Best Time to Buy and Sell Stock with Cooldown

Sharemy DP solution (By State Machine Thinking)Hi,I just come across this problem, and it's very frustating since I'm bad at DP.So I just draw all the actions that can be done.Here is the drawing (Feel like an elementary ...)There are three states, according to the action that you can take.Hence, from there, you can now the profit at a state at time i as:

s0[i] = max(s0[i - 1], s2[i - 1]); // Stay at s0, or rest from s2
s1[i] = max(s1[i - 1], s0[i - 1] - prices[i]); // Stay at s1, or buy from s0
s2[i] = s1[i - 1] + prices[i]; // Only one way from s1

Then, you just find the maximum of s0and s2, since they will be the maximum profit we need (No one can buy stock and left with more profit that sell right :) )Define base case:

s0[0] = 0; // At the start, you don't have any stock if you just rest
s1[0] = -prices[0]; // After buy, you should have -prices[0] profit. Be positive!

s2[0] = INT_MIN; // Lower base case

class Solution {public:int maxProfit(vector<int>& prices) {if (prices.size() <= 1)return 0;vector<int> s0(prices.size(), 0);vector<int> s1(prices.size(), 0);vector<int> s2(prices.size(), 0);s0[0] = 0;s1[0] = -prices[0];s2[0] = 0;for (int i=1; i<prices.size(); ++i) {s0[i] = max(s0[i-1], s2[i-1]);s1[i] = max(s1[i-1], s0[i-1] - prices[i]);s2[i] = s1[i-1] + prices[i];}return max(s0[prices.size() - 1], s2[prices.size() - 1]);}};

class Solution(object):
def maxProfit(self, prices):
"""
:type prices: List[int]
:rtype: int
"""
if len(prices) <= 1:
return 0
s0, s1, s2 = [0]*len(prices), [0]*len(prices), [0]*len(prices)
s1[0] = -prices[0]
for i in range(1, len(prices)):
s0[i] = max(s0[i - 1], s2[i - 1])
s1[i] = max(s1[i - 1], s0[i - 1] - prices[i])
s2[i] = s1[i - 1] + prices[i]

return max(s0[-1], s2[-1])

Share my thinking processThe series of problems are typical dp. The key for dp is to find the variables to represent the states and deduce the transition function.Of course one may come up with a O(1) space solution directly, but I think it is better to be generous when you think and be greedy when you implement.The natural states for this problem is the 3 possible transactions : 

buy

, 

sell

, 

rest

. Here 

rest

 means no transaction on that day (aka cooldown).Then the transaction sequences can end with any of these three states.For each of them we make an array, 

buy

, 

sell

 and 

rest

.

buy[i]

 means before day 

i

 what is the maxProfit for any sequence end with 

buy

.

sell[i]

 means before day 

i

 what is the maxProfit for any sequence end with sell.

rest[i]

 means before day 

i

 what is the maxProfit for any sequence end with 

rest

.Then we want to deduce the transition functions for 

buy

 

sell

 and 

rest

. By definition we have:

buy[i]  = max(rest[i-1]-price, buy[i-1])sell[i] = max(buy[i-1]+price, sell[i-1])rest[i] = max(sell[i-1], buy[i-1], rest[i-1])

Where 

price

 is the price of day 

i

. All of these are very straightforward. They simply represents :

(1) We have to `rest` before we `buy` and(2) we have to `buy` before we `sell`

One tricky point is how do you make sure you 

sell

 before you 

buy

, since from the equations it seems that 

[buy, rest, buy]

 is entirely possible.Well, the answer lies within the fact that 

buy[i] <= rest[i]

 which means 

rest[i] = max(sell[i-1], rest[i-1])

. That made sure 

[buy, rest, buy]

 is never occurred.A further observation is that and 

rest[i] <= sell[i]

 is also true therefore

rest[i] = sell[i-1]

Substitute this in to 

buy[i]

 we now have 2 functions instead of 3:

buy[i] = max(sell[i-2]-price, buy[i-1])sell[i] = max(buy[i-1]+price, sell[i-1])

This is better than 3, butwe can do even betterSince states of day 

i

 relies only on 

i-1

 and 

i-2

 we can reduce the O(n) space to O(1). And here we are at our final solution:Java

public int maxProfit(int[] prices) {int sell = 0, prev_sell = 0, buy = Integer.MIN_VALUE, prev_buy;for (int price : prices) {prev_buy = buy;buy = Math.max(prev_sell - price, prev_buy);prev_sell = sell;sell = Math.max(prev_buy + price, prev_sell);}return sell;}

C++

int maxProfit(vector<int> &prices) {int buy(INT_MIN), sell(0), prev_sell(0), prev_buy;for (int price : prices) {prev_buy = buy;buy = max(prev_sell - price, buy);prev_sell = sell;sell = max(prev_buy + price, sell);}return sell;}

For this problem it is ok to use 

INT_MIN

 as initial value, but in general we would like to avoid this. We can do the same as the following python:Python

def maxProfit(self, prices):if len(prices) < 2:return 0sell, buy, prev_sell, prev_buy = 0, -prices[0], 0, 0for price in prices:prev_buy = buybuy = max(prev_sell - price, prev_buy)prev_sell = sellsell = max(prev_buy + price, prev_sell)return sell