# 801. Minimum Swaps To Make Sequences Increasing

**<font color=red>难度: Medium</font>**

## 刷题内容

> 原题连接

* https://leetcode.com/problems/minimum-swaps-to-make-sequences-increasing/description/

> 内容描述

```

We have two integer sequences A and B of the same non-zero length.

We are allowed to swap elements A[i] and B[i].  Note that both elements are in the same index position in their respective sequences.

At the end of some number of swaps, A and B are both strictly increasing.  (A sequence is strictly increasing if and only if A[0] < A[1] < A[2] < ... < A[A.length - 1].)

Given A and B, return the minimum number of swaps to make both sequences strictly increasing.  It is guaranteed that the given input always makes it possible.

Example:
Input: A = [1,3,5,4], B = [1,2,3,7]
Output: 1
Explanation: 
Swap A[3] and B[3].  Then the sequences are:
A = [1, 3, 5, 7] and B = [1, 2, 3, 4]
which are both strictly increasing.
Note:

A, B are arrays with the same length, and that length will be in the range [1, 1000].
A[i], B[i] are integer values in the range [0, 2000].
```

## 解题方案

> 思路 1


参考大神[Luckypants](https://leetcode.com/problems/minimum-swaps-to-make-sequences-increasing/discuss/119830/Python-14-line-O(1)-space-O(n)-time-DP-solution)的思路


- dp[i][0] means the least swaps used to make A[:i+1] and B[:i+1] sorted having no swap at i-th position.
- dp[i][1] means the least swaps used to make A[:i+1] and B[:i+1] sorted having swap at i-th position.

```python
class Solution(object):
    def minSwap(self, A, B):
        """
        :type A: List[int]
        :type B: List[int]
        :rtype: int
        """
        n = len(A)
        pre = [0, 1]
        for i in range(1, n):
            cur = [sys.maxsize, sys.maxsize]
            if A[i]>A[i-1] and B[i]>B[i-1]:
                cur[0] = min(cur[0], pre[0])
                cur[1] = min(cur[1], pre[1]+1)
            if A[i]>B[i-1] and B[i]>A[i-1]:
                cur[0] = min(cur[0], pre[1])
                cur[1] = min(cur[1], pre[0]+1)
            pre = cur
        return min(pre)
```
