188. Best Time to Buy and Sell Stock IV

Say you have an array for which the ith 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.

// Same as 122. Best Time to Buy and Sell Stock II
int maxProfit2(vector<int>& prices) {
    int res = 0;
    for (int i = 1; i < prices.size(); ++i) {
        res += max(0, prices[i] - prices[i - 1]);
    }
    return res;
}
// DP + Greedy
int maxProfit(int k, vector<int>& prices) { // time: O(n * k); space: O(k)
    int n = prices.size();
    if (n <= 1) return 0;
    else if (k >= n / 2) return maxProfit2(prices);
    else {
        vector<int> sell(k + 1, 0), buy(k + 1, INT_MIN);
        for (int i = 0; i < n; ++i) {
            for (int j = k; j >= 1; --j) {
                sell[j] = max(sell[j], buy[j] + prices[i]);
                buy[j] = max(buy[j], sell[j - 1] - prices[i]);
            }
        }
        return sell[k];
    }
}

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