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# Best Time to Buy and Sell Stock III

Say you have an array 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 at most *two* transactions.

**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: [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
             Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 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.
```

```java
class Solution {
    public int maxProfit(int price[], int i, int n, int K, int dp[][][], int ongoing) {
        // Base cases
        if (i == n)
            return 0;
        if (K == 0) {
            dp[i][K][ongoing] = 0;
            return 0;
        }
        if (dp[i][K][ongoing] != -1)
            return dp[i][K][ongoing];
        int ans = maxProfit(price, i + 1, n, K, dp, ongoing);
        if (ongoing == 1) {
            int option = maxProfit(price, i + 1, n, K - 1, dp, 0) + price[i];
            ans = Math.max(ans, option);
        } else {
            int option = maxProfit(price, i + 1, n, K, dp, 1) - price[i];
            ans = Math.max(ans, option);
        }
        dp[i][K][ongoing] = ans;
        return ans;
    }

    public int maxProfit(int[] prices) {
        int n = prices.length;
        int dp[][][] = new int[n + 1][2 + 1][2];
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= 2; j++) {
                dp[i][j][0] = -1;
                dp[i][j][1] = -1;
            }
        }
        return maxProfit(prices, 0, n, 2, dp, 0);
    }
}
```