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++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define MAX 1000000bool 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 answervoid 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 codeint 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
// Java implementation of the approachclass 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
# 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#
// C# implementation of the approachusing 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); }} |
Javascript
<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 answerfunction 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> |
Output
10
complexity Analysis:
- Time Complexity: O(n + MAX3/2)
- Auxiliary Space: O(MAX)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
