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Algomap/Best_Time_to_Buy_and_Sell_Stock/README.md

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+# Best Time to Buy and Sell Stock (Easy)
+
+## Table of Contents
+
+- [Problem Statement](#problem-statement)
+- [Examples](#examples)
+- [Constraints](#constraints)
+- [Solutions](#solutions)
+  - [Approach 1: One-Pass](#approach-1-one-pass)
+  - [Approach 2: Two-Pointer](#approach-2-two-pointer)
+- [Code Explanation](#code-explanation)
+- [Complexity Analysis](#complexity-analysis)
+- [Related Resources](#related-resources)
+
+## Problem Statement
+
+[Best Time to Buy and Sell Stock](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/)
+
+You are given an array `prices` where `prices[i]` is the price of a given stock on the i-th 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.
+
+## Examples
+
+### 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
+
+## Solutions
+
+```python
+class Solution(object):
+    def maxProfit(self, prices):
+        """
+        :type prices: List[int]
+        :rtype: int
+        """
+        # # Step 1:
+        # cheapest_choice, maxP = float('inf'), 0
+        # for price in prices:
+        #     if price < cheapest_choice:
+        #         cheapest_choice = price
+        #     profit = price - cheapest_choice
+        #     if profit > maxP:
+        #         maxP = max(profit, maxP)
+        # return maxP
+
+        # # Step 2:
+        left, right = 0, 1
+        maxP = 0
+        while right < len(prices):
+            if prices[left] < prices[right]:
+                profit = prices[right] - prices[left]
+                maxP = max(profit, maxP)
+            else:
+                left = right
+            right += 1
+        return maxP
+```
+
+### Approach 1: One-Pass
+
+This approach involves iterating through the prices array once, keeping track of the minimum price seen so far and updating the maximum profit if a higher profit is found.
+
+### Approach 2: Two-Pointer
+
+This approach uses two pointers to keep track of the buying and selling days, moving the pointers based on the profit calculation.
+
+## Code Explanation
+
+### Approach 1: One-Pass (Commented Out)
+
+1. Initialize `cheapest_choice` to positive infinity and `maxP` (maximum profit) to 0.
+2. Iterate through each price in the `prices` array:
+   - If the current price is less than `cheapest_choice`, update `cheapest_choice`.
+   - Calculate the potential profit by subtracting `cheapest_choice` from the current price.
+   - Update `maxP` if the calculated profit is greater.
+3. Return the maximum profit `maxP`.
+
+### Approach 2: Two-Pointer (Active Code)
+
+1. Initialize two pointers: `left` (buying day) at index 0 and `right` (selling day) at index 1.
+2. Initialize `maxP` (maximum profit) to 0.
+3. While `right` is within the array bounds:
+   - If the price at `left` is less than the price at `right`, calculate the profit and update `maxP` if necessary.
+   - If the price at `left` is greater or equal to the price at `right`, move `left` to `right` (new potential buying day).
+   - Move `right` to the next day.
+4. Return the maximum profit `maxP`.
+
+## Complexity Analysis
+
+- Time Complexity: O(n), where n is the number of prices in the array.
+- Space Complexity: O(1), as we only use a constant amount of extra space.
+
+## Related Resources
+
+- [LeetCode Submission](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/submissions/1360165286)
+- [Detailed Explanation](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/solutions/5655153/solution/)
+- [AlgoMap YouTube Explanation](https://www.youtube.com/watch?v=kJZrMGpyWpk)
+- [NeetCode YouTube Explanation](https://www.youtube.com/watch?v=1pkOgXD63yU&t=426s)
+- [GitHub Implementation](https://github.com/gahogg/Leetcode-Solutions/blob/main/Best%20Time%20to%20Buy%20and%20Sell%20Stock%20-%20Leetcode%20121/Best%20Time%20to%20Buy%20and%20Sell%20Stock%20-%20Leetcode%20121.py)
+
+> [!NOTE]
+> This problem is part of a larger collection following the roadmap on [algomap.io](https://algomap.io/). For more details and related problems, please refer to the AlgoMap website.