Best Time to Buy and Sell Stock II

Greedy

15 min

Problem Statement

You are given an integer array prices where prices[i] is the price of a given stock on the ith day. On each day, you may decide to buy and/or sell the stock. You can only hold at most one share of the stock at any time. However, you can buy it then immediately sell it on the same day. Find and return the maximum profit you can achieve.

Example

Example 1:

Input: prices = [7,1,5,3,6,4]

prices: [7, 1, 5, 3, 6, 4]

Output: 7

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. Total profit is 4 + 3 = 7.

Solution

The greedy approach is to take every opportunity to make a profit. We can iterate through the prices and whenever the price of the next day is higher than the current day, we buy on the current day and sell on the next day.

Algorithm Steps

Initialize `maxProfit` to 0.
Iterate through the prices array from the second day.
If the price on the current day is greater than the price on the previous day, add the difference to `maxProfit`.
Return `maxProfit`.

Best Time to Buy and Sell Stock II

Greedy Approach

Input: prices = [7, 1, 5, 3, 6, 4]

Output: 0

Progress1 / 1

Ready to start the visualization

Prices

Total Profit

Best Time to Buy and Sell Stock II Solution


class Solution:
    def maxProfit(self, prices: list[int]) -> int:
        profit = 0
        for i in range(1, len(prices)):
            if prices[i] > prices[i-1]:
                profit += prices[i] - prices[i-1]
        return profit