<p>You are given two integer arrays of the same length <code>nums1</code> and <code>nums2</code>. In one operation, you are allowed to swap <code>nums1[i]</code> with <code>nums2[i]</code>.</p>

<ul>
	<li>For example, if <code>nums1 = [1,2,3,<u>8</u>]</code>, and <code>nums2 = [5,6,7,<u>4</u>]</code>, you can swap the element at <code>i = 3</code> to obtain <code>nums1 = [1,2,3,4]</code> and <code>nums2 = [5,6,7,8]</code>.</li>
</ul>

<p>Return <em>the minimum number of needed operations to make </em><code>nums1</code><em> and </em><code>nums2</code><em> <strong>strictly increasing</strong></em>. The test cases are generated so that the given input always makes it possible.</p>

<p>An array <code>arr</code> is <strong>strictly increasing</strong> if and only if <code>arr[0] &lt; arr[1] &lt; arr[2] &lt; ... &lt; arr[arr.length - 1]</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [1,3,5,4], nums2 = [1,2,3,7]
<strong>Output:</strong> 1
<strong>Explanation:</strong> 
Swap nums1[3] and nums2[3]. Then the sequences are:
nums1 = [1, 3, 5, 7] and nums2 = [1, 2, 3, 4]
which are both strictly increasing.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [0,3,5,8,9], nums2 = [2,1,4,6,9]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= nums1.length &lt;= 10<sup>5</sup></code></li>
	<li><code>nums2.length == nums1.length</code></li>
	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 2 * 10<sup>5</sup></code></li>
</ul>

class Solution:
    def minSwap(self, nums1: List[int], nums2: List[int]) -> int:
        """
        brute force 2^n * n
        for i, swap or not, then check if increasing

        if i is the desc place, only 2 ways to swap, itself or after 
        """

        visited = {}
        def dfs(prev1, prev2, i):
            if i >= len(nums1):
                return 0
            if (prev1, prev2, i) in visited:
                return visited[(prev1, prev2, i)]
            ans = float('inf')
            if prev1 < nums1[i] and prev2 < nums2[i]:
                ans = dfs(nums1[i], nums2[i], i+1)
            if prev1 < nums2[i] and prev2 < nums1[i]:
                ans = min(ans, 1 + dfs(nums2[i], nums1[i], i+1))
            visited[(prev1, prev2, i)] = ans
            return ans

        return dfs(-1, -1, 0)
    

        """ not working yet.. too complicated to handle edge cases
        # nums1.append(float('inf'))
        # nums2.append(float('inf'))
        n = len(nums1)
        ans = 0
        i = 0
        while i < n-1:
            if nums1[i]<nums1[i+1] and nums2[i]<nums2[i+1]:
                i += 1
            else:
                if i == n-2 and nums1[i]<nums2[i+1] and nums2[i]<nums1[i+1]:
                    nums1[i+1], nums2[i+1] = nums2[i+1], nums1[i+1]
                    ans += 1
                elif i < n-2 and nums1[i]<nums2[i+1]<nums1[i+2] and nums2[i]<nums1[i+1]<nums2[i+2]:
                    nums1[i+1], nums2[i+1] = nums2[i+1], nums1[i+1]
                    ans += 1
                elif i==0 and nums2[i]<nums1[i+1] and nums1[i]<nums2[i+1]:
                    nums1[i], nums2[i] = nums2[i], nums1[i]
                    ans += 1
                elif i != 0 and nums1[i-1]<nums2[i]<nums1[i+1] and nums2[i-1]<nums1[i]<nums2[i+1]:
                    nums1[i], nums2[i] = nums2[i], nums1[i]
                    ans += 1
                else:
                    # print(i)
                    return -1
                i += 1
            # print(nums1, nums2)
        return ans
        """


        
119976