[LeetCode] Best Time to Buy and Sell Stock III

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).

Find the maximum profit, it must be a DP problem.

We can use two pointers, one forward, and the other backward. The first pointer means we bought the stock some day before, and we sell it now; the second pointer means we buy the stock now, and will sell it some day later. So the only thing we need to do is build two arrays, and find one day that the sum of the profit in these two arrays is maximum.

Using DP, the code is really simple.

    int maxProfit(vector<int> &prices) {
        if (prices.size() == 0)
            return 0;
        vector<int> forward(prices.size());
        vector<int> backward(prices.size());
        int minV = prices[0];
        forward[0] = 0;
        for(int i = 1; i < prices.size(); i++){
            minV = min(minV, prices[i]);
            forward[i] = max(forward[i-1], prices[i] - minV);
        }
        int maxV = prices[prices.size()-1];
        backward[backward.size()-1] = 0;
        for(int i = prices.size() - 2; i >= 0; i--){
            maxV = max(prices[i], maxV);
            backward[i] = max(backward[i+1], maxV - prices[i]);
        }
        int sum = 0;
        for(int i = 0; i < prices.size(); i++)
            sum = max(sum, forward[i] + backward[i]);
        return sum;
    }

 

🙂

 

This entry was posted in Algorithm, Dynamic Programming, LeetCode and tagged , , , . Bookmark the permalink. Post a comment or leave a trackback: Trackback URL.

Post a Comment

Your email is never published nor shared. Required fields are marked *

You may use these HTML tags and attributes <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <s> <strike> <strong>

*
*