<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, nums2):
        
        merged = nums1 + nums2

        
        merged.sort()

        
        total = len(merged)

        if total % 2 == 1:
            
            return float(merged[total // 2])
        else:
            
            middle1 = merged[total // 2 - 1]
            middle2 = merged[total // 2]
            return (float(middle1) + float(middle2)) / 2.0
16800