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# Best Time to Buy and Sell Stock II

Say you have an array `prices` 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 as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).

**Note:** You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

**Example 1:**

```
Input: [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
             Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
```

**Example 2:**

```
Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.
```

**Example 3:**

```
Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.
```

**Constraints:**

* `1 <= prices.length <= 3 * 10 ^ 4`
* `0 <= prices[i] <= 10 ^ 4`

```java
class Solution {
    public int maxProfit(int[] prices) {
        int T_ik0 = 0, T_ik1 = Integer.MIN_VALUE;

        for (int price : prices) {
            int T_ik0_old = T_ik0;
            // If we want 0 stocks in our hand at end of ith day
            // Either take ans from 0 stock i-1th day
            // Or take ans from 1 stock i-1th day and sell it
            T_ik0 = Math.max(T_ik0, T_ik1 + price);
            // If we want 1 stock in our hand at end of ith day
            // Either take ans from 1 stock i-1th day
            // Or take ans from 0 stock i-1th day and but 1 stock at ith day
            T_ik1 = Math.max(T_ik1, T_ik0_old - price);
        }

        return T_ik0;
    }
}
```