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Median of Two Sorted Arrays

Certification: Advanced Level Accuracy: 0% Submissions: 1 Points: 15

Write a C program to find the median of two sorted arrays. The overall run time complexity should be O(log (m+n)). Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays.

Example 1
    
  
  • Input: nums1 = [1,3], nums2 = [2]
    • 
  
  • Output: 2.00000
    • 
  
  • Explanation: 
    • Merged array = [1,2,3]. 
    • The median is 2 (middle element). 
    • Therefore, the median is 2.00000
    •
Example 2
    
  
  • Input: nums1 = [1,2], nums2 = [3,4]
    • 
  
  • Output: 2.50000
    • 
  
  • Explanation: 
    • Merged array = [1,2,3,4]. 
    • The median is (2 + 3) / 2 = 2.5 (average of two middle elements). 
    • Therefore, the median is 2.50000
    •
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
    • 
  
  • Time Complexity: O(log(min(m,n)))
    • 
  
  • Space Complexity: O(1)
    •
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Hint
Solution

Solution Hints

    
  
  • Use binary search on the smaller array to partition both arrays
    • 
  
  • Ensure that elements on the left side are smaller than elements on the right side
    • 
  
  • Find the correct partition where left_max ≤ right_min for both arrays
    • 
  
  • If total length is odd, median is the maximum of left elements
    • 
  
  • If total length is even, median is the average of maximum left and minimum right elements
    •

Steps to solve by this approach:

 Step 1: Ensure nums1 is the smaller array to optimize binary search
 Step 2: Initialize binary search boundaries (low = 0, high = size of smaller array)
 Step 3: Calculate partition points for both arrays using binary search
 Step 4: Find maximum elements on left side and minimum elements on right side
 Step 5: Check if partition is valid (left_max ≤ right_min for both arrays)
 Step 6: If valid, calculate median based on total length (odd/even)
 Step 7: If not valid, adjust binary search boundaries and repeat

Submitted Code :