Sum of all the prime numbers in a given range

Given a range [l, r], the task is to find the sum of all the prime numbers within that range.

Examples:

Input : l=1 and r=6
Output : 10


Input : l=4 and r=13
Output : 36

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

Approach 1: (Naive Approach)
Iterate the loop from ‘l’ to ‘r’ and add all the numbers which are prime.

Below is the implementation of the above approach:

C++

// C++ Program to computer sum of prime number 
// in a given range 
#include <iostream> 
using namespace std; 
  
// Method to compute the prime number 
// Time Complexity is O(sqrt(N)) 
bool checkPrime(int numberToCheck) 
{ 
    if(numberToCheck == 1) { 
        return false; 
    } 
    for (int i = 2; i*i <= numberToCheck; i++) { 
        if (numberToCheck % i == 0) { 
            return false; 
        } 
    } 
    return true; 
} 
  
// Method to iterate the loop from l to r 
// If the current number is prime, sum the value 
int primeSum(int l, int r) 
{ 
    int sum = 0; 
    for (int i = r; i >= l; i--) { 
  
        // Check for prime 
        bool isPrime = checkPrime(i); 
        if (isPrime) { 
  
            // Sum the prime number 
            sum = sum + i; 
        } 
    } 
    return sum; 
} 
// Time Complexity is O(r x sqrt(N)) 
  
//Driver code 
int main() 
{ 
    int l = 4, r = 13; 
  
    // Call the method with l and r 
    cout << primeSum(l, r); 
} 

Java

// Java Program to computer sum of prime number 
// in a given range 
public class GFG { 
  
    // Method to compute the prime number 
    // Time Complexity is O(sqrt(N)) 
    static boolean checkPrime(int numberToCheck) 
    { 
        if(numberToCheck == 1) { 
            return false; 
        } 
        for (int i = 2; i*i <= numberToCheck; i++) { 
            if (numberToCheck % i == 0) { 
                return false; 
            } 
        } 
        return true; 
    } 
  
    // Method to iterate the loop from l to r 
    // If prime number detects, sum the value 
    static int primeSum(int l, int r) 
    { 
        int sum = 0; 
        for (int i = r; i >= l; i--) { 
  
            // Check for prime 
            boolean isPrime = checkPrime(i); 
            if (isPrime) { 
  
                // Sum the prime number 
                sum = sum + i; 
            } 
        } 
        return sum; 
    } 
    // Time Complexity is O(r x sqrt(N)) 
  
    // Driver code 
    public static void main(String[] args) 
    { 
        int l = 4, r = 13; 
  
        // Call the method with l and r 
        System.out.println(primeSum(l, r)); 
    } 
} 

Python 3

# Python3 Program to computer sum  
# of prime number in a given range 
  
# from math lib import sqrt method 
from math import sqrt 
  
# Function to compute the prime number  
# Time Complexity is O(sqrt(N)) 
def checkPrime(numberToCheck) : 
  
    if numberToCheck == 1 : 
        return False
  
    for i in range(2, int(sqrt(numberToCheck)) + 1) : 
  
        if numberToCheck % i == 0 : 
            return False
  
    return True
  
# Function to iterate the loop  
# from l to r. If the current 
# number is prime, sum the value  
def primeSum(l, r) : 
  
    sum = 0
  
    for i in range(r, (l - 1), -1) : 
  
        # Check for prime  
        isPrime = checkPrime(i) 
          
        if (isPrime) : 
  
            # Sum the prime number 
            sum += i 
  
    return sum
  
# Time Complexity is O(r x sqrt(N))  
  
# Driver code      
if __name__ == "__main__" : 
  
    l, r = 4, 13
  
    # Call the function with l and r  
    print(primeSum(l, r)) 
  
# This code is contributed  
# by ANKITRAI1 

C#

// C# Program to computer sum  
// of prime number in a given range 
using System; 
  
class GFG 
{ 
  
// Method to compute the prime  
// number Time Complexity is O(sqrt(N)) 
static bool checkPrime(int numberToCheck) 
{ 
    if(numberToCheck == 1)  
    { 
        return false; 
    } 
    for (int i = 2; 
             i * i <= numberToCheck; i++)  
    { 
        if (numberToCheck % i == 0)  
        { 
            return false; 
        } 
    } 
    return true; 
} 
  
// Method to iterate the loop from l to r 
// If prime number detects, sum the value 
static int primeSum(int l, int r) 
{ 
    int sum = 0; 
    for (int i = r; i >= l; i--) 
    { 
  
        // Check for prime 
        bool isPrime = checkPrime(i); 
        if (isPrime) 
        { 
  
            // Sum the prime number 
            sum = sum + i; 
        } 
    } 
    return sum; 
} 
// Time Complexity is O(r x sqrt(N)) 
  
// Driver code 
public static void Main() 
{ 
    int l = 4, r = 13; 
  
    // Call the method with l and r 
    Console.Write(primeSum(l, r)); 
} 
} 
  
// This code is contributed  
// by ChitraNayal 

PHP

<?php 
// PHP Program to computer sum of 
// prime number in a given range  
  
// Method to compute the prime number  
// Time Complexity is O(sqrt(N))  
function checkPrime($numberToCheck)  
{  
    if($numberToCheck == 1)  
    {  
        return false;  
    }  
    for ($i = 2; $i * $i <= $numberToCheck; $i++) 
    {  
        if ($numberToCheck % $i == 0)  
        {  
            return false;  
        }  
    }  
    return true;  
}  
  
// Method to iterate the loop from  
// l to r. If the current number 
// is prime, sum the value  
function primeSum($l, $r)  
{  
    $sum = 0;  
    for ($i = $r; $i >= $l; $i--)  
    {  
  
