<p>You are given two <strong>0-indexed</strong> integer arrays <code>nums1</code> and <code>nums2</code> of length <code>n</code>.</p>

<p>Let&#39;s define another <strong>0-indexed</strong> integer array, <code>nums3</code>, of length <code>n</code>. For each index <code>i</code> in the range <code>[0, n - 1]</code>, you can assign either <code>nums1[i]</code> or <code>nums2[i]</code> to <code>nums3[i]</code>.</p>

<p>Your task is to maximize the length of the <strong>longest non-decreasing subarray</strong> in <code>nums3</code> by choosing its values optimally.</p>

<p>Return <em>an integer representing the length of the <strong>longest non-decreasing</strong> subarray in</em> <code>nums3</code>.</p>

<p><strong>Note: </strong>A <strong>subarray</strong> is a contiguous <strong>non-empty</strong> sequence of elements within an array.</p>

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

<pre>
<strong>Input:</strong> nums1 = [2,3,1], nums2 = [1,2,1]
<strong>Output:</strong> 2
<strong>Explanation: </strong>One way to construct nums3 is: 
nums3 = [nums1[0], nums2[1], nums2[2]] =&gt; [2,2,1]. 
The subarray starting from index 0 and ending at index 1, [2,2], forms a non-decreasing subarray of length 2. 
We can show that 2 is the maximum achievable length.</pre>

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

<pre>
<strong>Input:</strong> nums1 = [1,3,2,1], nums2 = [2,2,3,4]
<strong>Output:</strong> 4
<strong>Explanation:</strong> One way to construct nums3 is: 
nums3 = [nums1[0], nums2[1], nums2[2], nums2[3]] =&gt; [1,2,3,4]. 
The entire array forms a non-decreasing subarray of length 4, making it the maximum achievable length.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [1,1], nums2 = [2,2]
<strong>Output:</strong> 2
<strong>Explanation:</strong> One way to construct nums3 is: 
nums3 = [nums1[0], nums1[1]] =&gt; [1,1]. 
The entire array forms a non-decreasing subarray of length 2, making it the maximum achievable length.
</pre>

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

<ul>
	<li><code>1 &lt;= nums1.length == nums2.length == n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums1[i], nums2[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def maxNonDecreasingLength(self, nums1: List[int], nums2: List[int]) -> int:
        n=len(nums1)
        break_ids = list()
        max_length = list()
        
        if n == 1:
            max_length.append(1)
        else:
            for i in range(1, n):
                if max(nums1[i], nums2[i]) < min(nums1[i-1], nums2[i-1]):
                    break_ids.append(i)
            
            if not break_ids:
                max_length.append(max(self.nonDecreasingLength(nums1, nums2)))
            else:
                break_ids = [0] + break_ids + [n-1]
                for j in range(1, len(break_ids)):
                    if max_length:
                        if max(max_length) > break_ids[j] - break_ids[j-1]:
                            continue

                    max_length.append(max(self.nonDecreasingLength(nums1[break_ids[j-1]:(break_ids[j]+1)], nums2[break_ids[j-1]:(break_ids[j]+1)])))

        return max(max_length)

    def nonDecreasingLength(self, nums1: List[int], nums2: List[int]) -> list:
        n=len(nums1)
        length = list()

        if n == 1:
            length.append(1)
        else:
            nums3, nonMinIdx = self.nonDecreasingArrayBeg(nums1, nums2)
            length.append(len(nums3))
            for i in nonMinIdx:
                if length:
                    if max(length) > n - 1 - i:
                        break
                nums3, _ = self.nonDecreasingArrayBeg(nums1[i:], nums2[i:])
                length.append(len(nums3))
                for j in _:
                    if i + j > max(nonMinIdx):
                        nonMinIdx.append(i + j)

        return length

    def nonDecreasingArrayBeg(self, nums1: List[int], nums2: List[int]) -> (list, list):
        n=len(nums1)
        nums3 = list()
        nonMinIdx = list()
    
        nums3.append(min(nums1[0], nums2[0]))
        for i in range(1, n):
            if min(nums1[i], nums2[i]) >= nums3[i-1]:
                nums3.append(min(nums1[i], nums2[i]))
            elif max(nums1[i], nums2[i]) >= nums3[i-1]:
                nums3.append(max(nums1[i], nums2[i]))
                nonMinIdx.append(i)
            else:
                break
        
        return nums3, nonMinIdx
        
896