4. Median of Two Sorted Array

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optimized Solution
if(nums2.size() < nums1.size()) return findMedianSortedArrays(nums2, nums1);
       int n1 = nums1.size();
       int n2 = nums2.size();
       int low = 0, high = n1;
       while(low <= high){
           int cut1 = (low + high) >> 1;
           int cut2 = (n1+n2+1)/2 - cut1;

           int left1 = cut1 ==0 ? INT_MIN : nums1[cut1-1];  //to much tough
           int left2 = cut2 ==0 ? INT_MIN : nums2[cut2 -1];

           int right1 = cut1 == n1 ? INT_MAX : nums1[cut1];
           int right2 = cut2 == n2 ? INT_MAX : nums2[cut2];

           if( left1 <= right2 && left2 <= right1){
               if((n1+n2) %2 == 0)
                    return(max(left1,left2) + min(right1, right2)) / 2.0;
                else
                    return max(left1, left2);
           }
           else if(left1 > right2){
               high = cut1 -1;
           }
           else{
               low = cut1 +1;
           }
       }
       return 0.0;
    }