Given a positive integer n, compute and return the sum of all prime numbers between 1 and n (inclusive).
Examples:
Input: n = 5
Output: 10
Explanation: 2, 3 and 5 are prime numbers between 1 and 5(inclusive), and their sum is 2 + 3 + 5 = 10.Input: n = 10
Output: 17
Explanation: 2, 3, 5 and 7 are prime numbers between 1 and 10(inclusive), and their sum is 2 + 3 + 5 + 7 = 17.
Try It Yourself
Table of Content
[Naive Approach] Trial Division Method - O(n^2) Time and O(1) Space
The idea is to check every number from 2 to n and find if it is prime by trying all possible divisors up to i / 2. If no divisor divides the number completely, it is prime, so we add it to the sum.
- Initialize sum = 0.
- Traverse every number i from 2 to n.
- Assume i is prime and check all divisors from 2 to i / 2.
- If any divisor divides i, mark it as non-prime.
- If i is still prime, add it to sum.
- Return the calculated sum.
#include <bits/stdc++.h>
using namespace std;
int primeSum(int n)
{
int sum = 0;
// Check every number from 2 to n
for (int i = 2; i <= n; i++)
{
bool isPrime = true;
// Check all possible divisors up to i / 2
for (int j = 2; j <= i / 2; j++)
{
if (i % j == 0)
{
isPrime = false;
break;
}
}
// Add the number if it is prime
if (isPrime)
sum += i;
}
return sum;
}
int main()
{
int n = 10;
int result = primeSum(n);
cout << result << endl;
return 0;
}
import java.util.*;
class GFG {
static int primeSum(int n)
{
int sum = 0;
// Check every number from 2 to n
for (int i = 2; i <= n; i++) {
boolean isPrime = true;
// Check all possible divisors up to i / 2
for (int j = 2; j <= i / 2; j++) {
if (i % j == 0) {
isPrime = false;
break;
}
}
// Add the number if it is prime
if (isPrime)
sum += i;
}
return sum;
}
public static void main(String[] args)
{
int n = 10;
int result = primeSum(n);
System.out.println(result);
}
}
def primeSum(n):
sum = 0
# Check every number from 2 to n
for i in range(2, n + 1):
isPrime = True
# Check all possible divisors up to i / 2
for j in range(2, i // 2 + 1):
if i % j == 0:
isPrime = False
break
# Add the number if it is prime
if isPrime:
sum += i
return sum
# Driver Code
if __name__ == "__main__":
n = 10
result = primeSum(n)
print(result)
using System;
class GFG {
static int primeSum(int n)
{
int sum = 0;
// Check every number from 2 to n
for (int i = 2; i <= n; i++) {
bool isPrime = true;
// Check all possible divisors up to i / 2
for (int j = 2; j <= i / 2; j++) {
if (i % j == 0) {
isPrime = false;
break;
}
}
// Add the number if it is prime
if (isPrime)
sum += i;
}
return sum;
}
public static void Main()
{
int n = 10;
int result = primeSum(n);
Console.WriteLine(result);
}
}
function primeSum(n)
{
let sum = 0;
// Check every number from 2 to n
for (let i = 2; i <= n; i++) {
let isPrime = true;
// Check all possible divisors up to i / 2
for (let j = 2; j <= Math.floor(i / 2); j++) {
if (i % j === 0) {
isPrime = false;
break;
}
}
// Add the number if it is prime
if (isPrime)
sum += i;
}
return sum;
}
// Driver Code
const n = 10;
const result = primeSum(n);
console.log(result);
Output
17
[Better Approach] Square Root Method - O(n * sqrt(n)) Time and O(1) Space
For each number, we try dividing it by possible divisors only up to its square root, because if a number has a factor larger than its square root, the corresponding smaller factor must already exist.
If no divisor is found, the number is prime, so we add it to the sum.
- Initialize sum = 0.
- Traverse every number i from 2 to n.
- Assume i is prime and check divisibility from 2 to sqrt(i).
