Question

Formatted question description: https://leetcode.ca/all/122.html

122. Best Time to Buy and Sell Stock II

Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit.
You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).

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

Example 1:

Input: [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.

Example 2:

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

@tag-array

Algorithm

From the second day on, if the current price is higher than the previous price, add the difference to the profit, because we can buy yesterday and sell today. If the price is higher tomorrow, we can buy today and sell again tomorrow. Then the maximum profit can be obtained after traversing the entire array.

Code

Java

public class Best_Time_to_Buy_and_Sell_Stock_II {


    class Solution {
        public int maxProfit(int[] prices) {

            if (prices == null || prices.length == 0) {
                return 0;
            }
            int result = 0;

            for (int i = 1; i < prices.length; i++) {
                int singleDayProfit = prices[i] - prices[i - 1];
                if (singleDayProfit > 0) {
                    result += singleDayProfit;
                }
            }

            return result;
        }
    }
}

// OJ: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int ans = 0;
        for (int i = 1; i < prices.size(); ++i) ans += max(prices[i] - prices[i - 1], 0);
        return ans;
    }
};

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        return sum(max(0, b - a) for a, b in pairwise(prices))

############

class Solution(object):
  def maxProfit(self, prices):
    """
    :type prices: List[int]
    :rtype: int
    """
    ans = 0
    for i in range(1, len(prices)):
      if prices[i] > prices[i - 1]:
        ans += prices[i] - prices[i - 1]
    return ans

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