Assign Mice to Holes

There are N Mice and N holes are placed in a straight line. Each hole can accomodate only 1 mouse. A mouse can stay at his position, move one step right from x to x + 1, or move one step left from x to x − 1. Any of these moves consumes 1 minute. Assign mice to holes so that the time when the last mouse gets inside a hole is minimized.

Example:

positions of mice are:
4 -4 2
positions of holes are:
4 0 5

Assign mouse at position x=4 to hole at position x=4 : Time taken is 0 minutes 
Assign mouse at position x=-4 to hole at position x=0 : Time taken is 4 minutes 
Assign mouse at position x=2 to hole at position x=5 : Time taken is 3 minutes 
After 4 minutes all of the mice are in the holes.

Since, there is no combination possible where the last mouse's time is less than 4, 
answer = 4.

Input:

A :  list of positions of mice
B :  list of positions of holes

Output:

single integer value

NOTE: The final answer will fit in a 32 bit signed integer.

public class Solution {
    public int mice(int[] A, int[] B) {
        Arrays.sort(A);
        Arrays.sort(B);
        int ans = Integer.MIN_VALUE;
        for (int i = 0; i < A.length; i++)
            ans = Math.max(ans, Math.abs(A[i] - B[i]));
        return ans;
    }
}