121. Best Time to Buy and Sell Stock

Easy LeetCode

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.

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.

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

How to solve the problem

• Brute Force O(n^2)

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        result = 0
        for buy in range(len(prices)):
            for sell in range(buy+1,len(prices)):
                if prices[sell] > prices[buy]:
                    profit = prices[sell] - prices[buy]
                    result = max(result, profit)
        return result

• Greedy

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        min_price = inf
        result = 0
        for i in range(len(prices)):
            if prices[i] < min_price:
                min_price = prices[i]
            profit = prices[i] - min_price
            result = max(profit,result)
        return result

Complexity

• Brute Force: Time complexity: O(n^2), Space complexity: O(1) - Where n is the length of the prices array. We check all possible buy-sell pairs.
• Greedy: Time complexity: O(n), Space complexity: O(1) - We traverse the array once and use constant extra space.