Median_of_Two_Sorted_Arrays.py

class Solution:
    def findMedianSortedArrays(
        self, nums1: List[int], nums2: List[int]
    ) -> float:
        m, n = len(nums1), len(nums2)
        p1, p2 = 0, 0

        # Get the smaller value between nums1[p1] and nums2[p2].
        def get_min():
            nonlocal p1, p2
            if p1 < m and p2 < n:
                if nums1[p1] < nums2[p2]:
                    ans = nums1[p1]
                    p1 += 1
                else:
                    ans = nums2[p2]
                    p2 += 1
            elif p2 == n:
                ans = nums1[p1]
                p1 += 1
            else:
                ans = nums2[p2]
                p2 += 1
            return ans

        if (m + n) % 2 == 0:
            for _ in range((m + n) // 2 - 1):
                _ = get_min()
            return (get_min() + get_min()) / 2
        else:
            for _ in range((m + n) // 2):
                _ = get_min()
            return get_min()
        


## Example usage:# sol = Solution()
# print(sol.findMedianSortedArrays([1, 3], [2]))  # Output: 2.0
# print(sol.findMedianSortedArrays([1, 2], [3, 4]))  # Output: 2.5

# time complexity: O(m + n) space complexity: O(1)