Product of every K ’ th prime number in an array | Java implementation of the approach ; 0 and 1 are not prime numbers ; If prime [ p ] is not changed , then it is a prime ; Update all multiples of p ; compute the answer ; count of primes ; product of the primes ; traverse the array ; if the number is a prime ; increase the count ; if it is the K 'th prime ; Driver code ; create the sieve
class GFG { static int MAX = 1000000 ; static boolean [ ] prime = new boolean [ MAX + 1 ] ; static void SieveOfEratosthenes ( ) { prime [ 1 ] = true ; prime [ 0 ] = true ; for ( int p = 2 ; p * p <= MAX ; p ++ ) { if ( prime [ p ] == false ) { for ( int i = p * 2 ; i <= MAX ; i += p ) prime [ i ] = true ; } } } static void productOfKthPrimes ( int arr [ ] , int n , int k ) { int c = 0 ; int product = 1 ; for ( int i = 0 ; i < n ; i ++ ) { if ( ! prime [ arr [ i ] ] ) { c ++ ; if ( c % k == 0 ) { product *= arr [ i ] ; c = 0 ; } } } System . out . println ( product ) ; } public static void main ( String [ ] args ) { SieveOfEratosthenes ( ) ; int n = 5 , k = 2 ; int [ ] arr = new int [ ] { 2 , 3 , 5 , 7 , 11 } ; productOfKthPrimes ( arr , n , k ) ; } }