Create 123_best_time_to_buy_and_sell_stock_III.md

Author: Joseph Luce

leetcode/hard/123_best_time_to_buy_and_sell_stock_III.md

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+# 123. Best Time to Buy and Sell Stock III
+
+## Dynamic Programming Solution
+- Runtime: O(N)
+- Space: O(N)
+- N = Number of elements in array
+
+By using divide in conquer, we can figure out the correct times to buy twice.
+If you did the question, Best Time to Buy and Sell Stock I, you may apply that algothrim here.
+
+The idea is to find the best transaction to the left of the price and again to the right of the price.
+We can use two arrays and traverse from left to right then again from right to left to find those transactions.
+Then its a matter of traversing a third time but on the two new arrays to find the two best transactions.
+
+```
+class Solution:
+    def maxProfit(self, prices: List[int]) -> int:
+        if len(prices) == 0:
+            return 0
+        left_dp, right_dp = [0] * len(prices), [0] * len(prices)
+        
+        high = low = prices[0]
+        left_max_profit = 0
+        for index, price in enumerate(prices): # left to right
+            if price < low:
+                high = low = price
+            else:
+                high = max(high, price)
+            left_dp[index] = left_max_profit = max(high - low, left_max_profit)
+            
+        high = low = prices[-1]
+        right_max_profit = 0
+        for index, price in enumerate(prices[::-1]): # right to left
+            index = len(prices)-index-1
+            if price > high:
+                high = low = price
+            else:
+                low = min(low, price)
+            right_dp[index] = right_max_profit = max(high - low, right_max_profit)
+
+        return max([left_dp[i]+right_dp[i] for i in range(len(prices))])
+```