<p>Given two sorted arrays <code>nums1</code> and <code>nums2</code> of size <code>m</code> and <code>n</code> respectively, return <strong>the median</strong> of the two sorted arrays.</p>

<p>The overall run time complexity should be <code>O(log (m+n))</code>.</p>

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

<pre>
<strong>Input:</strong> nums1 = [1,3], nums2 = [2]
<strong>Output:</strong> 2.00000
<strong>Explanation:</strong> merged array = [1,2,3] and median is 2.
</pre>

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

<pre>
<strong>Input:</strong> nums1 = [1,2], nums2 = [3,4]
<strong>Output:</strong> 2.50000
<strong>Explanation:</strong> merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.
</pre>

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

<ul>
	<li><code>nums1.length == m</code></li>
	<li><code>nums2.length == n</code></li>
	<li><code>0 &lt;= m &lt;= 1000</code></li>
	<li><code>0 &lt;= n &lt;= 1000</code></li>
	<li><code>1 &lt;= m + n &lt;= 2000</code></li>
	<li><code>-10<sup>6</sup> &lt;= nums1[i], nums2[i] &lt;= 10<sup>6</sup></code></li>
</ul>


    
class Solution:
    def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
        A, B = nums1, nums2
        m, n = len(A), len(B)

        # Ensure A is the smaller array to minimize the binary search range
        if m > n:
            A, B, m, n = B, A, n, m

        imin, imax, half_len = 0, m, (m + n + 1) // 2

        while imin <= imax:
            i = (imin + imax) // 2
            j = half_len - i

            # Increase i if A[i] is less than B[j - 1]
            if i < m and B[j - 1] > A[i]:
                imin = i + 1
            # Decrease i if A[i - 1] is greater than B[j]
            elif i > 0 and A[i - 1] > B[j]:
                imax = i - 1
            else:
                # Perfect partition found
                if i == 0:
                    max_of_left = B[j - 1]
                elif j == 0:
                    max_of_left = A[i - 1]
                else:
                    max_of_left = max(A[i - 1], B[j - 1])

                # If total length is odd
                if (m + n) % 2 == 1:
                    return float(max_of_left)

                # Calculate min_of_right for even total length
                if i == m:
                    min_of_right = B[j]
                elif j == n:
                    min_of_right = A[i]
                else:
                    min_of_right = min(A[i], B[j])

                # Return the average for even total length
                return (max_of_left + min_of_right) / 2.0

        # If the arrays are not sorted or inputs are invalid
        raise ValueError("Input arrays are not sorted or invalid.")
            
17000