188. Best Time to Buy and Sell Stock IV

/*
 * @lc app=leetcode id=188 lang=cpp
 *
 * [188] Best Time to Buy and Sell Stock IV
 *
 * https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/description/
 *
 * algorithms
 * Hard (45.01%)
 * Likes:    7593
 * Dislikes: 212
 * Total Accepted:    534.4K
 * Total Submissions: 1.2M
 * Testcase Example:  '2\n[2,4,1]'
 *
 * 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
 *
 *
 */

// @lc code=start
class Solution {
public:
    int maxProfit(int k, vector<int>& prices)
    {
        int n = prices.size();

        if (n == 0) return 0;

        if (k >= n) return getMaxProfit(prices);

        vector<int> curMaxProfit(k + 1, 0);
        vector<int> maxProfit(k + 1, 0);

        for (int i = 1; i < n; i++)
        {
            int diff = prices[i] - prices[i - 1];
            for (int j = k; j >= 1; j--)
            {
                curMaxProfit[j] = max(curMaxProfit[j] + diff, maxProfit[j - 1] + max(0, diff));
                maxProfit[j] = max(maxProfit[j], curMaxProfit[j]);
            }
        }
        return maxProfit[k];
    }
    int getMaxProfit(vector<int>& prices)
    {
        int maxProfit = 0;
        for (int i = 1; i < prices.size(); i++)
        {
            if (prices[i] > prices[i - 1])
            {
                maxProfit += prices[i] - prices[i - 1];
            }
        }
        return maxProfit;
    }
};
// @lc code=end

• T: $O(n \cdot k)$
• S: $O(n \cdot k)$