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Solution.py
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Solution.py
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"""
Say you have an array for which the i-th element is the price of a given stock on day i.
Design an algorithm to find the maximum profit. You may complete at most k transactions.
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
Example 1:
Input: [2,4,1], k = 2
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.
Example 2:
Input: [3,2,6,5,0,3], k = 2
Output: 7
Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4.
Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
"""
class Solution:
def maxProfit(self, k: int, prices: List[int]) -> int:
def get_transactions(prices):
if len(prices) < 2:
return []
ret = []
start = prices[0]
for i, price in enumerate(prices):
if i == 0:
continue
if price < prices[i - 1]:
if start < prices[i - 1]:
ret.append(start)
ret.append(prices[i - 1])
start = price
if prices[-1] > start:
ret.append(start)
ret.append(prices[-1])
return ret
def remove_one_transaction(prices):
merge_pos = 0
cost = None
for i, price in enumerate(prices):
if i == 0:
continue
if cost is None or cost > abs(prices[i] - prices[i - 1]):
merge_pos = i - 1
cost = abs(prices[i] - prices[i - 1])
del prices[merge_pos:merge_pos + 2]
def get_profit(prices):
ret = 0
for i in range(0, len(prices), 2):
ret = ret + prices[i + 1] - prices[i]
return ret
prices = get_transactions(prices)
while len(prices) // 2 > k:
remove_one_transaction(prices)
return get_profit(prices)