LeetCode 188 - Best Time to Buy and Sell Stock IV

Difficulty: hard

Stock Series:

Part 1: LeetCode 121 - Best Time to Buy and Sell Stock
Part 2: LeetCode 122 - Best Time to Buy and Sell Stock II
Part 3: LeetCode 123 - Best Time to Buy and Sell Stock III
Part 4: LeetCode 188 - Best Time to Buy and Sell Stock IV
Part 5: LeetCode 309 - Best Time to Buy and Sell Stock with Cooldown
Part 6: LeetCode 714 - Best Time to Buy and Sell Stock with Transaction Fee

Problem Description

English (Best Time to Buy and Sell Stock IV)

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

Chinese (买卖股票的最佳时机 IV)

给定一个整数数组 prices ，它的第 i 个元素 prices[i] 是一支给定的股票在第 i 天的价格，和一个整型 k 。

设计一个算法来计算你所能获取的最大利润。你最多可以完成 k 笔交易。也就是说，你最多可以买 k 次，卖 k 次。

注意： 你不能同时参与多笔交易（你必须在再次购买前出售掉之前的股票）。

示例 1：

输入：k = 2, prices = [2,4,1]
输出：2
解释：在第 1 天 (股票价格 = 2) 的时候买入，在第 2 天 (股票价格 = 4) 的时候卖出，这笔交易所能获得利润 = 4-2 = 2 。

示例 2：

输入：k = 2, prices = [3,2,6,5,0,3]
输出：7
解释：在第 2 天 (股票价格 = 2) 的时候买入，在第 3 天 (股票价格 = 6) 的时候卖出, 这笔交易所能获得利润 = 6-2 = 4 。
     随后，在第 5 天 (股票价格 = 0) 的时候买入，在第 6 天 (股票价格 = 3) 的时候卖出, 这笔交易所能获得利润 = 3-0 = 3 。

提示：

• 0 <= k <= 100
• 0 <= prices.length <= 1000
• 0 <= prices[i] <= 1000

Solution

C++

class Solution {
public:
    int maxProfit(int k, vector<int>& prices) {
        vector<int> buy(k + 1, INT_MIN), sell(k + 1, 0);

        for (int price : prices) {
            for (int i = 1; i <= k; ++i) {
                buy[i] = max(buy[i], sell[i - 1] - price);
                sell[i] = max(sell[i], buy[i] + price);
            }
        }

        return sell[k];
    }
};