Best Time to Buy and Sell Stock II

I joined leet code 30 days of code challenge and this is the 5th challenge. I would like to give some of the insight I found while solving this problem with swift. The problem as follows.

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).

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.

The first approach consist in taking advantage of the moment we find a valley and immediately after take into account its adjacent peak. now we add the difference.

class Solution {
    func maxProfit(_ prices: [Int]) -> Int {
        if prices.count == 0 {return 0}
        var maxProfit: Int = 0
        var valley: Int = prices[0]
        var peak: Int = prices[0]
        var i: Int = 0
        print(prices.count)
        while i < prices.count - 1{
            while i < prices.count - 1  && prices[i] >= prices[i+1]{
                i += 1
            }
            print("this is  a valley \(prices[i]) ")
            valley = prices[i]
            
            while i < prices.count - 1 && prices[i] < prices[i+1]{
                i += 1
            }
            print("this is a peak \(prices[i])")
            peak = prices[i]
            
            maxProfit += peak - valley
        }
}

The second solution takes advantage of the fact that the sum of every incremental value can be uses to simulated the highest peak. So the sum of small increment A, B and C are equals to the largest increment to the peak.

class Solution {
    func maxProfit(_ prices: [Int]) -> Int {
        var diff: Int = 0
        if prices.count == 0 || prices.count == 1 {return 0}
        for i in 1...prices.count-1{
            print(i)
            if prices[i] > prices[i-1]{
                diff += prices[i] - prices[i-1]
            }
        }
        
        return diff
    }
}

 

Author: enroblog

Computer science student at ITESM

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s