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Best Time to Buy and Sell Stock IV⚓︎

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Description⚓︎

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

Solution⚓︎

See reference (Chinese).

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class Solution {
private:
    int maxProfit_k_inf(vector<int>& prices) {
        int n = prices.size();
        vector<vector<int>> dp(n, vector<int>(2));
        for (int i = 0; i < n; i++) {
            if (!i) {
                dp[i][0] = 0;
                dp[i][1] = -prices[i];
                continue;
            }
            dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] + prices[i]);
            dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] - prices[i]);
        }
        return dp[n - 1][0];
    }

public:
    int maxProfit(int k, vector<int>& prices) {
        int n = prices.size();
        // if (n <= 0) return 0;
        if (k > n / 2) return maxProfit_k_inf(prices);

        vector<vector<vector<int>>> dp(n, vector<vector<int>>(k + 1, vector<int>(2)));
        for (int i = 0; i < n; i++) {
            dp[i][0][1] = INT_MIN;
            dp[i][0][0] = 0;
        }

        for (int i = 0; i < n; i++) {
            for (int _k = k; _k >= 1; _k--) {
                if (!i) {
                    dp[i][_k][0] = 0;
                    dp[i][_k][1] = -prices[i];
                    continue;
                }
                dp[i][_k][0] = max(dp[i - 1][_k][0], dp[i - 1][_k][1] + prices[i]);
                dp[i][_k][1] = max(dp[i - 1][_k][1], dp[i - 1][_k - 1][0] - prices[i]);
            }
        }

        return dp[n - 1][k][0];
    }
};