main.py

# Source: https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii
# Title: Best Time to Buy and Sell Stock II
# Difficulty: Medium
# Author: Mu Yang <http://muyang.pro>

################################################################################################################################
# You are given an integer array `prices` where `prices[i]` is the price of a given stock on the `i^th` 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 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.
# Total profit is 4 + 3 = 7.
# ```
#
# **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.
# Total profit is 4.
# ```
#
# **Example 3:**
#
# ```
# Input: prices = [7,6,4,3,1]
# Output: 0
# Explanation: There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0.
# ```
#
# **Constraints:**
#
# - `1 <= prices.length <= 3 * 10^4`
# - `0 <= prices[i] <= 10^4`
#
################################################################################################################################


from typing import List


# We can buy the stock everyday and sell it at the next day if the price is higher.
class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        n = len(prices)
        ans = 0
        for i in range(n - 1):
            ans += max(0, prices[i + 1] - prices[i])

        return ans