LeetCode 122. Best Time to Buy and Sell Stock II

Description

https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/

Say you have an array for which the i^th 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.

Explanation

This is a follow-up question for Best Time to Buy and Sell Stock. The difference is that it allows multiple transactions.

A single for loop could solve the problem. If a price is greater than the previous price, add their difference to the max profit.

Java Solution

public class Solution {
    public int maxProfit(int[] prices) {
        int maxProfit = 0;
        
        if (prices.length <= 1) {
            return maxProfit;
        }
        
        for (int i = 1; i < prices.length; i++) {
            if (prices[i] > prices[i - 1]) {
               maxProfit += prices[i] - prices[i - 1]; 
            }
        }
    
        return maxProfit;
    }
}

Python Solution

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        max_profit = 0
        
        if prices is None:
            return max_profit
        
        for i in range(1, len(prices)):
            if prices[i] > prices[i - 1]:
                max_profit += prices[i] - prices[i - 1]
                
        return max_profit
            

• Time complexity : O(N). 
• Space complexity : O(1).