Java Program to Count all distinct pairs with difference equal to k

Given an integer array and a positive integer k, count all distinct pairs with difference equal to k.

Example:-

arr[]:-{1, 5, 3, 4, 2}, k = 3
Output: 2
2 pairs({1, 4} and {5, 2}) with difference equal to 3.

Sample Program:-

package com.shc.ecom.payment.web.rest;  
 /**  
  * @author sdixit  
  *  
  */  
 public class Example {  
      /**  
       * @param args  
       */  
      public static void main(String[] args) {  
           int arr[] = { 1, 5, 3, 4, 2 };  
           int k = 3;  
           System.out.println("Count of pairs :-" + countPairsWithDiffK(arr, k));  
      }  
      private static int countPairsWithDiffK(int[] arr, int k) {  
           int count = 0;  
           int n = arr.length;  
           for (int i = 0; i < n; i++) {  
                for (int j = i + 1; j < n; j++)  
                     if (arr[i] - arr[j] == k || arr[j] - arr[i] == k)  
                          count++;  
           }  
           return count;  
      }  
 }  

Sample Output:-

 Count of pairs :-2  

Enjoy Programming.