Pairs


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
Returns

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 .

Constraints

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

3


Solution :



title-img


                            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;
        _a.push_back(_a_item);
    }
    
    res = pairs(_a,_k);
    cout << res;
    
    return 0;
}







In  Java  :







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

public class Solution {
    static int pairs(int[] a,int k) {
        Arrays.sort(a);
        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);
			j--;
			while((j < N) && (a[j++] - a[i]) == k)
				count++;			
		}

        return count;
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        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);
        System.out.println(res);
    }
}







In   C :





#include<stdio.h>
int get_num()
{
int num=0;
char c=getchar_unlocked();
while(!(c>='0' && c<='9'))
c=getchar_unlocked();
while(c>='0' && c<='9')
{
num=(num<<3)+(num<<1)+c-'0';
c=getchar_unlocked();
}
return num;
}
void quicksort(int x[],int first,int last)
{
    int pivot,j,temp,i;

     if(first<last)
    {
         pivot=first;
         i=first;
         j=last;

         while(i<j){
             while(x[i]<=x[pivot]&&i<last)
                 i++;
             while(x[j]>x[pivot])
                 j--;
             if(i<j){
                 temp=x[i];
                  x[i]=x[j];
                  x[j]=temp;
             }
         }

         temp=x[pivot];
         x[pivot]=x[j];
         x[j]=temp;
         quicksort(x,first,j-1);
         quicksort(x,j+1,last);

    }
}
int main()
{
int n=get_num();
int k=get_num();
int a[100000]={0};
int i=0;
while(i<n)
a[i++]=get_num();
quicksort(a,0,n-1);
int temp=0,count=0,flag=0;
for(i=0;i<n-1;i++)
{
int j=i+1;
temp=0;
for(;j<n;j++)
{
if(a[j]-a[i]==k)
count++;
else if(a[j]-a[i]>k)
break;
}
}
printf("%d\n",count);
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
	main()
                        




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