Median of Two Sorted Arrays(Leetcode)

Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays.

Follow up: 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.

Example 3:

Input: nums1 = [0,0], nums2 = [0,0]
Output: 0.00000

Example 4:

Input: nums1 = [], nums2 = [1]
Output: 1.00000

Example 5:

Input: nums1 = [2], nums2 = []
Output: 2.00000

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(int[] nums1, int[] nums2) {
        
        if(nums1.length>nums2.length){
            return findMedianSortedArrays(nums2,nums1);
        }
        int x = nums1.length;
        int y = nums2.length;
        
        int low =0;
        int high = x;
        
        while(low<=high){
            int partitionX = (low+high)/2;
            int partitionY = (x+y+1)/2 - partitionX;
            
            int maxLeftX = (partitionX == 0)?Integer.MIN_VALUE:nums1[partitionX-1];
            int minRightX = ( partitionX == x)?Integer.MAX_VALUE:nums1[partitionX];
            
            int maxLeftY = (partitionY == 0)?Integer.MIN_VALUE:nums2[partitionY-1];
            int minRightY =  (partitionY == y)?Integer.MAX_VALUE:nums2[partitionY];
            
            if((maxLeftX<=minRightY) &&(maxLeftY<=minRightX)){
                if((x+y)%2==0){
                    return (double)(Math.max(maxLeftX,maxLeftY)+Math.min(minRightX,minRightY))/2;
                }else{
                    return (double)(Math.max(maxLeftX,maxLeftY));
                }
            }else if(maxLeftX>minRightY){
                high=partitionX-1;
            }else{
                low=partitionX+1;
            }
        
        }
        throw new IllegalArgumentException();
    }
}

Thank you for Reading!!