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# 188. Best Time to Buy and Sell Stock IV (H)

You are given an integer array `prices` where `prices[i]` is the price of a given stock on the `ith` day, and an integer `k`.

Find the maximum profit you can achieve. You may complete at most `k` transactions.

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

 

**Example 1:**

```
Input: k = 2, prices = [2,4,1]
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.
```

**Example 2:**

```
Input: k = 2, prices = [3,2,6,5,0,3]
Output: 7
Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4. Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
```

 

**Constraints:**

* `0 <= k <= 100`
* `0 <= prices.length <= 1000`
* `0 <= prices[i] <= 1000`

### Solution:

有了上一题 `k = 2` 的铺垫，这题应该和上一题的第一个解法没啥区别。但是出现了一个超内存的错误，原来是传入的 `k` 值会非常大，`dp` 数组太大了。现在想想，交易次数 `k` 最多有多大呢？

一次交易由买入和卖出构成，至少需要两天。所以说有效的限制 `k` 应该不超过 `n/2`，如果超过，就没有约束作用了，相当于 `k = +infinity`。这种情况是之前解决过的。

直接把之前的代码重用：

```java
int maxProfit_k_any(int max_k, int[] prices) {
    int n = prices.length;
    if (n <= 0) {
        return 0;
    }
    if (max_k > n / 2) {
        // 交易次数 k 没有限制的情况
        return maxProfit_k_inf(prices);
    }

    // base case：
    // dp[-1][...][0] = dp[...][0][0] = 0
    // dp[-1][...][1] = dp[...][0][1] = -infinity
    int[][][] dp = new int[n][max_k + 1][2];
    // k = 0 时的 base case
    for (int i = 0; i < n; i++) {
        dp[i][0][1] = Integer.MIN_VALUE;
        dp[i][0][0] = 0;
    }

    for (int i = 0; i < n; i++) 
        for (int k = max_k; k >= 1; k--) {
            if (i - 1 == -1) {
                // 处理 i = -1 时的 base case
                dp[i][k][0] = 0;
                dp[i][k][1] = -prices[i];
                continue;
            }
            dp[i][k][0] = Math.max(dp[i-1][k][0], dp[i-1][k][1] + prices[i]);
            dp[i][k][1] = Math.max(dp[i-1][k][1], dp[i-1][k-1][0] - prices[i]);     
        }
    return dp[n - 1][max_k][0];
}
```