Add Median of Two Sorted Arrays Algorithm to Divide and Conquer Section

divide_and_conquer/median_of_two_sorted_arrays.py

@@ -0,0 +1,112 @@
+"""
+Median of Two Sorted Arrays
+
+Problem: Given two sorted arrays nums1 and nums2 of sizes m and n,
+return the median of the two sorted arrays.
+
+This implementation uses a partition-based divide-and-conquer approach
+to achieve optimal time complexity.
+
+Time Complexity: O(log(min(m, n)))
+Space Complexity: O(1)
+
+Reference: https://leetcode.com/problems/median-of-two-sorted-arrays/
+"""
+
+from __future__ import annotations
+
+
+def find_median_sorted_arrays(
+    nums1: list[int | float], nums2: list[int | float]
+) -> float:
+    """
+    Find the median of two sorted arrays using binary search.
+
+    The algorithm works by partitioning both arrays such that:
+    - All elements on the left side are smaller than elements on the right side
+    - The left side has the same number of elements as the right side (or one more)
+
+    Args:
+        nums1: First sorted array
+        nums2: Second sorted array
+
+    Returns:
+        The median of the two sorted arrays as a float
+
+    Raises:
+        ValueError: If both input arrays are empty
+
+    Examples:
+        >>> find_median_sorted_arrays([1, 3], [2])
+        2.0
+        >>> find_median_sorted_arrays([1, 2], [3, 4])
+        2.5
+        >>> find_median_sorted_arrays([], [1])
+        1.0
+        >>> find_median_sorted_arrays([2], [])
+        2.0
+        >>> find_median_sorted_arrays([1, 2, 3, 4, 5], [6, 7, 8])
+        4.5
+        >>> find_median_sorted_arrays([1.5, 2.5], [2.0, 3.0])
+        2.25
+        >>> find_median_sorted_arrays([1, 1, 1], [1, 1, 1])
+        1.0
+        >>> find_median_sorted_arrays([-5, -3, -1], [0, 2, 4])
+        -0.5
+        >>> find_median_sorted_arrays([1], [2, 3, 4, 5, 6])
+        3.5
+        >>> find_median_sorted_arrays([], [])
+        Traceback (most recent call last):
+            ...
+        ValueError: Both input arrays are empty
+    """
+    if not nums1 and not nums2:
+        raise ValueError("Both input arrays are empty")
+
+    # Ensure nums1 is the smaller array for efficiency
+    if len(nums1) > len(nums2):
+        nums1, nums2 = nums2, nums1
+
+    m, n = len(nums1), len(nums2)
+    total_left = (m + n + 1) // 2  # Number of elements on the left side
+
+    left, right = 0, m
+
+    while left <= right:
+        # Partition indices for nums1 and nums2
+        i = (left + right) // 2  # Partition index for nums1
+        j = total_left - i  # Partition index for nums2
+
+        # Get the boundary elements around the partitions
+        nums1_left_max = nums1[i - 1] if i > 0 else float("-inf")
+        nums1_right_min = nums1[i] if i < m else float("inf")
+
+        nums2_left_max = nums2[j - 1] if j > 0 else float("-inf")
+        nums2_right_min = nums2[j] if j < n else float("inf")
+
+        # Check if we found the correct partition
+        if nums1_left_max <= nums2_right_min and nums2_left_max <= nums1_right_min:
+            # Correct partition found
+            if (m + n) % 2 == 1:
+                # Odd total length: median is the max of left side
+                return float(max(nums1_left_max, nums2_left_max))
+            else:
+                # Even total length: median is average of max(left) and min(right)
+                left_max = max(nums1_left_max, nums2_left_max)
+                right_min = min(nums1_right_min, nums2_right_min)
+                return (left_max + right_min) / 2.0
+        elif nums1_left_max > nums2_right_min:
+            # We are too far right in nums1, move left
+            right = i - 1
+        else:
+            # We are too far left in nums1, move right
+            left = i + 1
+
+    # This should not be reached if inputs are valid sorted arrays
+    raise ValueError("Input arrays are not sorted or some unexpected error occurred")
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()