Leetcode 188: Best Time to Buy and Sell Stock IV

Say you have an array for which the i-th 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.

Problem Analysis:

Using dynamic programming to solve it.

Suppose dp[i][j] is maximum profit after i transactions in jth day. When in jth day, we can deal:

1. no transaction, so dp[i][j] = dp[i][j-1]
2. sell stock, for 0≤ k < j-1, iterating k, suppose previous transaction is in kth day, dp[i][j]=max(dp[i-1][k] + max profit of current transaction), the later is maximum profit that transaction from k+1 to jth day

Solution:

class Solution {
public:
    int maxProfit(int k, vector<int>& prices) {
        if (k == 0 || prices.size() < 2) return 0;
        if (k > prices.size() / 2) {
            int res = 0;
            for (int i = 1; i < prices.size(); ++i)
                res += max(prices[i] - prices[i - 1], 0);
            return res;
        }
        
        vector<vector<int>> dp(k + 1, vector<int>(prices.size(), 0));
        for (int i = 1; i <= k; i++) {
            int tmpMax =  -prices[0];
            for (int j = 1; j < prices.size(); j++) {
                dp[i][j] = max(dp[i][j - 1], prices[j] + tmpMax);
                tmpMax =  max(tmpMax, dp[i - 1][j - 1] - prices[j]);
            }
        }
    
        return dp[k][prices.size() - 1];
    }
};

Time Complexity: O(kn)

Space Complexity: O(kn)