LeetCode 121 - Best Time to Buy and Sell Stock

Difficulty: easy

Stock Series:

Part 1: LeetCode 121 - Best Time to Buy and Sell Stock
Part 2: LeetCode 122 - Best Time to Buy and Sell Stock II
Part 3: LeetCode 123 - Best Time to Buy and Sell Stock III
Part 4: LeetCode 188 - Best Time to Buy and Sell Stock IV
Part 5: LeetCode 309 - Best Time to Buy and Sell Stock with Cooldown
Part 6: LeetCode 714 - Best Time to Buy and Sell Stock with Transaction Fee

Problem Description

English (Best Time to Buy and Sell Stock)

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

Chinese (买卖股票的最佳时机)

给定一个数组 prices ，它的第 i 个元素 prices[i] 表示一支给定股票第 i 天的价格。

你只能选择 某一天 买入这只股票，并选择在 未来的某一个不同的日子 卖出该股票。设计一个算法来计算你所能获取的最大利润。

返回你可以从这笔交易中获取的最大利润。如果你不能获取任何利润，返回 0 。

示例 1：

输入：[7,1,5,3,6,4]
输出：5
解释：在第 2 天（股票价格 = 1）的时候买入，在第 5 天（股票价格 = 6）的时候卖出，最大利润 = 6-1 = 5 。
     注意利润不能是 7-1 = 6, 因为卖出价格需要大于买入价格；同时，你不能在买入前卖出股票。

示例 2：

输入：prices = [7,6,4,3,1]
输出：0
解释：在这种情况下, 没有交易完成, 所以最大利润为 0。

提示：

• 1 <= prices.length <= 10^5
• 0 <= prices[i] <= 10^4

Solution

C++

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int max_profit = 0;

        int min_price = INT_MAX;
        for (int price : prices) {
            max_profit = max(max_profit, price - min_price);
            min_price = min(min_price, price);
        }

        return max_profit;
    }
};