Median of Two Sorted Arrays

Find the median of two sorted arrays in O(log(m+n)).

This is my first hard rated LeetCode problem, and I solved it in 0ms!

• Find the total length of both arrays.
• Determine if the total length is even or odd. This will affect how we calculate the median.
• Use two pointers to traverse both arrays simultaneously.
• Keep track of the current and previous elements. We will need the previous element if the total length is even.
• If i has reached the end of nums1, we must take the next element from nums2.
• If j has reached the end of nums2, we must take the next element from nums1.
• If the current element in nums1 is less than or equal to the current element in nums2, we take the element from nums1.
• Otherwise, we take the element from nums2.
• Stop when we have traversed half of the total elements.

class Solution:
    def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
        n = len(nums1) + len(nums2)
        even = n % 2 == 0
        stop = n // 2

        i = 0
        j = 0
        prev = None
        curr = None

        while i < len(nums1) or j < len(nums2):
            if curr is not None:
                prev = curr

            if i == len(nums1):
                curr = nums2[j]
                j += 1
            elif j == len(nums2):
                curr = nums1[i]
                i += 1
            elif nums1[i] <= nums2[j]:
                curr = nums1[i]
                i += 1
            else:
                curr = nums2[j]
                j += 1

            if i + j > stop:
                break

        if even:
            return (prev + curr) / 2

        return curr