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 :
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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