Sum of every K’th prime number in an array
Given an integer k and an array of integers arr (less than 10^6), the task is to find the sum of every k’th prime number in the array.
Examples:
Input: arr[] = {2, 3, 5, 7, 11}, k = 2
Output: 10
All the elements of the array are prime. So, the prime numbers after every K (i.e. 2) interval are 3, 7 and their sum is 10.Input: arr[] = {11, 13, 15, 17, 19}, k = 2
Output: 32
Simple Approach: Traverse the array and find every K’th prime number in the array and calculate the running sum. 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 that 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 sum.
Below is the implementation of the above approach:
C++
// 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 SumOfKthPrimes( int arr[], int n, int k) { // count of primes int c = 0; // sum of the primes long long int sum = 0; // 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) { sum += arr[i]; c = 0; } } } cout << sum << endl; } // Driver code int main() { // create the sieve SieveOfEratosthenes(); int arr[] = { 2, 3, 5, 7, 11 }; int n = sizeof (arr) / sizeof (arr[0]); int k = 2; SumOfKthPrimes(arr, n, k); return 0; } |
Java
// Java implementation of the approach public 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. for ( int i = 0 ; i < prime.length; i++) { prime[i] = true ; } // 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 static void SumOfKthPrimes( int arr[], int n, int k) { // count of primes int c = 0 ; // sum of the primes long sum = 0 ; // 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 ) { sum += arr[i]; c = 0 ; } } } System.out.println(sum); } // Driver code public static void main(String[] args) { // create the sieve SieveOfEratosthenes(); int arr[] = { 2 , 3 , 5 , 7 , 11 }; int n = arr.length; int k = 2 ; SumOfKthPrimes(arr, n, k); } } |
Python3
# Python3 implementation of the approach MAX = 100000 ; 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]): # Update all multiples of p i = p * 2 ; while (i < = MAX ): prime[i] = False ; i + = p; p + = 1 ; # compute the answer def SumOfKthPrimes(arr, n, k): # count of primes c = 0 ; # sum of the primes sum = 0 ; # 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 ): sum + = arr[i]; c = 0 ; print ( sum ); # Driver code # create the sieve SieveOfEratosthenes(); arr = [ 2 , 3 , 5 , 7 , 11 ]; n = len (arr); k = 2 ; SumOfKthPrimes(arr, n, k); # This code is contributed by mits |
C#
// 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] = 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] == false ) { // Update all multiples of p for ( int i = p * 2; i <= MAX; i += p) prime[i] = true ; } } } // compute the answer static void SumOfKthPrimes( int [] arr, int n, int k) { // count of primes int c = 0; // sum of the primes long sum = 0; // 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) { sum += arr[i]; c = 0; } } } System.Console.WriteLine(sum); } // Driver code static void Main() { // create the sieve SieveOfEratosthenes(); int [] arr = new int [] { 2, 3, 5, 7, 11 }; int n = arr.Length; int k = 2; SumOfKthPrimes(arr, n, k); } } // This code is contributed by mits |
PHP
<?php // PHP implementation of the approach $MAX = 100000; $prime = array_fill (0, $MAX + 1, true); function SieveOfEratosthenes() { global $MAX , $prime ; // 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; for ( $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 ( $i = $p * 2; $i <= $MAX ; $i += $p ) $prime [ $i ] = false; } } } // compute the answer function SumOfKthPrimes( $arr , $n , $k ) { global $MAX , $prime ; // count of primes $c = 0; // sum of the primes $sum = 0; // traverse the array for ( $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) { $sum += $arr [ $i ]; $c = 0; } } } echo $sum . "\n" ; } // Driver code // create the sieve SieveOfEratosthenes(); $arr = array ( 2, 3, 5, 7, 11 ); $n = sizeof( $arr ); $k = 2; SumOfKthPrimes( $arr , $n , $k ); // This code is contributed by mits ?> |
Javascript
<script> // Javascript implementation of the approach let MAX = 100000; let prime = new Array(MAX + 1).fill( true ); 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. // 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 SumOfKthPrimes(arr, n, k) { // count of primes let c = 0; // sum of the primes let sum = 0; // 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) { sum += arr[i]; c = 0; } } } document.write(sum + "<br>" ); } // Driver code // create the sieve SieveOfEratosthenes(); let arr = new Array(2, 3, 5, 7, 11); let n = arr.length; let k = 2; SumOfKthPrimes(arr, n, k); // This code is contributed by gfgking </script> |
10
Complexity Analysis:
- Time Complexity: O(n + MAX2)
- Auxiliary Space: O(MAX)
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