Given an array. The task is to count the possible pairs which can be formed using the elements of the array where both of the elements in the pair are prime.
Note: Pairs such as (a, b) and (b, a) should not be considered different.
Examples:
Input: arr[] = {1, 2, 3, 5, 7, 9}
Output: 6
From the given array, prime pairs are
(2, 3), (2, 5), (2, 7), (3, 5), (3, 7), (5, 7)
Input: arr[] = {1, 4, 5, 9, 11}
Output: 1A naive approach is to count all possible pairs in the array and check if both the elements in the pair are prime.
An efficient approach is to count a number of primes in the array using Sieve of Eratosthenes. Let it be C. Now, total number of possible pairs is equal to C*(C-1)/2.
Below is the implementation of the above approach:
C++
// CPP program to find count of// prime pairs in given array.#include <bits/stdc++.h>using namespace std;
// Function to find count of prime pairsint pairCount(int arr[], int n)
{ // Find maximum value in the array
int max_val = *max_element(arr, arr + n);
// USE SIEVE TO FIND ALL PRIME NUMBERS LESS
// THAN OR EQUAL TO max_val
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
vector<bool> prime(max_val + 1, true);
// Remaining part of SIEVE
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= max_val; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * 2; i <= max_val; i += p)
prime[i] = false;
}
}
// Find all primes in arr[]
int count = 0;
for (int i = 0; i < n; i++)
if (prime[arr[i]])
count++;
// return the count of
// possible prime pairs
// Number of unique pairs
// with N elements is N*(N-1)/2
return (count * (count - 1)) / 2;
}// Driver codeint main()
{ int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << pairCount(arr, n);
return 0;
} |
Java
// Java program to find count of// prime pairs in given array.import java.util.*;
class GFG {
// Function to find count of prime pairs
static int pairCount(int arr[], int n)
{
// Find maximum value in the array
int max_val = Arrays.stream(arr).max().getAsInt();
// USE SIEVE TO FIND ALL PRIME NUMBERS LESS
// THAN OR EQUAL TO max_val
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
Vector<Boolean> prime = new Vector<>(max_val + 1);
for (int i = 0; i < max_val + 1; i++) {
prime.add(true);
}
// Remaining part of SIEVE
prime.add(0, Boolean.FALSE);
prime.add(1, Boolean.FALSE);
for (int p = 2; p * p <= max_val; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime.get(p) == true) {
// Update all multiples of p
for (int i = p * 2; i <= max_val; i += p) {
prime.add(i, Boolean.FALSE);
}
}
}
// Find all primes in arr[]
int count = 0;
for (int i = 0; i < n; i++) {
if (prime.get(arr[i])) {
count++;
}
}
// return the count of
// possible prime pairs
// Number of unique pairs
// with N elements is N*(N-1)/2
return (count * (count - 1)) / 2;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
int n = arr.length;
System.out.println(pairCount(arr, n));
}
}/* This code has been contributed by PrinciRaj1992*/ |
Python3
# Python 3 program to find count of# prime pairs in given array.from math import sqrt
# Function to find count of prime pairsdef pairCount(arr, n):
# Find maximum value in the array
max_val = arr[0]
for i in range(len(arr)):
if(arr[i] > max_val):
max_val = arr[i]
