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.
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.
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Example 2:
Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transactions are done and the max profit = 0.
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My Answer
# Time Limit Exceeded
class Solution:
def maxProfit(self, prices: List[int]) -> int:
max_profit = 0
for i in range(len(prices)):
for j in range(i,len(prices)):
max_profit = max(max_profit, prices[j]-prices[i])
return max_profit
# Time Limit Exceeded
class Solution:
def maxProfit(self, prices: List[int]) -> int:
max_profit = 0
for i in range(len(prices)):
local_max = max(prices[i:]) - prices[i]
max_profit = max(max_profit, local_max)
return max_profit
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Optimized Version
class Solution:
def maxProfit(self, prices: List[int]) -> int:
min_price = float('inf') # 현재까지의 최소 가격
max_profit = 0 # 최대 이익
for price in prices:
min_price = min(min_price, price) # 최솟값 갱신
max_profit = max(max_profit, price - min_price)
return max_profit
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내가 작성한 코드는 loop안에서 list에 대해 max연산을 수행하려면 O(n*n-i)가 걸리고,
따라서 O(n2)시간이 소요됨.
루프를 한번만 돌면서 처리해야한다.
두 숫자간의 min,max연산은 O(1)시간이므로 결과적으로 O(n)시간안에 해결된다.
