In this post, we are going to solve the 122. Best Time to Buy and Sell Stock II problem of Leetcode. This problem 122. Best Time to Buy and Sell Stock II is a Leetcode medium level problem. Let’s see code, 122. Best Time to Buy and Sell Stock II – Leetcode Solution.

Problem

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

Now, let’s see the code of 122. Best Time to Buy and Sell Stock II – Leetcode Solution.

Best Time to Buy and Sell Stock II – Leetcode Solution

122. Best Time to Buy and Sell Stock II – Solution in Java

class Solution {
    public int maxProfit(int[] prices) {
        
        int buyDay = 0;
        int sellDay = 0;
        
        int profit = 0;
        
        for(int i=1;i<prices.length;i++){
            if(prices[i] >= prices[i-1]){
                sellDay++;
            }else{
                profit += prices[sellDay] - prices[buyDay];
                buyDay = sellDay = i;
            }
        }
        profit += prices[sellDay] - prices[buyDay];
        return profit;
    }
}

122. Best Time to Buy and Sell Stock II – Solution in C++

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int buyDay = 0;
        int sellDay = 0;
        
        int profit = 0;
        
        for(int i=1;i<prices.size();i++){
            if(prices[i] >= prices[i-1]){
                sellDay++;
            }else{
                profit += prices[sellDay] - prices[buyDay];
                buyDay = sellDay = i;
            }
        }
        profit += prices[sellDay] - prices[buyDay];
        return profit;
    }
};

122. Best Time to Buy and Sell Stock II – Solution in Python

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        mx = 0 # max profit
        n = len(prices)
        for i in reversed(range(1, n)):
            mx = max(mx, prices[i]-prices[i-1]+mx)
        return mx

Note: This problem 122. Best Time to Buy and Sell Stock II is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.