"""
4. Median of Two Sorted Arrays
Difficulty: Hard
https://leetcode.com/problems/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 1:

Input: nums1 = [1,3], nums2 = [2]
Output: 2.00000
Explanation: merged array = [1,2,3] and median is 2.

Example 2:

Input: nums1 = [1,2], nums2 = [3,4]
Output: 2.50000
Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.



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
"""


class Solution:
    def findMedianSortedArrays(self, nums1: list[int], nums2: list[int]) -> float:
        if len(nums1) > len(nums2):
            return self.findMedianSortedArrays(nums2, nums1)

        m, n = len(nums1), len(nums2)
        left, right = 0, m

        while left <= right:
            partA = (left + right) // 2
            partB = (m + n + 1) // 2 - partA

            maxLeftA = float("-inf") if partA == 0 else nums1[partA - 1]
            minRightA = float("inf") if partA == m else nums1[partA]
            maxLeftB = float("-inf") if partB == 0 else nums2[partB - 1]
            minRightB = float("inf") if partB == n else nums2[partB]

            if maxLeftA <= minRightB and maxLeftB <= minRightA:
                if (m + n) % 2 == 0:
                    return (max(maxLeftA, maxLeftB) + min(minRightA, minRightB)) / 2
                else:
                    return max(maxLeftA, maxLeftB)
            elif maxLeftA > minRightB:
                right = partA - 1
            else:
                left = partA + 1