Find the median of two sorted arrays in logarithmic time.

Interview question: Median of Two Sorted Arrays

Given two sorted arrays `nums1` and `nums2` of size `m` and `n` respectively, return the median of the two sorted arrays.

The overall run time complexity should be `O(log (m+n))`.

## Example
```
Input: nums1 = [1,3], nums2 = [2]
Output: 2.0
Explanation: merged array = [1,2,3] and median is 2.
```

## Constraints
* `nums1.length == m`, `nums2.length == n`
* `0 <= m <= 1000`, `0 <= n <= 1000`
* `1 <= m + n <= 2000`
* `-10^6 <= nums1[i], nums2[i] <= 10^6`

- The median splits the combined sorted values into a left half and a right half.
- Every value on the left must be less than or equal to every value on the right.
- If we can find that split without merging, we get logarithmic time.

Think in Terms of a Partition

- Choose how many values come from `nums1` on the left side, and infer how many must come from `nums2`.
- Searching the smaller array keeps the binary search range as small as possible and avoids invalid indexes.
- The left side should contain `(m + n + 1) // 2` values so odd lengths put the median on the left.

Binary Search the Smaller Array

- For a partition `i` in `nums1`, use `j = half - i` in `nums2`.
- Compare only four boundary values: left and right around each partition.
- Sentinels `-inf` and `inf` make empty sides easy to handle.

Check the Partition Boundaries

- A partition is valid when `left1 <= right2` and `left2 <= right1`.
- If `left1 > right2`, too many values were taken from `nums1`, so move left.
- Otherwise, too few values were taken from `nums1`, so move right.

Move Toward a Valid Split

- For odd total length, the median is the largest value on the left side.
- For even total length, average the largest left value and smallest right value.
- Time complexity is O(log(min(m, n))), and extra space complexity is O(1).