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71 lines
1.9 KiB
Python
71 lines
1.9 KiB
Python
"""
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4. Median of Two Sorted Arrays
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Difficulty: Hard
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https://leetcode.com/problems/median-of-two-sorted-arrays/
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──────────────────────────────────────────────────
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Given two sorted arrays nums1 and nums2 of size m and n respectively,
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return the median of the two sorted arrays.
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The overall run time complexity should be O(log (m+n)).
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Example 1:
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Input: nums1 = [1,3], nums2 = [2]
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Output: 2.00000
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Explanation: merged array = [1,2,3] and median is 2.
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Example 2:
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Input: nums1 = [1,2], nums2 = [3,4]
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Output: 2.50000
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Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.
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Constraints:
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• nums1.length == m
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• nums2.length == n
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• 0 <= m <= 1000
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• 0 <= n <= 1000
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• 1 <= m + n <= 2000
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• -10^6 <= nums1[i], nums2[i] <= 10^6
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"""
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class Solution:
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def findMedianSortedArrays(self, nums1: list[int], nums2: list[int]) -> float:
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if len(nums1) > len(nums2):
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return self.findMedianSortedArrays(nums2, nums1)
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m, n = len(nums1), len(nums2)
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left, right = 0, m
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while left <= right:
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partA = (left + right) // 2
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partB = (m + n + 1) // 2 - partA
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maxLeftA = float("-inf") if partA == 0 else nums1[partA - 1]
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minRightA = float("inf") if partA == m else nums1[partA]
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maxLeftB = float("-inf") if partB == 0 else nums2[partB - 1]
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minRightB = float("inf") if partB == n else nums2[partB]
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if maxLeftA <= minRightB and maxLeftB <= minRightA:
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if (m + n) % 2 == 0:
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return (max(maxLeftA, maxLeftB) + min(minRightA, minRightB)) / 2
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else:
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return max(maxLeftA, maxLeftB)
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elif maxLeftA > minRightB:
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right = partA - 1
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else:
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left = partA + 1
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