Count pairs in an array such that at least one element is prime
Last Updated :
29 Nov, 2022
Given an array arr[] of distinct elements, the task is to count the total number of distinct pairs in which at least one element is prime.
Examples:
Input: arr[] = {1, 3, 10, 7, 8}
Output: 7
Pairs with at least one prime are (1, 3), (1, 7),
(3, 1), (3, 7), (3, 8), (10, 7), (7, 8).
Input:arr[]={4, 6, 8, 2, 9};
Output: 4
Approach: First precompute all the prime till the Maximum element of array using Sieve . Traverse each and every possible pair and check if any of the elements in the pair is prime. If yes, then increment the count.
Below is the implementation of above approach:
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find primes
void sieve(int maxm, int prime[])
{
prime[0] = prime[1] = 1;
for (int i = 2; i * i <= maxm; i++)
if (!prime[i])
for (int j = 2 * i; j <= maxm; j += i)
prime[j] = 1;
}
// Function to count the pair
int countPair(int a[], int n)
{
// Find the maximum element of the array
int maxm = *max_element(a, a + n);
int prime[maxm + 1];
memset(prime, 0, sizeof(prime));
// Find primes upto maximum
sieve(maxm, prime);
// Count pairs with at least prime
int count = 0;
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
if (prime[a[i]] == 0 || prime[a[j]] == 0)
count++;
return count;
}
// Driver code
int main()
{
int arr[] = { 2, 3, 5, 4, 7 };
int n = 5;
cout << countPair(arr, n);
return 0;
}
// Java implementation of the above approach
class GFG
{
// Function to find primes
static void sieve(int maxm, int prime[])
{
prime[0] = prime[1] = 1;
for (int i = 2; i * i <= maxm; i++)
if (prime[i] == 0)
for (int j = 2 * i; j <= maxm; j += i)
prime[j] = 1;
}
// Function to count the pair
static int countPair(int a[], int n)
{
// Find the maximum element of the array
int maxm = a[0];
for(int i = 1; i < n; i++)
if(a[i] > maxm)
maxm = a[i];
int [] prime = new int[maxm + 1];
for(int i = 0; i < maxm + 1; i++)
prime[i] = 0;
// Find primes upto maximum
sieve(maxm, prime);
// Count pairs with at least prime
int count = 0;
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
if (prime[a[i]] == 0 || prime[a[j]] == 0)
count++;
return count;
}
// Driver code
public static void main(String []args)
{
int arr[] = { 2, 3, 5, 4, 7 };
int n = arr.length;
System.out.println(countPair(arr, n));
}
}
// This code is contributed by ihritik
# Python 3 implementation of the above approach
from math import sqrt
# Function to count the pair
def countPair(a, n):
# Find the maximum element of the array
maxm = a[0]
for i in range(len(a)):
if(a[i] > maxm):
maxm = a[i]
prime = [0 for i in range(maxm + 1)]
# Find primes upto maximum
prime[0] = prime[1] = 1;
for i in range(2, int(sqrt(maxm)) + 1, 1):
if (prime[i] == 0):
for j in range(2 * i, maxm + 1, i):
prime[j] = 1
# Count pairs with at least prime
count = 0
for i in range(n):
for j in range(i + 1, n, 1):
if (prime[a[i]] == 0 or
prime[a[j]] == 0):
count += 1
return count
# Driver code
if __name__ == '__main__':
arr = [2, 3, 5, 4, 7]
n = 5
print(countPair(arr, n))
# This code is contributed by
# Sanjit_Prasad
// C# implementation of the above approach
using System;
class GFG
{
// Function to find primes
static void sieve(int maxm, int []prime)
{
prime[0] = prime[1] = 1;
for (int i = 2; i * i <= maxm; i++)
if (prime[i] == 0)
for (int j = 2 * i; j <= maxm; j += i)
prime[j] = 1;
}
// Function to count the pair
static int countPair(int []a, int n)
{
// Find the maximum element of the array
int maxm = a[0];
for(int i = 1; i < n; i++)
if(a[i] > maxm)
maxm = a[i];
int [] prime = new int[maxm + 1];
for(int i = 0; i < maxm + 1; i++)
prime[i] = 0;
// Find primes upto maximum
sieve(maxm, prime);
// Count pairs with at least prime
int count = 0;
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
if (prime[a[i]] == 0 || prime[a[j]] == 0)
count++;
return count;
}
// Driver code
public static void Main()
{
int []arr = { 2, 3, 5, 4, 7 };
int n = arr.Length;
Console.WriteLine(countPair(arr, n));
}
}
