121. Best Time to Buy and Sell Stock

1. Description

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.

2. Example

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.

3. Constraints

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

4. Solutions

Greedy

n = prices.size()
Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        const int n = prices.size();
        int buy = numeric_limits<int>::max(), profit = 0;
        for (int i = 0; i < n; ++i) {
            buy = min(buy, prices[i]);
            profit = max(profit, prices[i] - buy);
        }

        return profit;
    }
};