Best Time to Buy and Sell Stock III [LeetCode]

Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete at most two transactions.

Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

Solution: DP,  caculate max profit both forward and backward, then return the max sum of them.

 1     int maxProfit(vector<int> &prices) {

 2         if(prices.size() <= 1)

 3             return 0;

 4             

 5         vector<int> max_forward;

 6         int lowest = prices[0];

 7         int max_profit = 0;

 8         max_forward.push_back(0);

 9         for(int i = 1; i < prices.size(); i ++ ){

10             int profit = prices[i] - lowest;

11             if(profit > max_profit)

12                 max_profit = profit;

13             

14             max_forward.push_back(max_profit);

15             lowest = min(prices[i], lowest);

16         }

17         

18         int result = max_forward[max_forward.size() - 1];

19         vector<int> max_backward;

20         int largest = prices[prices.size() - 1];

21         max_profit = 0;

22         for(int i = prices.size() - 2; i >= 0; i --) {

23             int profit = largest - prices[i];

24             max_profit = max(profit, max_profit);

25             

26             result = max(result, max_profit + max_forward[i]);

27             

28             largest = max(largest, prices[i]);

29         }

30         return result;

31     }