Problem Statement :

Given an array of integers and a target value, determine the number of pairs of array elements that have a difference equal to the target value.

Function Description

Complete the pairs function below.

pairs has the following parameter(s):

int k: an integer, the target difference
int arr[n]: an array of integers

int: the number of pairs that satisfy the criterion
Input Format

The first line contains two space-separated integers n and k, the size of arr and the target value.
The second line contains n space-separated integers of the array arr .


2   <=   n   <=   10^5
0   <=   k   <=   10^9
0   <=   arr[ i ]  <=  2^31 -1
each integer arr[ i ]  will be unique

Sample Input

STDIN          Function
-----                 --------
5 2              arr[] size n = 5, k =2
1 5 3 4 2    arr = [1, 5, 3, 4, 2]

Sample Output


Solution :


                            Solution in C :

In   C++  :

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <cstdlib>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
using namespace std;
/* Head ends here */

int pairs(vector <int> a,int k) {
    int ans = 0;
    set<long long> s;
    for(int i = 0; i < a.size(); i++) s.insert(a[i]);
    for(int i = 0; i < a.size(); i++){
        long long b = a[i] - k;
        if(s.count(b)) ans ++;
    return ans;

/* Tail starts here */
int main() {
    int res;
    int _a_size,_k;
    cin >> _a_size>>_k;
    cin.ignore (std::numeric_limits<std::streamsize>::max(), '\n'); 
    vector<int> _a;
    int _a_item;
    for(int _a_i=0; _a_i<_a_size; _a_i++) {
        cin >> _a_item;
    res = pairs(_a,_k);
    cout << res;
    return 0;

In  Java  :

import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {
    static int pairs(int[] a,int k) {
        int N = a.length;
		int count = 0;
		for (int i = 0; i < N - 1; i++)
			int j = i + 1;
			while((j < N) && (a[j++] - a[i]) < k);
			while((j < N) && (a[j++] - a[i]) == k)

        return count;

    public static void main(String[] args) {
        Scanner in = new Scanner(;
        int res;
        String n = in.nextLine();
        String[] n_split = n.split(" ");
        int _a_size = Integer.parseInt(n_split[0]);
        int _k = Integer.parseInt(n_split[1]);
        int[] _a = new int[_a_size];
        int _a_item;
        String next = in.nextLine();
        String[] next_split = next.split(" ");
        for(int _a_i = 0; _a_i < _a_size; _a_i++) {
            _a_item = Integer.parseInt(next_split[_a_i]);
            _a[_a_i] = _a_item;
        res = pairs(_a,_k);

In   C :

int get_num()
int num=0;
char c=getchar_unlocked();
while(!(c>='0' && c<='9'))
while(c>='0' && c<='9')
return num;
void quicksort(int x[],int first,int last)
    int pivot,j,temp,i;




int main()
int n=get_num();
int k=get_num();
int a[100000]={0};
int i=0;
int temp=0,count=0,flag=0;
int j=i+1;
else if(a[j]-a[i]>k)
return 0;

In  Python3 :

def main():
	N, K = (int(x) for x in sys.stdin.readline().split())
	A = [int(x) for x in sys.stdin.readline().split()]
	setA = set(A)
	count = 0
	for x in A:
		if (x-K) in setA:
			count = count +1
	print (count)

if __name__ == '__main__':
	import sys

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