Best Time to Buy and Sell Stock II in Python

Suppose we have an array A, here A[i] is indicating the price of a given stock on day i. We have to find the maximum profit. We can complete as many transactions as we like. (Transaction means to buy and sell stocks). But we have to keep in mind that we may not engage in multiple transactions at the same time. So we have to sell the stock before buying the new one.

Suppose the array is like A = [7, 1, 5, 3, 6, 4], then the result will be 7. As we can see, if we buy on day 2 (index 1), then it will take 1 as a buying price. Then if we sell on day 3, the profit will be 5 – 1 = 4. Then buy on day 4, and sell on day 5, so profit will be 6 – 3 = 3

To solve this, follow these steps −

• let answer = 0
• for i in range 0 to n – 1 (n is the number of elements in A) −
  • if A[i] – A[i – 1] > 0, then
    • answer := answer + A[i] – A[i – 1]
• return answer

Example

Let us see the implementation to get a better understanding

 Live Demo

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] >0:
            ans+=(prices[i] - prices[i-1])
      return ans
ob1 = Solution()
print(ob1.maxProfit([7,2,5,8,6,3,1,4,5,4,7]))

Input

print(ob1.maxProfit([7,2,5,8,6,3,1,4,5,4,7]))

Output

13