Leetcode - 121: Best Time to Buy and Sell Stock

What's the question?

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.

Let's check an example

Input: prices = [7,1,5,3,6,4]
Output: 5

Here, we are given prices for 6 days we need to tell how can we earn a maximum profit by buy a stock on some day and selling it on some another day For the prices list given above we see that the minimum price to buy the stock when going from left to right is 1 and after the price 1 the maximum price which the stock can be sold for is 6 so the profit becomes 6 - 1 = 5.

Pretty Easy right! Let's translate this into some code

So, at the start we can initialize two variables minPrice which will be price at the 0^th index and profit which will be 0 at the start.

Now as we iterate through the prices list we will have two steps,

🔵 Step - 1: Check if price at current index is lesser than our minPrice

🟦 If True at Step - 1: Update the minPrice with the current price

🔵 Step - 2: Update the profit

To update the profit we can find the maximum between the current profit and difference between the current price and minPrice. If the minPrice in the first step would be changed then the differene between the current price and the minPrice will be 0. And if the profit is greater than 0 it will remain unchanged and even if it's 0 it will remain 0 only.

Once the end of the for loop is reached we return the profit

Let's code it out

def maxPrice(prices):

    minPrice = min(prices)
    profit = 0

    for ix, price in enumerate(prices[1:]):

        if price < minPrice:
            minPrice = price

        profit = max(profit, prices[ix] - minPrice)

    return profit

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