Leetcode - 121. Best Time to Buy and Sell Stock

Easy

Dec 22, 2025

#sliding-window

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

Constraints:

Notes

Intuition

Since we must buy before we sell, we want to find the lowest price followed by the highest price after it. When we encounter a new low, that becomes our new potential buy point since any future profit would be better calculated from this lower price.

Implementation

Use two pointers: buy starting at index 0 and sell at index 1. Iterate while sell is in bounds. Calculate the current profit—if positive, update the max profit. If the profit is negative or zero, we've found a lower price, so move buy to the current sell position. Always increment sell to continue scanning forward.

Edge-cases

If the array has fewer than 2 elements, no transaction is possible, so return 0 immediately.

Time: O(n) — single pass through the array
Space: O(1) — only tracking pointers and max profit

Complexity

Solution

from typing import List

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        if len(prices) <= 1:
            return 0
        max_p = 0
        b, s = 0, 1
        while s < len(prices):
            profit = prices[s] - prices[b]
            if profit > 0:
                max_p = max(max_p, profit)
            else:
                b = s
            s += 1
        return max_p
        



if __name__ == "__main__":
    # Include one-off tests here or debugging logic that can be run by running this file
    # e.g. print(solution.two_sum([1, 2, 3, 4], 3))
    solution = Solution()