188. Best Time to Buy and Sell Stock IV

• Difficulty: Hard

• Topics: 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 III

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 k transactions.

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

Example 1:

Input: [2,4,1], k = 2
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.

Example 2:

Input: [3,2,6,5,0,3], k = 2
Output: 7
Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4.
             Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.

Solutions:

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

        k = min(k, n/2);

        vector<vector<int>> hold (n , vector<int> (2, 0));
        vector<vector<int>> empty(n , vector<int> (2, 0));

        for (int i = 0; i < n; ++i) {
            empty[i][0] = 0;
        }

        for (int j = 1; j <= k; ++j) {
            hold[0][j%2] = -prices[0];
            empty[0][j%2] = 0;
        }


        for (int j = 1; j <= k; ++j) {
            for (int i = 1; i < n; ++i) {
                hold[i][j%2] = max(hold[i-1][j%2], empty[i-1][(j-1)%2] - prices[i]);
                empty[i][j%2] = max(empty[i-1][j%2], hold[i-1][j%2] + prices[i]);        
            }
        }  

        return empty[n-1][k%2];
    }
};