README_EN.md

123. Best Time to Buy and Sell Stock III

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Description

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

Find the maximum profit you can achieve. You may complete at most two transactions.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

 

Example 1:

Input: prices = [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.

Example 2:

Input: prices = [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

Example 4:

Input: prices = [1]
Output: 0

 

Constraints:

• 1 <= prices.length <= 105
• 0 <= prices[i] <= 105

Solutions

Python3

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        f1, f2, f3, f4 = -prices[0], 0, -prices[0], 0
        for price in prices[1:]:
            f1 = max(f1, -price)
            f2 = max(f2, f1 + price)
            f3 = max(f3, f2 - price)
            f4 = max(f4, f3 + price)
        return f4

Java

class Solution {
    public int maxProfit(int[] prices) {
        int f1 = -prices[0], f2 = 0, f3 = -prices[0], f4 = 0;
        for (int i = 1; i < prices.length; ++i) {
            f1 = Math.max(f1, -prices[i]);
            f2 = Math.max(f2, f1 + prices[i]);
            f3 = Math.max(f3, f2 - prices[i]);
            f4 = Math.max(f4, f3 + prices[i]);
        }
        return f4;
    }
}

C++

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int f1 = -prices[0], f2 = 0, f3 = -prices[0], f4 = 0;
        for (int i = 1; i < prices.size(); ++i) {
            f1 = max(f1, -prices[i]);
            f2 = max(f2, f1 + prices[i]);
            f3 = max(f3, f2 - prices[i]);
            f4 = max(f4, f3 + prices[i]);
        }
        return f4;
    }
};

Go

func maxProfit(prices []int) int {
    f1, f2, f3, f4 := -prices[0], 0, -prices[0], 0
    for i := 1; i < len(prices); i++ {
        f1 = max(f1, -prices[i])
        f2 = max(f2, f1 + prices[i])
        f3 = max(f3, f2 - prices[i])
        f4 = max(f4, f3 + prices[i])
    }
    return f4
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}

C#

public class Solution {
    public int MaxProfit(int[] prices) {
        int f1 = -prices[0], f2 = 0, f3 = -prices[0], f4 = 0;
        for (int i = 1; i < prices.Length; ++i)
        {
            f1 = Math.Max(f1, -prices[i]);
            f2 = Math.Max(f2, f1 + prices[i]);
            f3 = Math.Max(f3, f2 - prices[i]);
            f4 = Math.Max(f4, f3 + prices[i]);
        }
        return f4;
    }
}

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