]) {
+      let profit = prices[right] - prices[left]; // our current profit
+
+      maxProfitValue = Math.max(maxProfitValue, profit);
+    } else {
+      left = right;
+    }
+    right++;
+  }
+  return maxProfitValue;
+};
+
+module.exports.maxProfit = maxProfit;

LeetcodeProblemsTests/Algorithms/easy/Best_Time_to_buy_and_sell_stock_Test.js

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+const assert = require("assert");
+const maxProfitEqual = require("../../../LeetcodeProblems/Algorithms/easy/Best_Time_to_buy_and_sell_stock").maxProfit;
+
+function test() {
+  assert.deepEqual(
+    maxProfit([7,1,5,3,6,4]),
+    5
+  );
+  assert.deepEqual(
+    maxProfit([7,6,4,3,1]),
+    0
+  );
+}
+
+module.exports.test = test;

README.md

@@ -72,6 +72,7 @@ The solutions are located under `/LeetcodeProblems`. Each problem has a test fil
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 | [BestTimeToBuy](LeetcodeProblems/Algorithms/easy/Best_Time_To_Buy_And_Sell_Stock_II.js)                                                                        | Medium | https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii                                                    |
+| [Best Time to Buy and Sell Stock](/LeetcodeProblems/Algorithms/easy/Best_Time_to_buy_and_sell_stock.js)                                                        | Easy   | https://leetcode.com/problems/best-time-to-buy-and-sell-stock/
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