main.py

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

################################################################################################################################
# You are given an array `prices` where `prices[i]` is the price of a given stock on the `i^th` day.
#
# You want to maximize your profit by choosing a **single day** to buy one stock and choosing a **different day in the future** to sell that stock.
#
# Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return `0`.
#
# **Example 1:**
#
# ```
# Input: prices = [7,1,5,3,6,4]
# Output: 5
# Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
# Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.
# ```
#
# **Example 2:**
#
# ```
# Input: prices = [7,6,4,3,1]
# Output: 0
# Explanation: In this case, no transactions are done and the max profit = 0.
# ```
#
# **Constraints:**
#
# - `1 <= prices.length <= 10^5`
# - `0 <= prices[i] <= 10^4`
#
################################################################################################################################


from typing import List


# It is always better to buy it at lowest price and sell it at highest price.
# We can loop through the prices and track the previous minimal value.
class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        ans = 0
        minPrice = int(1e5)

        for price in prices:
            ans = max(ans, price - minPrice)
            minPrice = min(minPrice, price)

        return ans