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<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;
}
                        


                        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;
}
                    


                        Solution in Java :

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);
    }
}
                    


                        Solution in Python : 
                            
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()
                    


View More Similar Problems

Polynomial Division

Consider a sequence, c0, c1, . . . , cn-1 , and a polynomial of degree 1 defined as Q(x ) = a * x + b. You must perform q queries on the sequence, where each query is one of the following two types: 1 i x: Replace ci with x. 2 l r: Consider the polynomial and determine whether is divisible by over the field , where . In other words, check if there exists a polynomial with integer coefficie

View Solution →

Costly Intervals

Given an array, your goal is to find, for each element, the largest subarray containing it whose cost is at least k. Specifically, let A = [A1, A2, . . . , An ] be an array of length n, and let be the subarray from index l to index r. Also, Let MAX( l, r ) be the largest number in Al. . . r. Let MIN( l, r ) be the smallest number in Al . . .r . Let OR( l , r ) be the bitwise OR of the

View Solution →

The Strange Function

One of the most important skills a programmer needs to learn early on is the ability to pose a problem in an abstract way. This skill is important not just for researchers but also in applied fields like software engineering and web development. You are able to solve most of a problem, except for one last subproblem, which you have posed in an abstract way as follows: Given an array consisting

View Solution →

Self-Driving Bus

Treeland is a country with n cities and n - 1 roads. There is exactly one path between any two cities. The ruler of Treeland wants to implement a self-driving bus system and asks tree-loving Alex to plan the bus routes. Alex decides that each route must contain a subset of connected cities; a subset of cities is connected if the following two conditions are true: There is a path between ever

View Solution →

Unique Colors

You are given an unrooted tree of n nodes numbered from 1 to n . Each node i has a color, ci. Let d( i , j ) be the number of different colors in the path between node i and node j. For each node i, calculate the value of sum, defined as follows: Your task is to print the value of sumi for each node 1 <= i <= n. Input Format The first line contains a single integer, n, denoti

View Solution →

Fibonacci Numbers Tree

Shashank loves trees and math. He has a rooted tree, T , consisting of N nodes uniquely labeled with integers in the inclusive range [1 , N ]. The node labeled as 1 is the root node of tree , and each node in is associated with some positive integer value (all values are initially ). Let's define Fk as the Kth Fibonacci number. Shashank wants to perform 22 types of operations over his tree, T

View Solution →