**source:**https://leetcode.com/problems/best-time-to-buy-and-sell-stock**C/C++**

**Solution to LeetCode**problem

**121**.

**Best Time to Buy and Sell Stock**.

## Problem

You are given an array `prices`

where `prices[i]`

is the price of a given stock on the `i`

day.^{th}

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`

.

## Examples

**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}

## Solution

This is a *greedy problem*. To solve it:

- Iterate throught every value.
- If current value is lower than a previous lower, then we have a new buy price (lower).
- If the profit of selling at current value (
`current - lower`

) is greater than the current profit, we have a new maximum.

- Return the max profit found.

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class Solution {
public:
int maxProfit(vector<int>& prices) {
int lowestPrice = INT_MAX;
int maxProfit = 0;
int cProfit = 0;
for (int i = 0; i < prices.size(); i++) {
if (prices[i] < lowestPrice)
lowestPrice = prices[i];
cProfit = prices[i] - lowestPrice;
if (cProfit > maxProfit)
maxProfit = cProfit;
}
return maxProfit;
}
};