Solution Explanation: Finding the Median of Two Sorted Arrays

This problem challenges us to find the median of two sorted arrays with a time complexity of O(log(m+n)), where 'm' and 'n' are the lengths of the input arrays. A naive approach of merging and sorting would be O(m+n), exceeding the requirement. Therefore, a divide and conquer strategy using binary search is necessary.

Core Idea:

The solution leverages a recursive helper function f(i, j, k) to efficiently find the k-th smallest element in the combined arrays. This function cleverly avoids explicitly merging the arrays, thus achieving the logarithmic time complexity.

• f(i, j, k): This function searches for the k-th smallest element, starting from index i in nums1 and index j in nums2.

• Base Cases:

  • If i reaches the end of nums1, the k-th smallest element is found directly in nums2 (starting from j).
  • Similarly, if j reaches the end of nums2, the element is in nums1 (starting from i).
  • If k is 1, the k-th smallest is simply the minimum of the current elements in nums1 and nums2.

• Recursive Step:

  • We divide k in half (p = k // 2). We then identify potential k-th smallest candidates: x from nums1 and y from nums2 at the positions i + p - 1 and j + p - 1, respectively. If either index is out of bounds, we use a large placeholder value (like 1 << 30) to represent infinity, guaranteeing that it won't be the k-th smallest.
  • We compare x and y:
    • If x <= y, x cannot be the k-th smallest, so we recursively call f excluding the elements from i to i + p in nums1.
    • If x > y, y cannot be the k-th smallest, so we recursively call f excluding the elements from j to j + p in nums2.

• Median Calculation: The median is found by calling f twice: once to get the floor((m+n+1)/2)th smallest and again for the floor((m+n+2)/2)th smallest. The average of these two values is the median.

Time Complexity Analysis:

Each recursive call to f effectively halves the search space (k). This is akin to a binary search, resulting in O(log(m+n)) time complexity.

Space Complexity Analysis:

The space complexity is determined by the recursive call stack. In the worst case, the depth of the recursion is O(log(m+n)), leading to O(log(m+n)) space complexity.

Code Examples (Python):

import sys
 
class Solution:
    def findMedianSortedArrays(self, nums1: list[int], nums2: list[int]) -> float:
        m, n = len(nums1), len(nums2)
 
        def f(i: int, j: int, k: int) -> int:
            if i >= m:
                return nums2[j + k - 1]
            if j >= n:
                return nums1[i + k - 1]
            if k == 1:
                return min(nums1[i], nums2[j])
            p = k // 2
            x = nums1[i + p - 1] if i + p - 1 < m else float('inf')
            y = nums2[j + p - 1] if j + p - 1 < n else float('inf')
            return f(i + p, j, k - p) if x <= y else f(i, j + p, k - p)
            
 
        a = f(0, 0, (m + n + 1) // 2)
        b = f(0, 0, (m + n + 2) // 2)
        return (a + b) / 2

This Python code directly reflects the algorithm described above. Note the use of float('inf') to handle cases where array indices go out of bounds. The other languages provided follow a very similar structure and logic.