122. Best Time to Buy and Sell Stock II #

Problem #

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.

Problem Summary #

Given an array, where its i-th element is the price of a given stock on day i. Design an algorithm to calculate the maximum profit you can obtain. You may complete as many transactions as possible (buy and sell one stock multiple times). Note: You cannot participate in multiple transactions at the same time (you must sell the previous stock before buying again).

Solution Approach #

• This problem is an enhanced version of Problem 121. It asks for the maximum profit. In this problem, you can buy and sell more than once, and buying and selling cannot be performed on the same day.
• The maximum profit must come from buying whenever the price drops and selling when it rises to a peak. As long as there is an upward peak, start calculating the money earned. For consecutive increases, you can calculate it by accumulating pairwise differences. Accumulating pairwise differences is equivalent to subtracting the valley value from the maximum value at the peak. Once you understand this point, the problem is very simple.

Code #

package leetcode

func maxProfit122(prices []int) int {
	profit := 0
	for i := 0; i < len(prices)-1; i++ {
		if prices[i+1] > prices[i] {
			profit += prices[i+1] - prices[i]
		}
	}
	return profit
}