Best Time to Buy and Sell Stock with Cooldown
Detailed guide and Python implementation for the 'Best Time to Buy and Sell Stock with Cooldown' problem.
1. 学ぶ
The 'Best Time to Buy and Sell Stock with Cooldown' problem is a key challenge in the 2D DP section.
This implementation focuses on medium-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Best Time to Buy and Sell Stock with Cooldown.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Best Time to Buy and Sell Stock with Cooldown carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
問題提起
You are given an array prices where prices[i] is the price of a given stock on the ith day.
Find the maximum profit you can achieve. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times) with the following restrictions:
- After you sell your stock, you cannot buy stock on the next day (i.e., cooldown one day).
Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
Write a function maxProfit(prices: List[int]) -> int.
- •1 <= len(prices) <= 5000
- •0 <= prices[i] <= 1000
例
prices = [1,2,3,0,2]
3
Transactions = [buy, sell, cooldown, buy, sell].
prices = [1]
0
No transaction can be made.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
解決する準備はできましたか?
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インタビューの洞察とバリエーション
複雑さの分析の内訳
なぜ時間がかかるのか: Directly evaluates all possibilities.
なぜ宇宙なのか: Uses standard local memory.
なぜ時間がかかるのか: Optimized paths reduce total operations.
なぜ宇宙なのか: May trade memory for speed.
最適化されたソリューションの Python コード
最適化されたソリューションの Python コード
def max_profit_opt(prices):
dp = {} # (i, buying)
def dfs(i, buying):
if i >= len(prices): return 0
if (i, buying) in dp: return dp[(i, buying)]
if buying:
buy = dfs(i + 1, not buying) - prices[i]
skip = dfs(i + 1, buying)
dp[(i, buying)] = max(buy, skip)
else:
sell = dfs(i + 2, not buying) + prices[i]
skip = dfs(i + 1, buying)
dp[(i, buying)] = max(sell, skip)
return dp[(i, buying)]
return dfs(0, True)ブルート フォース コード (スポイラーガード付き)
ブルート フォース コード (スポイラーガード付き)
def max_profit_brute(prices):
def dfs(i, buying):
if i >= len(prices): return 0
if buying:
buy = dfs(i + 1, not buying) - prices[i]
skip = dfs(i + 1, buying)
return max(buy, skip)
else:
sell = dfs(i + 2, not buying) + prices[i]
skip = dfs(i + 1, buying)
return max(sell, skip)
return dfs(0, True)Algorithm Pattern Checklist
When dealing with 2D DP data patterns.
- Are constraints clear?
- Is there a linear or logarithmic optimization possible?
Key Revision Notes
Standard 2D DP problem properties apply.
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