Best Time to Buy and Sell Stock
Post
Cancel

# LeetCode #121: Best Time to Buy and Sell Stock (C/C++).

easy

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 ith 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.

## 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 <= 105
• 0 <= prices[i] <= 104

## 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.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 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; } };