// This code is contributed by ihritik
<?php
// PHP implementation of the above approach
// Function to find primes
function sieve($maxm, $prime)
{
$prime[0] = $prime[1] = 1;
for ($i = 2; $i * $i <= $maxm; $i++)
if (!$prime[$i])
for ($j = 2 * $i;
$j <= $maxm; $j += $i)
$prime[$j] = 1;
}
// Function to count the pair
function countPair($a, $n)
{
// Find the maximum element of the array
$maxm = max($a);
$prime = array();
$prime = array_fill(0, $maxm + 1, 0);
// Find primes upto maximum
sieve($maxm, $prime);
// Count pairs with at least prime
$count = 0;
for ($i = 0; $i < $n; $i++)
for ($j = $i + 1; $j < $n; $j++)
if ($prime[$a[$i]] == 0 ||
$prime[$a[$j]] == 0)
$count++;
return $count;
}
// Driver code
$arr = array( 2, 3, 5, 4, 7 );
$n = 5;
echo countPair($arr, $n);
// This code is contributed by Ryuga
?>
<script>
// Javascript implementation of the above approach
// Function to find primes
function sieve(maxm,prime)
{
prime[0] = prime[1] = 1;
for (let i = 2; i * i <= maxm; i++)
if (prime[i] == 0)
for (let j = 2 * i; j <= maxm; j += i)
prime[j] = 1;
}
// Function to count the pair
function countPair(a,n)
{
// Find the maximum element of the array
let maxm = a[0];
for(let i = 1; i < n; i++)
if(a[i] > maxm)
maxm = a[i];
let prime = new Array(maxm + 1);
for(let i = 0; i < maxm + 1; i++)
prime[i] = 0;
// Find primes upto maximum
sieve(maxm, prime);
// Count pairs with at least prime
let count = 0;
for (let i = 0; i < n; i++)
for (let j = i + 1; j < n; j++)
if (prime[a[i]] == 0 || prime[a[j]] == 0)
count++;
return count;
}
// Driver code
let arr=[2, 3, 5, 4, 7];
let n = arr.length;
document.write(countPair(arr, n));
// This code is contributed by unknown2108
</script>
Time Complexity: O(max(n2, m*log(log(m)))), where n is the size of the given array and m is the maximum element in the array.
Auxiliary Space: O(m), where m is the maximum element in the array.
Efficient Approach:
Approach: First precompute all the prime till the Maximum element of array using Sieve . Maintain the count of primes and non-primes. Then count pairs with single prime i.e. nonPrimes * Primes and count pairs with both primes (Primes * (Primes - 1)) / 2. Return the sum of both counts.
Below is the implementation of above approach:
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
// Function to find primes
void sieve(int maxm, int prime[])
{
prime[0] = prime[1] = 1;
for (int i = 2; i * i <= maxm; i++)
if (!prime[i])
for (int j = 2 * i; j <= maxm; j += i)
prime[j] = 1;
}
ll countPair(int a[], int n)
{
// Find the maximum element of the array
int maxm = *max_element(a, a + n);
int prime[maxm + 1];
memset(prime, 0, sizeof(prime));
// Find primes upto maximum
sieve(maxm, prime);
// Count number of primes
int countPrimes = 0;
for (int i = 0; i < n; i++)
if (prime[a[i]] == 0)
countPrimes++;
int nonPrimes = n - countPrimes;
ll pairswith1Prime = nonPrimes * countPrimes;
ll pairsWith2Primes = (countPrimes * (countPrimes - 1)) / 2;
return pairswith1Prime + pairsWith2Primes;
}
// Driver code
int main()
{
int arr[] = { 2, 3, 5, 4, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << countPair(arr, n);
return 0;
}
// Java implementation of the above approach
class GFG
{
// Function to find primes
static void sieve(int maxm, int []prime)
{
prime[0] = prime[1] = 1;
for (int i = 2; i * i <= maxm; i++)
if (prime[i]==0)
for (int j = 2 * i; j <= maxm; j += i)
prime[j] = 1;
}
static long countPair(int []a, int n)
{
// Find the maximum element of the array
int maxm = a[0];
int i;
for( i = 1; i < n ; i++)
if(a[i] > maxm)
maxm = a[i];
int [] prime = new int[maxm + 1];
for( i = 0; i < maxm + 1 ;i++)
prime[i] = 0;
// Find primes upto maximum
sieve(maxm, prime);
// Count number of primes
int countPrimes = 0;
for ( i = 0; i < n; i++)
if (prime[a[i]] == 0)
countPrimes++;
int nonPrimes = n - countPrimes;
long pairswith1Prime = nonPrimes *
countPrimes;
long pairsWith2Primes = (countPrimes *
(countPrimes - 1)) / 2;
return pairswith1Prime + pairsWith2Primes;
}
// Driver code
