Method findMedianSortedArrays

MedianOfSortedArrays.java:3–39

(int[] nums1, int[] nums2)

class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
    
// When length are different then apply binary search on shortest array to avoid index out of bound
   if(nums1.length > nums2.length){
        return(findMedianSortedArrays(nums2, nums1));
    }
    int n1=nums1.length;
    int n2=nums2.length;
    int lo=0,hi=n1;
    while(lo<=hi)
    {
//         Initialize the cuts or partitions
        int cut1=lo+(hi-lo)/2;
//         Total required - already present
        int cut2=((n1+n2)/2)-cut1;
        
//         Initialize l1,l2,r1,r2
        int l1= (cut1==0)?Integer.MIN_VALUE:nums1[cut1-1];
        int l2= (cut2==0)?Integer.MIN_VALUE:nums2[cut2-1];
        int r1= (cut1==n1)?Integer.MAX_VALUE:nums1[cut1];
        int r2= (cut2==n2)?Integer.MAX_VALUE:nums2[cut2];
        
//         Shift element to the left
        if(l1>r2) hi=cut1-1;
        else if(l2>r1) lo=cut1+1;
        else
        {
//             Check for even length
            if((n1+n2)%2==0)
                return (double)(Math.max(l1,l2)+Math.min(r1,r2))/2;
            else
                return (double)(Math.min(r1,r2));
        }
//         To avoid error  
    } 
    return 0.0; 
    }
}