<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:
        
        nums1 = [-1] + nums1
        nums2 = [-1] + nums2

        @cache
        def dp(idx, f):
            if idx == len(nums1):
                return 0

            n1v, n2v = nums1[idx], nums2[idx]
            l1v, l2v = nums1[idx-1], nums2[idx-1]
            if f:
                l1v, l2v = l2v, l1v
            
            lst = [inf]
            if n1v > l1v and n2v > l2v:
                lst.append(dp(idx+1, 0))
            
            if n2v > l1v and n1v > l2v:
                lst.append(1 + dp(idx+1, 1))
            
            return min(lst)
        
        return dp(1, 0)

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