Best Time to Buy and Sell Stock with Cooldown

Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times) with the following restrictions:

• You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
• After you sell your stock, you cannot buy stock on next day. (ie, cooldown 1 day)

Example:

prices = [1, 2, 3, 0, 2]
maxProfit = 3
transactions = [buy, sell, cooldown, buy, sell]

分析：

因为当前的选择对后面的选择会产生影响，我们可以倒着来分析。

我们用maximumProfit[i]来表示从i 到 length - 1 可以获取的最大profit. 怎么计算maximumProfit[i]呢？那么对于i来讲，最坏情况是maximumProfit[i] = maximumProfit[i + 1].

然后我们把price[i]作为当前最低价，如果后面发现更高的价格price[j]，我们就做出更新。

if (j < length - 2) {
　　maximumProfit[i] = Math.max(prices[j] - low + maximumProfit[j + 2], maximumProfit[i]);
} else {
　　maximumProfit[i] = Math.max(prices[j] - low, maximumProfit[i]);
}

 public int maxProfit(int[] prices) {
     if (prices == null || prices.length <= )
         return ;

     int length = prices.length;
     int[] maximumProfit = new int[length];
     maximumProfit[length - ] = ;
     maximumProfit[length - ] = Math.max(, prices[length - ] - prices[length - ]);

     for (int i = length - ; i >= ; i--) {
         maximumProfit[i] = maximumProfit[i + ];
         int low = prices[i];
         for (int j = i + ; j < length; j++) {
             if (prices[j] > low) {
                 if (j < length - ) {
                     maximumProfit[i] = Math.max(prices[j] - low + maximumProfit[j + ], maximumProfit[i]);
                 } else {
                     maximumProfit[i] = Math.max(prices[j] - low, maximumProfit[i]);
                 }
             } else {
                 low = prices[j];
             }
         }
     }
     return maximumProfit[];
 }

很明显，时间总复杂度为O(n^2).

另一个方法：

Analysis: https://discuss.leetcode.com/topic/30431/easiest-java-solution-with-explanations

1. Define States

To represent the decision at index i:

• buy[i]: Max profit till index i. The series of transaction is ending with a buy.
• sell[i]: Max profit till index i. The series of transaction is ending with a sell.

To clarify:

• Till index i, the buy / sell action must happen and must be the last action. It may not happen at index i. It may happen at i - 1, i - 2, ... 0.
• In the end n - 1, return sell[n - 1]. Apparently we cannot finally end up with a buy. In that case, we would rather take a rest at n - 1.
• For special case no transaction at all, classify it as sell[i], so that in the end, we can still return sell[n - 1]. Thanks @alex153 @kennethliaoke @anshu2.
  

  ---

2. Define Recursion

• buy[i]: To make a decision whether to buy at i, we either take a rest, by just using the old decision at i - 1, or sell at/beforei - 2, then buy at i, We cannot sell at i - 1, then buy at i, because of cooldown.
• sell[i]: To make a decision whether to sell at i, we either take a rest, by just using the old decision at i - 1, or buy at/before i - 1, then sell at i.

So we get the following formula:

buy[i] = Math.max(buy[i - 1], sell[i - 2] - prices[i]);
sell[i] = Math.max(sell[i - 1], buy[i - 1] + prices[i]);

---

3. Optimize to O(1) Space

DP solution only depending on i - 1 and i - 2 can be optimized using O(1) space.

• Let b2, b1, b0 represent buy[i - 2], buy[i - 1], buy[i]
• Let s2, s1, s0 represent sell[i - 2], sell[i - 1], sell[i]

Then arrays turn into Fibonacci like recursion:

b0 = Math.max(b1, s2 - prices[i]);
s0 = Math.max(s1, b1 + prices[i]);

---

4. Write Code in 5 Minutes

First we define the initial states at i = 0:

• We can buy. The max profit at i = 0 ending with a buy is -prices[0].
• We cannot sell. The max profit at i = 0 ending with a sell is 0.

 public int maxProfit(int[] prices) {
     if(prices == null || prices.length <= ) return ;

     int b0 = -prices[], b1 = b0;
     int s0 = , s1 = , s2 = ;

     for(int i = ; i < prices.length; i++) {
         b0 = Math.max(b1, s2 - prices[i]);
         s0 = Math.max(s1, b1 + prices[i]);
         b1 = b0; s2 = s1; s1 = s0;
     }
     return s0;
 }