mice.java

/*

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 int mice(ArrayList<Integer> a, ArrayList<Integer> b) {

        if(a.isEmpty() || a==null || b.isEmpty() || b==null)
            return 0;

        Collections.sort(a);
        Collections.sort(b);

        int result =Integer.MIN_VALUE;
        for( int i =0; i<a.size(); i++)
        {
            result=Math.max(Math.abs(a.get(i)-b.get(i)),result);
        }
        return result;
    }