Sum of every K’th prime number in an array
Last Updated :
06 Sep, 2022
Given an array of integers (less than 10^6), the task is to find the sum of all the prime numbers which appear after every (k-1) prime number
i.e. every K’th prime number in the array.
Examples:
Input : Array : 2, 3, 5, 7, 11 ; n=5; k=2
Output : Sum = 10
Explanation: All the elements of the array are prime. So,
the prime numbers after every K intervals are 3, 7 and their sum is 10.
Input : Array : 41, 23, 12, 17, 18, 19 ; n=6; k=2
Output : Sum = 42
A simple approach: We have to traverse the array and find the prime numbers after every (k-1) prime numbers. 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: We will 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++
#include <bits/stdc++.h>
using namespace std;
#define MAX 1000000
bool prime[MAX + 1];
void SieveOfEratosthenes()
{
memset (prime, true , sizeof (prime));
prime[1] = false ;
prime[0] = false ;
for ( int p = 2; p * p <= MAX; p++) {
if (prime[p] == true ) {
for ( int i = p * 2; i <= MAX; i += p)
prime[i] = false ;
}
}
}
void solve( int arr[], int n, int k)
{
int c = 0;
long long int sum = 0;
for ( int i = 0; i < n; i++) {
if (prime[arr[i]]) {
c++;
if (c % k == 0) {
sum += arr[i];
c = 0;
}
}
}
cout << sum << endl;
}
int main()
{
SieveOfEratosthenes();
int n = 5, k = 2;
int arr[n] = { 2, 3, 5, 7, 11 };
solve(arr, n, k);
return 0;
}
|
Java
class GFG
{
static final int MAX= 1000000 ;
static boolean []prime= new boolean [MAX + 1 ];
static void SieveOfEratosthenes()
{
for ( int i= 0 ;i<=MAX;i++)
prime[i]= true ;
prime[ 1 ] = false ;
prime[ 0 ] = false ;
for ( int p = 2 ; p * p <= MAX; p++) {
if (prime[p] == true ) {
for ( int i = p * 2 ; i <= MAX; i += p)
prime[i] = false ;
}
}
}
static void solve( int []arr, int n, int k)
{
int c = 0 ;
long sum = 0 ;
for ( int i = 0 ; i < n; i++) {
if (prime[arr[i]]) {
c++;
if (c % k == 0 ) {
sum += arr[i];
c = 0 ;
}
}
}
System.out.println(sum);
}
public static void main(String []args)
{
SieveOfEratosthenes();
int n = 5 , k = 2 ;
int []arr = { 2 , 3 , 5 , 7 , 11 };
solve(arr, n, k);
}
}
|
Python3
def SieveOfEratosthenes():
prime[ 1 ] = False
prime[ 0 ] = False
p = 2
while p * p < = MAX :
if prime[p] = = True :
for i in range (p * 2 , MAX + 1 , p):
prime[i] = False
p + = 1
def solve(arr, n, k):
c = 0
Sum = 0
for i in range ( 0 , n):
if prime[arr[i]]:
c + = 1
if c % k = = 0 :
Sum + = arr[i]
c = 0
print ( Sum )
if __name__ = = "__main__" :
MAX = 1000000
prime = [ True ] * ( MAX + 1 )
SieveOfEratosthenes()
n, k = 5 , 2
arr = [ 2 , 3 , 5 , 7 , 11 ]
solve(arr, n, k)
|
C#
using System;
class GFG
{
static int MAX=1000000;
static bool []prime= new bool [MAX + 1];
static void SieveOfEratosthenes()
{
for ( int i=0;i<=MAX;i++)
prime[i]= true ;
prime[1] = false ;
prime[0] = false ;
for ( int p = 2; p * p <= MAX; p++) {
if (prime[p] == true ) {
for ( int i = p * 2; i <= MAX; i += p)
prime[i] = false ;
}
}
}
static void solve( int []arr, int n, int k)
{
int c = 0;
long sum = 0;
for ( int i = 0; i < n; i++) {
if (prime[arr[i]]) {
c++;
if (c % k == 0) {
sum += arr[i];
c = 0;
}
}
}
Console.WriteLine(sum);
}
public static void Main()
{
SieveOfEratosthenes();
int n = 5, k = 2;
int []arr = { 2, 3, 5, 7, 11 };
solve(arr, n, k);
}
}
|
Javascript
<script>
MAX = 1000000;
prime = new Array(MAX + 1);
function SieveOfEratosthenes()
{
prime.fill( true );
prime[1] = false ;
prime[0] = false ;
for ( var p = 2; p * p <= MAX; p++)
{
if (prime[p] == true )
{
for ( var i = p * 2; i <= MAX; i += p)
prime[i] = false ;
}
}
}
function solve(arr, n, k)
{
var c = 0;
var sum = 0;
for ( var i = 0; i < n; i++) {
if (prime[arr[i]]) {
c++;
if (c % k == 0) {
sum += arr[i];
c = 0;
}
}
}
document.write( sum + "<br>" )
}
SieveOfEratosthenes();
var n = 5, k = 2;
var arr = [ 2, 3, 5, 7, 11 ];
solve(arr, n, k);
</script>
|
complexity Analysis:
- Time Complexity: O(n + MAX3/2)
- Auxiliary Space: O(MAX)
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