Oct 20, 2024 · Oct 20, 2024 · Oct 20, 2024 · Oct 20, 2024
diff --git a/LeetcodeProblems/Algorithms/easy/Best_Time_to_buy_and_sell_stock.js b/LeetcodeProblems/Algorithms/easy/Best_Time_to_buy_and_sell_stock.js
 /*
 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;
diff --git a/LeetcodeProblemsTests/Algorithms/easy/Best_Time_to_buy_and_sell_stock_Test.js b/LeetcodeProblemsTests/Algorithms/easy/Best_Time_to_buy_and_sell_stock_Test.js
 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;
diff --git a/README.md b/README.md
 | [Gas Station](/LeetcodeProblems/Algorithms/medium/GasStation/index.js)                                                                                         | Medium | https://leetcode.com/problems/gas-station/description/                                                              |
 | [K Closest Points to Origin](/LeetcodeProblems/Algorithms/medium/K_Closest_Points_to_Origin.js/)                                                               | Medium | https://leetcode.com/problems/k-closest-points-to-origin/
 | [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/
 | [Flood Fill ](/LeetcodeProblems/Algorithms/easy/Flood_Fill.js)                                                                                                 | Easy   | https://leetcode.com/problems/flood-fill/                                                                           |
 | [Implement stack using queues ](/LeetcodeProblems/Algorithms/easy/Implement_stack_using_queues.js)                                                             | Easy   | https://leetcode.com/problems/implement-stack-using-queues/                                                         |
 | [Number of Segments in a String ](/LeetcodeProblems/Algorithms/easy/Number_of_Segments_in_a_String.js)                                                         | Easy   | https://leetcode.com/problems/number-of-segments-in-a-string/                                                       |
Original file line number	Diff line number	Diff line change
		@@ -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;
Original file line number	Diff line number	Diff line change
		@@ -0,0 +1,15 @@
		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;
Original file line number	Diff line number	Diff line change
Expand Up		@@ -72,6 +72,7 @@ The solutions are located under `/LeetcodeProblems`. Each problem has a test fil
		\| [Gas Station](/LeetcodeProblems/Algorithms/medium/GasStation/index.js) \| Medium \| https://leetcode.com/problems/gas-station/description/ \|
		\| [K Closest Points to Origin](/LeetcodeProblems/Algorithms/medium/K_Closest_Points_to_Origin.js/) \| Medium \| https://leetcode.com/problems/k-closest-points-to-origin/
		\| [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/
		\| [Flood Fill ](/LeetcodeProblems/Algorithms/easy/Flood_Fill.js) \| Easy \| https://leetcode.com/problems/flood-fill/ \|
		\| [Implement stack using queues ](/LeetcodeProblems/Algorithms/easy/Implement_stack_using_queues.js) \| Easy \| https://leetcode.com/problems/implement-stack-using-queues/ \|
		\| [Number of Segments in a String ](/LeetcodeProblems/Algorithms/easy/Number_of_Segments_in_a_String.js) \| Easy \| https://leetcode.com/problems/number-of-segments-in-a-string/ \|
Expand Down