(LeetCode) 122. Best Time to Buy and Sell Stock II

Best Time to Buy and Sell Stock II

• Explore : Interview > Top Interveiw Questions > Easy Collection
• 분류 : Array
• 난이도 : Easy

Problem

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

Find the maximum profit you can achieve. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).

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.

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

Constraints

• 1 <= prices.length <= 3 * 10^4
• 0 <= prices[i] <= 10^4

Solution

Exapnder

class Solution {
    fun maxProfit(prices: IntArray): Int {
        var sum = 0
        for (i in 0 until prices.size - 1) {
            if ((prices[i + 1] - prices[i]) > 0) {
                sum += prices[i + 1] - prices[i]
            }
        }
        return sum
    }
}

Point of Thinking

• 날짜를 특정해서 사고 파는 것과 오늘과 내일의 차이가 0 이상인, 즉 이득인 날짜마다 사고 파는 것의 총합은 동일하다.
  • 즉 [i + 1]과 [i]의 차이가 0 이상인 것의 총합이 최적해