123. Best Time to Buy and Sell Stock III

• Difficulty: Hard

• Topics: Array, Dynamic Programming

• Similar Questions:

  • Best Time to Buy and Sell Stock

  • Best Time to Buy and Sell Stock II

  • Best Time to Buy and Sell Stock IV

  • Maximum Sum of 3 Non-Overlapping Subarrays

Problem:

Say you have an array for which the i^th 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 (i.e., you must sell the stock before you buy again).

Example 1:

Input: [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
             Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.

Example 2:

Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

Solutions:

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        if (n == 0) return 0;

        vector<int> forward(n, 0);
        vector<int> backward(n, 0);

        int minVal = prices[0];
        for (int i = 1; i < n; ++i) {
            forward[i] = max(forward[i - 1], prices[i] - minVal);
            minVal = min(minVal, prices[i]);
        }

        int maxVal = prices[n-1];
        for (int i = n - 2; i >= 0; --i) {
            backward[i] = max(backward[i + 1], maxVal - prices[i]);
            maxVal = max(maxVal, prices[i]);
        }

        int ret = 0;
        for (int i = 0; i < n; ++i) {
            ret = max(ret, forward[i] + backward[i]);
        }

        return ret;
    }
};