/*

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.

Examples:

Input: arr[] = {1, 5, 3, 4, 2}, k = 3

Output: 2

There are 2 pairs with difference 3, the pairs are {1, 4} and {5, 2}

Input: arr[] = {8, 12, 16, 4, 0, 20}, k = 4

Output: 5

There are 5 pairs with difference 4, the pairs are {0, 4}, {4, 8},

{8, 12}, {12, 16} and {16, 20}

*/

//http://www.geeksforgeeks.org/count-pairs-difference-equal-k/

//The method2 in that page is more efficient than this one

import java.util.ArrayList;

import java.util.List;

public class CountDistinctPair {

public void count(int[] arr, int k){

List <Integer> list = new ArrayList<Integer>();

int[] arr2 = arr;

for(int i:arr){

for(int j:arr2){

if(i>j){

if(i-j==k){

list.add(i);

list.add(j);

}

}

}

}

System.out.println("Output Num: "+list.size()/2);

for(int z=0;z<list.size();z++){

System.out.println("{"+list.get(z)+", "+list.get(++z)+"}");

}

}

/**

* @param args

*/

public static void main(String[] args) {

// TODO Auto-generated method stub

CountDistinctPair cdp = new CountDistinctPair();

int[] arr = {8, 12, 16, 4, 0, 20};

int k = 4;

cdp.count(arr, k);

}

}