Arrays_median-of-two-sorted-arrays

https://leetcode.com/problems/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:

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
-106 <= nums1[i], nums2[i] <= 106


class Solution {
public:
    double findMedianSortedArrays(vector<int>& A, vector<int>& B)
{
    int n = A.size();
    int m = B.size();
    if (n > m)
        return findMedianSortedArrays(B, A); // Swapping to make A smaller
 
    int start = 0;
    int end = n;
    int realmidinmergedarray = (n + m + 1) / 2;
 
    while (start <= end) {
        int mid = (start + end) / 2;
        int leftAsize = mid;
        int leftBsize = realmidinmergedarray - mid;
        int leftA
            = (leftAsize > 0)
                  ? A[leftAsize - 1]
                  : INT_MIN; // checking overflow of indices
        int leftB
            = (leftBsize > 0) ? B[leftBsize - 1] : INT_MIN;
        int rightA
            = (leftAsize < n) ? A[leftAsize] : INT_MAX;
        int rightB
            = (leftBsize < m) ? B[leftBsize] : INT_MAX;
 
        // if correct partition is done
        if (leftA <= rightB and leftB <= rightA) {
            if ((m + n) % 2 == 0)
                return (max(leftA, leftB)
                        + min(rightA, rightB))
                       / 2.0;
            return max(leftA, leftB);
        }
        else if (leftA > rightB) {
            end = mid - 1;
        }
        else
            start = mid + 1;
    }
    return 0.0;
}
};