188. Best Time to Buy and Sell Stock IV

1. Description

You are given an integer array prices where prices[i] is the price of a given stock on the ith day, and an integer k.
Find the maximum profit you can achieve. You may complete at most k transactions: i.e. you may buy at most k times and sell at most k times.
Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

2. Example

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

Example 2:
Input: k = 2, prices = [3,2,6,5,0,3]
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.

3. Constraints

• 1 <= k <= 100
• 1 <= prices.length <= 1000
• 0 <= prices[i] <= 1000

4. Solutions

Dynamic Programming

n = prices.size()
Time complexity: O(nk)
Space complexity: O(k)

class Solution {
public:
    int maxProfit(int k, vector<int> &prices) {
        vector<vector<int>> buy(min(int((prices.size()) / 2), k) + 1, vector<int>(2));
        for (int i = 0; i < buy.size(); ++i) {
            buy[i][0] = -prices[0];
        }
        vector<vector<int>> sell(min(int((prices.size()) / 2), k) + 1, vector<int>(2));

        for (int i = 0; i < prices.size(); ++i) {
            for (int j = 1; j < buy.size(); ++j) {
                buy[j][1] = max(sell[j - 1][0] - prices[i], buy[j][0]);
                sell[j][1] = max(buy[j][0] + prices[i], sell[j][0]);
            }

            for (int j = 0; j < buy.size(); ++j) {
                swap(buy[j][0], buy[j][1]);
                swap(sell[j][0], sell[j][1]);
            }
        }

        return max(buy.back()[0], sell.back()[0]);
    }
};