Sum of all Primes in a given range using Sieve of Eratosthenes

Given a range [L, R]. The task is to find the sum of all the prime numbers in the given range from L to R both inclusive.

Examples:

Input : L = 10, R = 20
Output : Sum = 60
Prime numbers between [10, 20] are:
11, 13, 17, 19
Therefore, sum = 11 + 13 + 17 + 19 = 60

Input : L = 15, R = 25
Output : Sum = 59

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A Simple Solution is to traverse from L to R, check if the current number is prime. If yes, add it to $sum$ . Finally, print the sum.

An Efficient Solution is to use Sieve of Eratosthenes to find all primes up to a given limit. Then, compute a prefix sum array to store sum till every value before the limit. Once we have prefix array, We just need to return prefix[R] – prefix[L-1].

Note: prefix[i] will store the sum of all prime numbers from 1 to $i$ .

Below is the implementation of the above approach:

C++

// CPP program to find sum of primes 
// in range L to R 
#include <bits/stdc++.h> 
using namespace std; 
  
const int MAX = 10000; 
  
// prefix[i] is going to store sum of primes 
// till i (including i). 
int prefix[MAX + 1]; 
  
// Function to build the prefix sum array 
void buildPrefix() 
{ 
    // Create a boolean array "prime[0..n]". A 
    // value in prime[i] will finally be false 
    // if i is Not a prime, else true. 
    bool prime[MAX + 1]; 
    memset(prime, true, sizeof(prime)); 
  
    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; 
        } 
    } 
  
    // Build prefix array 
    prefix[0] = prefix[1] = 0; 
    for (int p = 2; p <= MAX; p++) { 
        prefix[p] = prefix[p - 1]; 
        if (prime[p]) 
            prefix[p] += p; 
    } 
} 
  
// Function to return sum of prime in range 
int sumPrimeRange(int L, int R) 
{ 
    buildPrefix(); 
  
    return prefix[R] - prefix[L - 1]; 
} 
  
// Driver code 
int main() 
{ 
    int L = 10, R = 20; 
  
    cout << sumPrimeRange(L, R) << endl; 
  
    return 0; 
} 

Java

// Java program to find sum of primes 
// in range L to R 
  
import java.util.*; 
import java.lang.*; 
import java.io.*; 
  
class GFG{ 
  
static final int MAX = 10000; 
  
// prefix[i] is going to store sum of primes 
// till i (including i). 
static int prefix[]=new int[MAX + 1]; 
  
// Function to build the prefix sum array 
static void buildPrefix() 
{ 
    // Create a boolean array "prime[0..n]". A 
    // value in prime[i] will finally be false 
    // if i is Not a prime, else true. 
    boolean prime[] = new boolean[MAX + 1]; 
      
    for(int i = 0; i < MAX+1; i++) 
    prime[i] = true; 
  
    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; 
        } 
    } 
  
    // Build prefix array 
    prefix[0] = prefix[1] = 0; 
    for (int p = 2; p <= MAX; p++) { 
        prefix[p] = prefix[p - 1]; 
        if (prime[p] == true) 
            prefix[p] += p; 
    } 
} 
  
// Function to return sum of prime in range 
static int sumPrimeRange(int L, int R) 
{ 
    buildPrefix(); 
  
    return prefix[R] - prefix[L - 1]; 
} 
  
// Driver code 
public static void main(String args[]) 
{ 
    int L = 10, R = 20; 
  
    System.out.println (sumPrimeRange(L, R)); 
  
} 
} 

Python3

# Python 3 program to find sum of primes 
# in range L to R 
from math import sqrt 
  
MAX = 10000
  
# prefix[i] is going to store sum of primes 
# till i (including i). 
prefix = [0 for i in range(MAX + 1)] 
  
# Function to build the prefix sum array 
def buildPrefix(): 
      
    # Create a boolean array "prime[0..n]". A 
    # value in prime[i] will finally be false 
    # if i is Not a prime, else true. 
    prime = [True for i in range(MAX + 1)] 
  
    for p in range(2,int(sqrt(MAX)) + 1, 1): 
          
        # 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
  
    # Build prefix array 
    prefix[0] = 0
    prefix[1] = 0
    for p in range(2, MAX + 1, 1): 
        prefix[p] = prefix[p - 1] 
        if (prime[p]): 
            prefix[p] += p 
  
# Function to return sum of prime in range 
def sumPrimeRange(L, R): 
    buildPrefix() 
  
    return prefix[R] - prefix[L - 1] 
  
# Driver code 
if __name__ == '__main__': 
    L = 10
    R = 20
    print(sumPrimeRange(L, R)) 
  
# This code is contributed by 
# Surendra_Gangwar 

C#

// C# program to find sum of  
// primes in range L to R 
using System; 
class GFG 
{ 
  
static int MAX = 10000; 
  
// prefix[i] is going to 
// store sum of primes 
// till i (including i). 
static int []prefix = new int[MAX + 1]; 
  
// Function to build the 
// prefix sum array 
static void buildPrefix() 
{ 
    // Create a boolean array "prime[0..n]". 
    // A value in prime[i] will finally  
    // be false if i is Not a prime,  
    // else true. 
    bool []prime = new bool[MAX + 1]; 
      
    for(int i = 0; i < MAX+1; i++) 
    prime[i] = true; 
  
    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; 
        } 
    } 
  
    // Build prefix array 
    prefix[0] = prefix[1] = 0; 
    for (int p = 2; p <= MAX; p++)  
    { 
        prefix[p] = prefix[p - 1]; 
        if (prime[p] == true) 
            prefix[p] += p; 
    } 
} 
  
// Function to return sum 
// of prime in range 
static int sumPrimeRange(int L, int R) 
{ 
    buildPrefix(); 
  
    return prefix[R] - prefix[L - 1]; 
} 
  
// Driver code 
public static void Main() 
{ 
    int L = 10, R = 20; 
  
    Console.WriteLine(sumPrimeRange(L, R)); 
} 
} 
  
// This code is contributed  
// by anuj_67 

Output:

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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