<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:
        # choices: swap some set of indices
        # naive: try random swaps until works
        # 2^10^5. not good at all
        # since min increasing is independent of previous elements in any window [i:]
        # we can track min to make a window increasing
        # [1,3,5,4]
        # [1,2,3,7] 
        # 

        # since we know it is bound to work
        # we find the elements which are out of place
        # match with the elements in the other array which are also out of place
        # those MUST be swapped
        # but then once swapped, the bigger element might make the next item out of order, in which case we must
        # swap again
        # this means a single pass like this won't do
        # if out of order in one, swap (it should decrease) as you walk. 
        # comparing previous

        # we can either swap the current element or the next element
        # better to limit it to the 
        # 
        # track whether previous was swapped and the number swaps here
        n = len(nums1)
        @lru_cache(None)
        def r(swapped, idx):

            if idx == n:
                return 0 # 0 swaps needed
            best = float('inf')
            pn1,pn2 = -1,-1
            if swapped and idx > 0:
                pn1,pn2 = nums2[idx-1], nums1[idx-1]
            elif idx > 0:
                pn1,pn2 = nums1[idx-1], nums2[idx-1]
            
            n1,n2 = nums1[idx], nums2[idx]

            # we might swap as long as the previous elements we can't touch are maintained
            if (n2 > pn1 and n1 > pn2):
                best = r(True, idx+1)+1 # if swapping this one works try it
            if (n1 > pn1 and n2 > pn2):
                best = min(best, r(False, idx+1))
                
                
            return best
        
        ans = r(False, 0)
        return ans
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