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++ 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 solve( 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 n = 5, k = 2;
int arr[n] = { 2, 3, 5, 7, 11 };
solve(arr, n, k);
return 0;
} |
// Java implementation of the approach class GFG
{ static final 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<=MAX;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 solve( 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 n = 5 , k = 2 ;
int []arr = { 2 , 3 , 5 , 7 , 11 };
solve(arr, n, k);
}
} |
# Python3 implementation of the approach def SieveOfEratosthenes():
# 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 solve(arr, n, k):
# count of primes
c = 0
# sum of the primes
Sum = 0
# Traverse the array
for i in range ( 0 , 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 if __name__ = = "__main__" :
MAX = 1000000
prime = [ True ] * ( MAX + 1 )
# Create the sieve
SieveOfEratosthenes()
n, k = 5 , 2
arr = [ 2 , 3 , 5 , 7 , 11 ]
solve(arr, n, k)
# This code is contributed by Rituraj Jain |
// C# implementation of the approach using System;
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.
for ( int i=0;i<=MAX;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 solve( 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;
}
}
}
Console.WriteLine(sum);
}
// Driver code
public static void Main()
{
// create the sieve
SieveOfEratosthenes();
int n = 5, k = 2;
int []arr = { 2, 3, 5, 7, 11 };
solve(arr, n, k);
}
} |
<script> // Javascript program to find next // greater number than N MAX = 1000000; 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 ( var 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 ( var i = p * 2; i <= MAX; i += p)
prime[i] = false ;
}
}
} // compute the answer function solve(arr, n, k)
{ // count of primes
var c = 0;
// sum of the primes
var sum = 0;
// traverse the array
for ( var 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>" )
} SieveOfEratosthenes(); var n = 5, k = 2;
var arr = [ 2, 3, 5, 7, 11 ];
solve(arr, n, k); // This code is contributed by SoumikMondal </script> |
10
complexity Analysis:
- Time Complexity: O(n + MAX3/2)
- Auxiliary Space: O(MAX)