public static void main(String []args)
{
int [] arr = { 2, 3, 5, 4, 7 };
int n = arr.length;
System.out.println(countPair(arr, n));
}
}
// This code is contributed by ihritik
# Python3 implementation of the above approach
# Function to find primes
def sieve(maxm, prime):
prime[0] = prime[1] = 1;
i = 2;
while (i * i <= maxm):
if (prime[i] == 0):
for j in range(2 * i, maxm + 1, i):
prime[j] = 1;
i += 1;
def countPair(a, n):
# Find the maximum element
# of the array
maxm = max(a);
prime = [0] * (maxm + 1);
# Find primes upto maximum
sieve(maxm, prime);
# Count number of primes
countPrimes = 0;
for i in range(n):
if (prime[a[i]] == 0):
countPrimes += 1;
nonPrimes = n - countPrimes;
pairswith1Prime = nonPrimes * countPrimes;
pairsWith2Primes = (countPrimes *
(countPrimes - 1)) // 2;
return pairswith1Prime + pairsWith2Primes;
# Driver code
arr = [ 2, 3, 5, 4, 7 ];
n = len(arr);
print(countPair(arr, n));
# This code is contributed by mits
// C# implementation of the above approach
using System;
class GFG
{
// Function to find primes
static void sieve(int maxm, int []prime)
{
prime[0] = prime[1] = 1;
for (int i = 2; i * i <= maxm; i++)
if (prime[i] == 0)
for (int j = 2 * i; j <= maxm; j += i)
prime[j] = 1;
}
static long countPair(int []a, int n)
{
// Find the maximum element of the array
int maxm = a[0];
int i;
for( i = 1; i < n ;i++)
if(a[i] > maxm)
maxm = a[i];
int [] prime = new int[maxm + 1];
for( i = 0; i < maxm + 1 ;i++)
prime[i] = 0;
// Find primes upto maximum
sieve(maxm, prime);
// Count number of primes
int countPrimes = 0;
for ( i = 0; i < n; i++)
if (prime[a[i]] == 0)
countPrimes++;
int nonPrimes = n - countPrimes;
long pairswith1Prime = nonPrimes *
countPrimes;
long pairsWith2Primes = (countPrimes *
(countPrimes - 1)) / 2;
return pairswith1Prime + pairsWith2Primes;
}
// Driver code
public static void Main()
{
int [] arr = { 2, 3, 5, 4, 7 };
int n = arr.Length;
Console.WriteLine(countPair(arr, n));
}
}
// This code is contributed by ihritik
<?php
// PHP implementation of the above approach
// Function to find primes
function sieve($maxm, $prime)
{
$prime[0] = $prime[1] = 1;
for ($i = 2; $i * $i <= $maxm; $i++)
if (!$prime[$i])
for ($j = 2 * $i;
$j <= $maxm; $j += $i)
$prime[$j] = 1;
}
function countPair($a, $n)
{
// Find the maximum element
// of the array
$maxm = max($a);
$prime = array_fill(0, $maxm + 1,0);
// Find primes upto maximum
sieve($maxm, $prime);
// Count number of primes
$countPrimes = 0;
for ($i = 0; $i < $n; $i++)
if ($prime[$a[$i]] == 0)
$countPrimes++;
$nonPrimes = $n - $countPrimes;
$pairswith1Prime = $nonPrimes * $countPrimes;
$pairsWith2Primes = ($countPrimes *
($countPrimes - 1)) / 2;
return $pairswith1Prime + $pairsWith2Primes;
}
// Driver code
$arr = array( 2, 3, 5, 4, 7 );
$n = count($arr);
echo countPair($arr, $n);
// This code is contributed by mits
?>
<script>
// JavaScript implementation of the above approach
// Function to find primes
function sieve(maxm, prime)
{
prime[0] = prime[1] = 1;
for (let i = 2; i * i <= maxm; i++)
if (prime[i] == 0)
for (let j = 2 * i; j <= maxm; j += i)
prime[j] = 1;
}
function countPair(a, n)
{
// Find the maximum element of the array
let maxm = a[0];
let i;
for( i = 1; i < n ;i++)
if(a[i] > maxm)
maxm = a[i];
let prime = new Array(maxm + 1);
for( i = 0; i < maxm + 1 ;i++)
prime[i] = 0;
// Find primes upto maximum
sieve(maxm, prime);
// Count number of primes
let countPrimes = 0;
for ( i = 0; i < n; i++)
if (prime[a[i]] == 0)
countPrimes++;
let nonPrimes = n - countPrimes;
let pairswith1Prime = nonPrimes * countPrimes;
let pairsWith2Primes = parseInt((countPrimes *
(countPrimes - 1)) / 2, 10);
return (pairswith1Prime + pairsWith2Primes);
}
let arr = [ 2, 3, 5, 4, 7 ];
let n = arr.length;
document.write(countPair(arr, n));
</script>
Time Complexity: O(max(n, m*log(log(m)))), where n is the size of the given array and m is the maximum element in the array.
Auxiliary Space: O(m), where m is the maximum element in the array.
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