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Add median of two sorted arrays implementation #13717 #13742
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            anshmittal2004
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              | Original file line number | Diff line number | Diff line change | 
|---|---|---|
| @@ -0,0 +1,88 @@ | ||
| """ | ||
| 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)). | ||
|  | ||
| Examples: | ||
| 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 | ||
|  | ||
| Implementation: Divide and Conquer (Binary Search on partitions). | ||
| Time Complexity: O(log(min(m, n))) | ||
| Space Complexity: O(1) | ||
| """ | ||
|  | ||
| from typing import List | ||
|  | ||
|  | ||
| def find_median_sorted_arrays(nums1: List[int], nums2: List[int]) -> float: | ||
| Check failure on line 35 in divide_and_conquer/median_of_two_sorted_arrays.py 
     | ||
| """ | ||
| Find the median of two sorted arrays using binary search on partitions. | ||
|  | ||
| Args: | ||
| nums1 (List[int]): First sorted array | ||
| nums2 (List[int]): Second sorted array | ||
|  | ||
| Returns: | ||
| float: Median of the two arrays | ||
| """ | ||
| # Ensure nums1 is the smaller array | ||
| if len(nums1) > len(nums2): | ||
| nums1, nums2 = nums2, nums1 | ||
|  | ||
| m, n = len(nums1), len(nums2) | ||
| left, right = 0, m | ||
|  | ||
| # Binary search for the partition in nums1 | ||
| while left <= right: | ||
| i = (left + right) // 2 # Partition in nums1 | ||
| j = (m + n + 1) // 2 - i # Partition in nums2 | ||
|  | ||
| # Edge cases for partitions | ||
| left1 = float("-inf") if i == 0 else nums1[i - 1] | ||
| right1 = float("inf") if i == m else nums1[i] | ||
| left2 = float("-inf") if j == 0 else nums2[j - 1] | ||
| right2 = float("inf") if j == n else nums2[j] | ||
|  | ||
| # Check if this partition is correct | ||
| if left1 <= right2 and left2 <= right1: | ||
| # Correct partition found | ||
| if (m + n) % 2 == 0: | ||
| # Even length: average of max(lefts) and min(rights) | ||
| return (max(left1, left2) + min(right1, right2)) / 2 | ||
| else: | ||
| # Odd length: max of lefts | ||
| return max(left1, left2) | ||
| elif left1 > right2: | ||
| # Move partition left in nums1 | ||
| right = i - 1 | ||
| else: | ||
| # Move partition right in nums1 | ||
| left = i + 1 | ||
|  | ||
| # Should not reach here if inputs are valid | ||
| raise ValueError("Input arrays are not sorted or invalid") | ||
|  | ||
|  | ||
| # Optional: Add simple tests | ||
| if __name__ == "__main__": | ||
| assert abs(find_median_sorted_arrays([1, 3], [2]) - 2.0) < 1e-5 | ||
| assert abs(find_median_sorted_arrays([1, 2], [3, 4]) - 2.5) < 1e-5 | ||
| print("All tests passed!") | ||
      
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As there is no test file in this pull request nor any test function or class in the file
divide_and_conquer/median_of_two_sorted_arrays.py, please provide doctest for the functionfind_median_sorted_arrays