# USE SIEVE TO FIND ALL PRIME NUMBERS
# LESS THAN OR EQUAL TO max_val
# Create a boolean array "prime[0..n]".
# A value in prime[i] will finally be
# false if i is Not a prime, else true.
prime = [True for i in range(max_val + 1)]
# Remaining part of SIEVE
prime[0] = False
prime[1] = False
k = int(sqrt(max_val)) + 1
for p in range(2, k, 1):
# If prime[p] is not changed,
# then it is a prime
if (prime[p] == True):
# Update all multiples of p
for i in range(p * 2, max_val + 1, p):
prime[i] = False
# Find all primes in arr[]
count = 0
for i in range(0, n, 1):
if (prime[arr[i]]):
count += 1
# return the count of possible prime
# pairs. Number of unique pairs
# with N elements is N*(N-1)/2
return (count * (count - 1)) / 2
# Driver codeif __name__ =='__main__':
arr = [1, 2, 3, 4, 5, 6, 7]
n = len(arr)
print(int(pairCount(arr, n)))
# This code is contributed by# Shahshank_Sharma |
C#
// C# program to find count of// prime pairs in given array.using System;
using System.Linq;
class GFG {
// Function to find count of prime pairs
static int pairCount(int[] arr, int n)
{
// Find maximum value in the array
int max_val = arr.Max();
// USE SIEVE TO FIND ALL PRIME NUMBERS LESS
// THAN OR EQUAL TO max_val
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
bool[] prime = new bool[max_val + 1];
for (int i = 0; i < max_val + 1; i++) {
prime[i] = true;
}
// Remaining part of SIEVE
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= max_val; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * 2; i <= max_val; i += p) {
prime[i] = false;
}
}
}
// Find all primes in arr[]
int count = 0;
for (int i = 0; i < n; i++) {
if (prime[arr[i]]) {
count++;
}
}
// return the count of
// possible prime pairs
// Number of unique pairs
// with N elements is N*(N-1)/2
return (count * (count - 1)) / 2;
}
// Driver code
public static void Main()
{
int[] arr = { 1, 2, 3, 4, 5, 6, 7 };
int n = arr.Length;
Console.WriteLine(pairCount(arr, n));
}
}// This code has been contributed by ihritik |
PHP
<?php// PHP program to find count of // prime pairs in given array. // Function to find count of prime pairs function pairCount($arr, $n)
{ // Find maximum value in the array
$max_val = max($arr);
// USE SIEVE TO FIND ALL PRIME NUMBERS
// LESS THAN OR EQUAL TO max_val
// Create a boolean array "prime[0..n]".
// A value in prime[i] will finally be
// false if i is Not a prime, else true.
// vector<bool> prime(max_val + 1, true);
$prime = array_fill(0, $max_val + 1, true);
// Remaining part of SIEVE
$prime[0] = false;
$prime[1] = false;
for ($p = 2; $p * $p <= $max_val; $p++)
{
// If prime[p] is not changed,
// then it is a prime
if ($prime[$p] == true)
{
// Update all multiples of p
for ($i = $p * 2;
$i <= $max_val; $i += $p)
$prime[$i] = false;
}
}
// Find all primes in arr[]
$count = 0;
for ($i = 0; $i < $n; $i++)
if ($prime[$arr[$i]])
$count++;
// return the count of
// possible prime pairs
// Number of unique pairs
// with N elements is N*(N-1)/2
return ($count * ($count - 1)) / 2;
} // Driver code $arr = array (1, 2, 3, 4, 5, 6, 7 );
$n = sizeof($arr);
echo pairCount($arr, $n);
// This code is contributed by ajit...?> |
Javascript
<script> // Javascript program to find count of
// prime pairs in given array.
// Function to find count of prime pairs
function pairCount(arr, n)
{
// Find maximum value in the array
let max_val = 0;
for (let i = 0; i < n; i++)
{
max_val = Math.max(max_val, arr[i]);
}
// USE SIEVE TO FIND ALL PRIME NUMBERS LESS
// THAN OR EQUAL TO max_val
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
let prime = new Array(max_val + 1);
for (let i = 0; i < max_val + 1; i++) {
prime[i] = true;
}
// Remaining part of SIEVE
prime[0] = false;
prime[1] = false;
for (let p = 2; p * p <= max_val; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (let i = p * 2; i <= max_val; i += p)
{
prime[i] = false;
}
}
}
// Find all primes in arr[]
let count = 0;
for (let i = 0; i < n; i++) {
if (prime[arr[i]]) {
count++;
}
}
// return the count of
// possible prime pairs
// Number of unique pairs
// with N elements is N*(N-1)/2
return (count * (count - 1)) / 2;
}
let arr = [ 1, 2, 3, 4, 5, 6, 7 ];
let n = arr.length;
document.write(pairCount(arr, n));
</script> |
Output
6