<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 {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        // Get the sizes of both input arrays.
        int n = nums1.size();
        int m = nums2.size();

        // Merge the arrays into a single sorted array.
        vector<int> merged;
        for (int i = 0; i < n; i++) {
            merged.push_back(nums1[i]);
        }
        for (int i = 0; i < m; i++) {
            merged.push_back(nums2[i]);
        }

        // Sort the merged array.
        sort(merged.begin(), merged.end());

        // Calculate the total number of elements in the merged array.
        int total = merged.size();

        if (total % 2 == 1) {
            // If the total number of elements is odd, return the middle element as the median.
            return static_cast<double>(merged[total / 2]);
        } else {
            // If the total number of elements is even, calculate the average of the two middle elements as the median.
            int middle1 = merged[total / 2 - 1];
            int middle2 = merged[total / 2];
            return (static_cast<double>(middle1) + static_cast<double>(middle2)) / 2.0;
        }
    }
};
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