LeetCode 121. Best Time to Buy and Sell Stock



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.
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.


  • 1 <= prices.length <= 105
  • 0 <= prices[i] <= 104


The problem is seeking the max profit in buying and selling stocks.

This is a simple modeling stock trading in real life. The essence of the problem is looking for the max value of prices[j] – prices[i], where j > i.

It’s easy to solve the problem by keep tracking of the minimum price and max profit when iterating the array.

Java Solution

public class Solution { 
    public int maxProfit(int[] prices) { 
        int maxProfit = 0; 
        int minPrice = Integer.MAX_VALUE; 
        for (int i = 0; i < prices.length; i++) { 
            minPrice = Math.min(minPrice, prices[i]); 
            maxProfit = Math.max(maxProfit, prices[i] - minPrice); 
        return maxProfit; 

Python Solution

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        if not prices:
            return 0
        max_profit = 0
        min_price = prices[0]
        for price in prices[1:]:            
            profit = price - min_price
            max_profit = max(max_profit, profit)
            min_price = min(min_price, price)
        return max_profit
  • Time complexity: O(N).
  • Space complexity: O(1).

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