LeetCode //C - 4. Median of Two Sorted Arrays

4. 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)).
 

Example 1:

Example 2:

Constraints:

• nums1.length == m
• nums2.length == n
• 0 <= m <= 1000
• 0 <= n <= 1000
• 1 <= m + n <= 2000
• − 1 0 6 < = n u m s 1 [ i ] , n u m s 2 [ i ] < = 1 0 6 -10^6 <= nums1[i], nums2[i] <= 10^6 −106<=nums1[i],nums2[i]<=106

From: LeetCode
Link: 4. Median of Two Sorted Arrays

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Solution:

Ideas：

In this code, findMedianSortedArrays function takes two sorted arrays nums1 and nums2, along with their sizes nums1Size and nums2Size, and returns the median of the combined array.

The algorithm first ensures that nums1 is the smaller array to optimize the binary search. It then performs a binary search on the smaller array, trying to find a partition such that the elements on the left side of the partition from both arrays are smaller than the elements on the right side of the partition from both arrays. Once the correct partition is found, it calculates the median based on the combined size of both arrays being even or odd.

Code:

double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size) {
    // Ensure nums1 is the smaller array
    if (nums1Size > nums2Size) {
        return findMedianSortedArrays(nums2, nums2Size, nums1, nums1Size);
    }
    
    int x = nums1Size;
    int y = nums2Size;

    int low = 0;
    int high = x;
    
    while (low <= high) {
        int partitionX = (low + high) / 2;
        int partitionY = (x + y + 1) / 2 - partitionX;
        
        int maxX = (partitionX == 0) ? INT_MIN : nums1[partitionX - 1];
        int maxY = (partitionY == 0) ? INT_MIN : nums2[partitionY - 1];
        
        int minX = (partitionX == x) ? INT_MAX : nums1[partitionX];
        int minY = (partitionY == y) ? INT_MAX : nums2[partitionY];
        
        if (maxX <= minY && maxY <= minX) {
            // We have partitioned array at correct place
            if ((x + y) % 2 == 0) {
                return ((double) fmax(maxX, maxY) + fmin(minX, minY)) / 2;
            } else {
                return (double) fmax(maxX, maxY);
            }
        } else if (maxX > minY) {
            high = partitionX - 1;
        } else {
            low = partitionX + 1;
        }
    }
    
    // If input arrays are not sorted or not valid
    return -1.0;
}