README_EN.md

122. Best Time to Buy and Sell Stock II

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Description

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

On each day, you may decide to buy and/or sell the stock. You can only hold at most one share of the stock at any time. However, you can buy it then immediately sell it on the same day.

Find and return the maximum profit you can achieve.

 

Example 1:

Input: prices = [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
Total profit is 4 + 3 = 7.

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.
Total profit is 4.

Example 3:

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0.

 

Constraints:

• 1 <= prices.length <= 3 * 104
• 0 <= prices[i] <= 104

Solutions

Greedy or Dynamic Programming.

Python3

Greedy:

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        res = 0
        for i in range(1, len(prices)):
            t = prices[i] - prices[i - 1]
            res += max(t, 0)
        return res

Dynamic Programming:

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

Java

Greedy:

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

Dynamic Programming:

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

TypeScript

function maxProfit(prices: number[]): number {
    let ans = 0;
    for (let i = 1; i < prices.length; i++) {
        ans += Math.max(0, prices[i] - prices[i - 1]);
    }
    return ans;
}

function maxProfit(prices: number[]): number {
    const n = prices.length;
    let res = 0;
    let max = prices[0];
    let min = prices[0];
    for (let i = 1; i < n; i++) {
        const price = prices[i];
        if (price < max) {
            res += max - min;
            max = price;
            min = price;
        } else {
            max = price;
        }
    }
    if (min < max) {
        res += max - min;
    }
    return res;
}

C++

Greedy:

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

Dynamic Programming:

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

Go

Greedy:

func maxProfit(prices []int) int {
	res := 0
	for i := 1; i < len(prices); i++ {
		t := prices[i] - prices[i-1]
		if t > 0 {
			res += t
		}
	}
	return res
}

Dynamic Programming:

func maxProfit(prices []int) int {
	f1, f2 := -prices[0], 0
	for _, price := range prices[1:] {
		f1 = max(f1, f2-price)
		f2 = max(f2, f1+price)
	}
	return f2
}

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

C#

Greedy:

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

Dynamic Programming:

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

Rust

impl Solution {
    pub fn max_profit(prices: Vec<i32>) -> i32 {
        let mut res = 0;
        for i in 1..prices.len() {
            res += 0.max(prices[i] - prices[i - 1]);
        }
        res
    }
}

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