leetcode 309. The best time to buy and sell stocks includes a freezing period
Topic link: 309. The best time to buy and sell stocks includes a freezing period - LeetCode
topic overview
Given an array of integers, where the i-th represents the stock price on the i-th day.
Design an algorithm to calculate the maximum profit. You can complete as many transactions as possible (buy and sell a stock multiple times) subject to the following constraints:
- After selling the stock, you cannot buy the stock the next day (that is, the freezing period is 1 day).
Note: You cannot participate in multiple transactions at the same time (you must sell the previous stock before buying again).
Example 1:
Input: prices = [1,2,3,0,2]
Output: 3
Explanation: The corresponding transaction status is: [buy, sell, freeze period, buy, sell]
Example 2:
Input: prices = [1]
Output: 0
train of thought
It's still a five-part series
1. Determine the meaning of the dp array
dp[i][j]: The status on the i-th day is j, and the most remaining cash is dp[i][j].
If we follow the thinking of the previous questions, then stocks will be divided into two states: holding stocks and not holding stocks . However, because there is a freezing period in this question, the state of not holding stocks will be further subdivided .
(1) dp[i][0]: stock holding status (bought today or before)
No stock status:
(2) dp[i][1]: Keep selling stocks (sold stocks in the previous two days, passed the one-day freezing period or sold stocks in the previous day)
(3) dp[i][2]: sell stocks today
(4)dp[i][3]: freezing period (only one day)
2. Determine the recursive formula
(1) dp[i][0]: stock holding status
1) Bought the day before: dp[i - 1][0]
2) I bought it today, but the previous day was a freezing period: dp[i - 1][3] - prices[i]
3) I bought it today, but I kept selling the stock the day before: dp[i - 1][1] - prices[i]
所以dp[i][0] = max(dp[i - 1][0], dp[i - 1][3] - prices[i], dp[i - 1][1] - prices[i])
(2) dp[i][1]: Keep selling stocks
1) Keep selling stocks the day before: dp[i - 1][1]
2) The previous day is the freezing period: dp[i - 1][3]
所以dp[i][1] = max(dp[i - 1][1], dp[i - 1][3])
(3) dp[i][2]: sell stocks today
1) Sell today: dp[i][2] = dp[i - 1][0] + prices[i]
(4) dp[i][3]: freezing period
1) The stock was sold yesterday: dp[i][3] = dp[i - 1][2]
3. Array initialization
dp[0][0] = -prices[0] (the stock must be bought on the same day)
dp[0][1]=0
dp[0][2]=0
dp[0][3]=0
4. Determine the traversal order
front to back
5. Print the array (take 1, 2, 3, 0, 2 as an example)
Code
class Solution {
public:
int maxProfit(vector<int>& prices) {
int len = prices.size();
if(len == 0) return 0;
vector<vector<int>> dp(len,vector<int>(4,0));
dp[0][0] -= prices[0];
for(int i = 1;i < len;i++) {
dp[i][0] = max(dp[i - 1][0],max(dp[i - 1][3] - prices[i],dp[i - 1][1] - prices[i]));
dp[i][1] = max(dp[i - 1][1],dp[i - 1][3]);
dp[i][2] = dp[i - 1][0] + prices[i];
dp[i][3] = dp[i - 1][2];
}
return max(dp[len - 1][3],max(dp[len - 1][1],dp[len - 1][2]));
}
};Leetcode 714. The best time to buy and sell stocks includes handling fees
Topic link: 714. The best time to buy and sell stocks includes handling fees - LeetCode
topic overview
Given an integer array prices, the i-th element represents the stock price on the i-th day; the non-negative integer fee represents the handling fee for trading stocks.
You can complete transactions an unlimited number of times, but you need to pay a transaction fee for each transaction. If you have already bought a stock, you cannot continue to buy another stock until you sell it.
Returns the maximum value of profit.
Note: A transaction here refers to the entire process of buying, holding and selling stocks, and you only need to pay a handling fee for each transaction.
Example 1:
Input: prices = [1, 3, 2, 8, 4, 9], fee = 2
output: 8
Explanation: The maximum profit that can be achieved:
buy here prices[0] = 1
sell here prices[3] = 8
buy here prices[4] = 4
sell here prices[5] = 9
Total profit: ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
train of thought
This question is almost the same as Question 122 (The best time to buy and sell stocks||), the only difference is that this question needs to subtract the handling fee.
Code
class Solution {
public:
int maxProfit(vector<int>& prices, int fee) {
int len = prices.size();
if(len == 0) return 0;
vector<vector<int>> dp(len,vector<int>(2,0));
dp[0][0] -= prices[0];
for(int i = 1;i < prices.size();i++) {
dp[i][0] = max(dp[i - 1][0],dp[i - 1][1] - prices[i]);
dp[i][1] = max(dp[i - 1][1],dp[i - 1][0] + prices[i] - fee);
}
return max(dp[len - 1][0],dp[len - 1][1]);
}
};