Description
https://leetcode.com/problems/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.
Constraints:
1 <= prices.length <= 105
0 <= prices[i] <= 104
Explanation
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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