Count distinct pairs with difference k Java

PROGRAM TO COUNT ALL DISTINCT PAIRS WITH DIFFERENCE EQUAL TO K

import java.util.*;

 

class MAIN {

 

/* Returns count of pairs with

difference k in arr[] of size n. */

static int countPairsWithDiffK(int arr[], int n,

                                          int k)

{

    int count = 0;

    Arrays.sort(arr); // Sort array elements

 

    int l = 0;

    int r = 0;

    while(r < n)

    {

        if(arr[r] - arr[l] == k)

        {

            count++;

            l++;

            r++;

        }

        else if(arr[r] - arr[l] > k)

            l++;

        else // arr[r] - arr[l] < sum

            r++;

    }

    return count;

}

 

// Driver program to test above function

public static void main(String[] args)

{

    int arr[] = {1, 5, 3, 4, 2};

    int n = arr.length;

    int k = 3;

    System.out.println("Count of pairs with given diff is " +

                        countPairsWithDiffK(arr, n, k));

}

}







OUTPUT

Count of pairs with given diff is 2

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