        // Check for prime  
        $isPrime = checkPrime($i);  
        if ($isPrime)  
        {  
  
            // Sum the prime number  
            $sum = $sum + $i;  
        }  
    }  
    return $sum;  
}  
  
// Time Complexity is O(r x sqrt(N))  
  
// Driver code  
$l = 4; $r = 13;  
  
// Call the method with l and r  
echo primeSum($l, $r);  
  
// This code is contributed by ajit 
?> 

Output:

Time Complexity: $O(n\sqrt{n})$
Space Complexity: $O(1)$

Approach 2: (Dynamic Programming)

Declare an array dp and arr
Fill the array arr to 0
Iterate the loop till sqrt(N) and if arr[i] = 0 (marked as prime), then set all of it’s multiples as non-prime by marking the respective location as 1
Update the dp array with the running prime numbers sum, where each location ‘dp[i]’ holds the sum of all the prime numbers withing the range [1, i]

Image Representation

Java

// Java Program to computer sum of prime number 
// in a given range 
public class GFG { 
  
    // Suppose the constraint is N<=1000 
    static int N = 1000; 
  
    // Declare an array for dynamic approach 
    static long dp[] = new long[N + 1]; 
  
    // Method to compute the array 
    static void sieve() 
    { 
        // Declare an extra array as arr 
        int arr[] = new int[N + 1]; 
        arr[0] = 1; 
        arr[1] = 1; 
  
        // Iterate the loop till sqrt(n) 
        // Time Complexity is O(log(n) X sqrt(n)) 
        for (int i = 2; i <= Math.sqrt(N); i++) 
  
            // if ith element of arr is 0 i.e. marked as prime 
            if (arr[i] == 0) 
  
                // mark all of it's multiples till N as non-prime 
                // by setting the locations to 1 
                for (int j = i * i; j <= N; j += i) 
                    arr[j] = 1; 
  
        long runningPrimeSum = 0; 
  
        // Update the array 'dp' with the running sum 
        // of prime numbers within the range [1, N] 
        // Time Complexity is O(n) 
        for (int i = 1; i <= N; i++) { 
            if (arr[i] == 0) 
                runningPrimeSum += i; 
  
            //Here, dp[i] is the sum of all the prime numbers 
            //within the range [1, i] 
            dp[i] = runningPrimeSum; 
        } 
    } 
  
    // Driver code 
    public static void main(String[] args) 
    { 
        int l = 4, r = 13; 
  
        // Compute dp 
        sieve(); 
        System.out.println(dp[r] - dp[l - 1]); 
    } 
} 

Python 3

# Python3 Program to computer sum of prime number
# in a given range
import math # for sqrt and ceil function

# Suppose the constraint is N<=1000 N = 1000 # Declare an array for dynamic approach dp = [0] * (N + 1) # Method to compute the array def seive(): # Declare an extra array as array array = [0] * (N + 1) array[0] = 1 array[1] = 1 # Iterate the loop till sqrt(N) # Time Complexity is O(log(n) X sqrt(N)) for i in range(2, math.ceil(math.sqrt(N) + 1)): # if ith element of arr is 0 # i.e. marked as prime if array[i] == 0: # mark all of it's multiples till N as # non-prime by setting the locations to 1 for j in range(i * i, N + 1, i): array[j] = 1 runningPrimeSum = 0 # Update the array 'dp' with the running sum # of prime numbers within the range [1, N] # Time Complexity is O(n) for i in range(1, N + 1): if array[i] == 0: runningPrimeSum += i # Here, dp[i] is the sum of all the prime numbers # within the range [1, i] dp[i] = runningPrimeSum # Driver Code l = 4 r = 13 seive() print(dp[r] - dp[l - 1]) # This code is contributed by Vivek Kumar Singh [tabby title="C#"]

// C# Program to computer sum of prime number  
// in a given range  
using System; 
public class GFG {  
  
  
    // Suppose the constraint is N<=1000  
    static int N = 1000;  
  
    // Declare an array for dynamic approach  
    static long[] dp = new long[N + 1];  
  
    // Method to compute the array  
    static void sieve()  
    {  
        // Declare an extra array as arr  
        int []arr = new int[N + 1];  
        arr[0] = 1;  
        arr[1] = 1;  
  
        // Iterate the loop till sqrt(n)  
        // Time Complexity is O(log(n) X sqrt(n))  
        for (int i = 2; i <= Math.Sqrt(N); i++)  
  
            // if ith element of arr is 0 i.e. marked as prime  
            if (arr[i] == 0)  
  
                // mark all of it's multiples till N as non-prime  
                // by setting the locations to 1  
                for (int j = i * i; j <= N; j += i)  
                    arr[j] = 1;  
  
        long runningPrimeSum = 0;  
  
        // Update the array 'dp' with the running sum  
        // of prime numbers within the range [1, N]  
        // Time Complexity is O(n)  
        for (int i = 1; i <= N; i++) {  
            if (arr[i] == 0)  
                runningPrimeSum += i;  
  
            //Here, dp[i] is the sum of all the prime numbers  
            //within the range [1, i]  
            dp[i] = runningPrimeSum;  
        }  
    }  
  
    // Driver code  
    public static void Main()  
    {  
        int l = 4, r = 13;  
  
        // Compute dp  
        sieve();  
        Console.WriteLine(dp[r] - dp[l - 1]);  
    }  
}  
  
  
/*This code is contributed by 29AjayKumar*/

Output:

Time Complexity: $O(n)$
Space Complexity: $O(n)$

My Personal Notes

Suggest a Topic

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

C++

Java

Python 3

C#

PHP

Java

Python 3

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