Best Time to Buy and Sell Stock IV - LeetCode

Description

You are given an integer array prices where prices[i] is the price of a given stock on the ith 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

Code

Time Complexity: $O(n)$, Space Complexity: $O(n)$

一脈相承 Most consistent ways of dealing with the series of stock problems & Best Time to Buy and Sell Stock，此題是 Best Time to Buy and Sell Stock III 的 generalization。

class Solution {
public:
    int maxProfit(int k, vector<int>& prices) {
        int days = prices.size();
        int trades = k;
        int T[days + 1][trades + 1][2];
 
        // On day 0, no way ending up with stock in hand
        for(int i = 0; i < trades + 1; i++) {
            T[0][i][0] = 0;
            T[0][i][1] = -1e4;
        }
 
        // no trade, no way ending up with stock in hand
        for(int i = 0; i < days + 1; i++) {
            T[i][0][0] = 0;
            T[i][0][1] = -1e4;
        }
 
        for(int k = 1; k < trades + 1; k++) {
            for(int i = 1; i < days + 1; i++) {
                T[i][k][0] = max(T[i - 1][k][0], T[i - 1][k][1] + prices[i - 1]);
                T[i][k][1] = max(T[i - 1][k][1], T[i - 1][k - 1][0] - prices[i - 1]);
            }
        }
        
        return T[days][trades][0];
    }
};

Source

• Best Time to Buy and Sell Stock IV - LeetCode