Given an integer ‘k’ and an array of integers ‘arr’ (less than 10^6), the task is to find the product of every K’th prime number in the array.
Examples:
Input: arr = {2, 3, 5, 7, 11}, k = 2
Output: 21
All the elements of the array are prime. So, the prime numbers after every K (i.e. 2) interval are 3, 7 and their product is 21.Input: arr = {41, 23, 12, 17, 18, 19}, k = 2
Output: 437
A simple approach: Traverse the array and find every K’th prime number in the array and calculate the running product. In this way, we’ll have to check every element of the array whether it is prime or not which will take more time as the size of the array increases.
Efficient approach: Create a sieve which will store whether a number is prime or not. Then, it can be used to check a number against prime in O(1) time. In this way, we only have to keep track of every K’th prime number and maintain the running product.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
#define MAX 1000000 bool prime[MAX + 1];
void SieveOfEratosthenes()
{ // Create a boolean array "prime[0..n]"
// and initialize all the entries as true.
// A value in prime[i] will finally be false
// if i is Not a prime, else true.
memset (prime, true , sizeof (prime));
// 0 and 1 are not prime numbers
prime[1] = false ;
prime[0] = false ;
for ( int p = 2; p * p <= MAX; 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; i += p)
prime[i] = false ;
}
}
} // compute the answer void productOfKthPrimes( int arr[], int n, int k)
{ // count of primes
int c = 0;
// product of the primes
long long int product = 1;
// traverse the array
for ( int i = 0; i < n; i++) {
// if the number is a prime
if (prime[arr[i]]) {
// increase the count
c++;
// if it is the K'th prime
if (c % k == 0) {
product *= arr[i];
c = 0;
}
}
}
cout << product << endl;
} // Driver code int main()
{ // create the sieve
SieveOfEratosthenes();
int n = 5, k = 2;
int arr[n] = { 2, 3, 5, 7, 11 };
productOfKthPrimes(arr, n, k);
return 0;
} |
// Java implementation of the approach class GFG
{ static int MAX= 1000000 ;
static boolean [] prime= new boolean [MAX + 1 ];
static void SieveOfEratosthenes()
{ // Create a boolean array "prime[0..n]"
// and initialize all the entries as true.
// A value in prime[i] will finally be false
// if i is Not a prime, else true.
//memset(prime, true, sizeof(prime));
// 0 and 1 are not prime numbers
prime[ 1 ] = true ;
prime[ 0 ] = true ;
for ( int p = 2 ; p * p <= MAX; p++) {
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == false ) {
// Update all multiples of p
for ( int i = p * 2 ; i <= MAX; i += p)
prime[i] = true ;
}
}
} // compute the answer static void productOfKthPrimes( int arr[], int n, int k)
{ // count of primes
int c = 0 ;
// product of the primes
int product = 1 ;
// traverse the array
for ( int i = 0 ; i < n; i++) {
// if the number is a prime
if (!prime[arr[i]]) {
// increase the count
c++;
// if it is the K'th prime
if (c % k == 0 ) {
product *= arr[i];
c = 0 ;
}
}
}
System.out.println(product);
} // Driver code public static void main(String[] args)
{ // create the sieve
SieveOfEratosthenes();
int n = 5 , k = 2 ;
int [] arr= new int []{ 2 , 3 , 5 , 7 , 11 };
productOfKthPrimes(arr, n, k);
} } // This code is contributed by mits |
# Python 3 implementation of the approach MAX = 1000000
prime = [ True ] * ( MAX + 1 )
def SieveOfEratosthenes():
# Create a boolean array "prime[0..n]"
# and initialize all the entries as true.
# A value in prime[i] will finally be false
# if i is Not a prime, else true.
# 0 and 1 are not prime numbers
prime[ 1 ] = False ;
prime[ 0 ] = False ;
p = 2
while p * p < = MAX :
# 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 + 1 , p):
prime[i] = False
p + = 1
# compute the answer def productOfKthPrimes(arr, n, k):
# count of primes
c = 0
# product of the primes
product = 1
# traverse the array
for i in range ( n):
# if the number is a prime
if (prime[arr[i]]):
# increase the count
c + = 1
# if it is the K'th prime
if (c % k = = 0 ) :
product * = arr[i]
c = 0
print (product)
# Driver code if __name__ = = "__main__" :
# create the sieve
SieveOfEratosthenes()
n = 5
k = 2
arr = [ 2 , 3 , 5 , 7 , 11 ]
productOfKthPrimes(arr, n, k)
# This code is contributed by ChitraNayal |
// C# implementation of the approach class GFG
{ static int MAX = 1000000;
static bool [] prime = new bool [MAX + 1];
static void SieveOfEratosthenes()
{ // Create a boolean array "prime[0..n]"
// and initialize all the entries as
// true. A value in prime[i] will
// finally be false if i is Not a prime,
// else true.
// 0 and 1 are not prime numbers
prime[1] = true ;
prime[0] = true ;
for ( int p = 2; p * p <= MAX; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == false )
{
// Update all multiples of p
for ( int i = p * 2;
i <= MAX; i += p)
prime[i] = true ;
}
}
} // compute the answer static void productOfKthPrimes( int [] arr,
int n, int k)
{ // count of primes
int c = 0;
// product of the primes
int product = 1;
// traverse the array
for ( int i = 0; i < n; i++)
{
// if the number is a prime
if (!prime[arr[i]])
{
// increase the count
c++;
// if it is the K'th prime
if (c % k == 0)
{
product *= arr[i];
c = 0;
}
}
}
System.Console.WriteLine(product);
} // Driver code static void Main()
{ // create the sieve
SieveOfEratosthenes();
int n = 5, k = 2;
int [] arr= new int []{ 2, 3, 5, 7, 11 };
productOfKthPrimes(arr, n, k);
} } // This code is contributed by mits |
<script> // Javascript implementation of the approach let MAX = 1000000; let prime = new Array(MAX + 1);
function SieveOfEratosthenes() {
// Create a boolean array "prime[0..n]"
// and initialize all the entries as true.
// A value in prime[i] will finally be false
// if i is Not a prime, else true.
prime.fill( true )
// 0 and 1 are not prime numbers
prime[1] = false ;
prime[0] = false ;
for (let p = 2; p * p <= MAX; 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; i += p)
prime[i] = false ;
}
}
} // compute the answer function productOfKthPrimes(arr, n, k) {
// count of primes
let c = 0;
// product of the primes
let product = 1;
// traverse the array
for (let i = 0; i < n; i++) {
// if the number is a prime
if (prime[arr[i]]) {
// increase the count
c++;
// if it is the K'th prime
if (c % k == 0) {
product *= arr[i];
c = 0;
}
}
}
document.write(product + "<br>" );
} // Driver code // create the sieve SieveOfEratosthenes(); let n = 5, k = 2; let arr = [2, 3, 5, 7, 11]; productOfKthPrimes(arr, n, k); // This code is contributed by gfgking. </script> |
21
Complexity Analysis:
- Time Complexity : O(n)
- Auxiliary Space: O(MAX)