Question : Given three ints, a b c, print true if one of b or c is "close" (differing from a by at most 1), while the other is "far", differing from both other values by 2 or more. 
Note : Math.abs(num) computes the absolute value of a number.
Output:
1, 2, 10 -> true
1, 2, 3  -> false
4, 1, 3  -> true

Answer:

import java.util.*;
import java.lang.Math;
public class CloseFar
{
 public static void main(String[] args) {
 Scanner sc=new Scanner(System.in); 
 System.out.println("Enter num 1 : "); 
 int a=sc.nextInt(); 
 System.out.println("Enter num 2 : "); 
 int b=sc.nextInt(); 
 System.out.println("Enter num 3 : "); 
 int c=sc.nextInt(); 
 if(Math.abs(b-a)<=1 || Math.abs(c-a)<=1)
 {
 System.out.println("true"); 
 }
 else 
 {
 System.out.println("false"); 
 }
 }
 }

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