# Best Time to Buy and Sell Stock -- Basic

Published: Jul 26, 2022

Easy Array DP

## Problem 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.

Constraints:

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

## Examples

``````example 1:
Input: prices = [7,1,5,3,6,4]
Explanation: 5 is the best profit which is from buying on the index 1 an selling on the index 4.
``````
``````example 2:
Input: prices = [7,6,4,3,1]
Explanation: the array value is ever descreasing. There is no profit available.
``````

## How to Solve

The problems of the best time to buy and sell stock have variations. Each has a slight different criteria to find the answer. This is the most basic and easy buy-sell-stock problem.

We are allowed to buy and sell only once. The minimum price so far should be updated in each array index if necessary. If the price at the current index is not the minimum, calculate profit so far. Then, update the grand max profit comparing with the current profit.

## Solution

``````#include <vector>
#include <iostream>
using namespace std;

{
public:
int maxProfit(vector<int> &prices)
{
int min_v = INT_MAX;
int max_profit = 0;
for (int i = 0; i < prices.size(); ++i)
{
if (prices[i] < min_v)
{
min_v = prices[i];
}
else
{
max_profit = max(max_profit, prices[i] - min_v);
}
}
return max_profit;
}
};
``````
``````class BestTimeToBuyAndSellStock {
public int maxProfit(int[] prices) {
int min_price = Integer.MAX_VALUE;
int max_profit = 0;
for (int i = 0; i < prices.length; ++i) {
if (min_price > prices[i]) {
min_price = prices[i];
} else {
max_profit = Math.max(max_profit, prices[i] - min_price);
}
}
return max_profit;
}
}
``````
``````/**
* @param {number[]} prices
* @return {number}
*/
var maxProfit = function(prices) {
let min_price = Number.MAX_SAFE_INTEGER, max_profit = 0;
for (let i = 0; i < prices.length; i++) {
if (min_price > prices[i]) {
min_price = prices[i];
} else {
max_profit = Math.max(max_profit, prices[i] - min_price)
}
}
return max_profit;
};
``````
``````from typing import List

def maxProfit(self, prices: List[int]) -> int:
max_profit, min_price = 0, float('inf')
if not prices:
return 0
min_price = prices
for i in range(1, len(prices)):
if prices[i] < min_price:
min_price = prices[i]
else:
max_profit = max(max_profit, prices[i] - min_price)
return max_profit
``````
``````# @param {Integer[]} prices
# @return {Integer}
def max_profit(prices)
min_price = 10**4 + 1
max_profit = 0
prices.each do |price|
if min_price > price
min_price = price
else
max_profit = [max_profit, price - min_price].max
end
end
max_profit
end
``````

## Complexities

• Time: O(n) – n: length of the input array.
• Space: O(1)