diff --git a/0121_best-time-to-buy-and-sell-stock/README.md b/0121_best-time-to-buy-and-sell-stock/README.md
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+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 <= 105`
+*   `0 <= prices[i] <= 104`
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diff --git a/0121_best-time-to-buy-and-sell-stock/python3/solution.py b/0121_best-time-to-buy-and-sell-stock/python3/solution.py
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+# Time: O(N)
+# Space: O(1)
+class Solution:
+    def maxProfit(self, prices: List[int]) -> int:
+        if len(prices) < 1: return 0
+        
+        i, j = 0, 1
+        max_profit = 0
+        
+        while i <= j < len(prices):
+            max_profit = max(prices[j] - prices[i], max_profit)
+            
+            # Buy low, sell high — so we need to keep moving i (like
+            # a buy pointer) if jth price is smaller than what ith is.
+            if prices[j] < prices[i]:
+                i = j
+            else:
+                j += 1
+        
+        return max_profit