188. Best Time to Buy and Sell Stock IV

📋 Đề Bài

You are given an integer array prices where prices[i] is the price of a given stock on the i^th 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).

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.

Constraints:

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

🧠 Thuật Toán & Kỹ Thuật

Dynamic Programming (Quy hoạch động)

⏱️ Thời gian O(n×m)

💾 Không gian O(n×m)

💻 Lời Giải

C++ 0188-best-time-to-buy-and-sell-stock-iv.cpp

class Solution {
public:
    int maxProfit(int k, vector<int>& prices) {
        int dp[1001][101][2];
        memset(dp, 0, sizeof(dp));
        const int n = prices.size();
        for (int i = n - 1; i >= 0; --i) {
            for (int transaction = k; transaction >= 1; --transaction) {
                for (int buy = 0; buy <= 1; ++buy) {
                    if (buy == 1) {
                        dp[i][transaction][buy] = max(
                            -prices[i] + dp[i + 1][transaction][0],
                            dp[i + 1][transaction][1]
                        );
                    }
                    else {
                        dp[i][transaction][buy] = max(
                            prices[i] + dp[i + 1][transaction - 1][1],
                            dp[i + 1][transaction][0]
                        );
                    }
                }
            }
        }
        return dp[0][k][1];
    }
};