- If any divisor is found, mark i as composite.
- If i is still prime, add it to sum.
- Return the final sum.
#include <bits/stdc++.h>
using namespace std;
int primeSum(int n)
{
int sum = 0;
// Check every number from 2 to n
for (int i = 2; i <= n; i++)
{
bool isPrime = true;
// Check divisors only up to sqrt(i)
for (int j = 2; j * j <= i; j++)
{
if (i % j == 0)
{
isPrime = false;
break;
}
}
// Add the number if it is prime
if (isPrime)
sum += i;
}
return sum;
}
int main()
{
int n = 10;
int result = primeSum(n);
cout << result << endl;
return 0;
}
import java.util.*;
class GFG {
static int primeSum(int n)
{
int sum = 0;
// Check every number from 2 to n
for (int i = 2; i <= n; i++) {
boolean isPrime = true;
// Check divisors only up to sqrt(i)
for (int j = 2; j * j <= i; j++) {
if (i % j == 0) {
isPrime = false;
break;
}
}
// Add the number if it is prime
if (isPrime)
sum += i;
}
return sum;
}
public static void main(String[] args)
{
int n = 10;
int result = primeSum(n);
System.out.println(result);
}
}
def primeSum(n):
sum = 0
# Check every number from 2 to n
for i in range(2, n + 1):
isPrime = True
# Check divisors only up to sqrt(i)
j = 2
while j * j <= i:
if i % j == 0:
isPrime = False
break
j += 1
# Add the number if it is prime
if isPrime:
sum += i
return sum
# Driver Code
if __name__ == "__main__":
n = 10
result = primeSum(n)
print(result)
using System;
class GFG {
static int primeSum(int n)
{
int sum = 0;
// Check every number from 2 to n
for (int i = 2; i <= n; i++) {
bool isPrime = true;
// Check divisors only up to sqrt(i)
for (int j = 2; j * j <= i; j++) {
if (i % j == 0) {
isPrime = false;
break;
}
}
// Add the number if it is prime
if (isPrime)
sum += i;
}
return sum;
}
public static void Main()
{
int n = 10;
int result = primeSum(n);
Console.WriteLine(result);
}
}
function primeSum(n)
{
let sum = 0;
// Check every number from 2 to n
for (let i = 2; i <= n; i++) {
let isPrime = true;
// Check divisors only up to sqrt(i)
for (let j = 2; j * j <= i; j++) {
if (i % j === 0) {
isPrime = false;
break;
}
}
// Add the number if it is prime
if (isPrime)
sum += i;
}
return sum;
}
// Driver Code
const n = 10;
const result = primeSum(n);
console.log(result);
Output
17
[Expected Approach] Sieve of Eratosthenes - O(n * loglog(n)) Time and O(n) Space
The idea is to find all prime numbers up to n efficiently using the Sieve of Eratosthenes. We initially consider every number from 2 to n as prime, then mark the multiples of each prime as non-prime. Finally, we traverse the remaining prime numbers and add them to the sum.
- Create a boolean array vis of size n + 1 and mark all numbers as prime.
- Mark 0 and 1 as non-prime.
- For every i from 2 to sqrt(n), if i is prime, mark all its multiples starting from i * i as non-prime.
- Traverse all numbers from 2 to n.
- If vis[i] is true, add i to the sum.
- Return the calculated sum.
#include <bits/stdc++.h>
using namespace std;
int primeSum(int n)
{
vector<bool> vis(n + 1, true);
int sum = 0;
// 0 and 1 are not prime numbers
if (n >= 0)
vis[0] = false;
if (n >= 1)
vis[1] = false;
// Find all prime numbers using Sieve of Eratosthenes
for (int i = 2; i * i <= n; i++)
{
if (vis[i])
{
// Mark all multiples of i as non-prime
for (int j = i * i; j <= n; j += i)
{
vis[j] = false;
}
}
}
// Calculate the sum of prime numbers
for (int i = 2; i <= n; i++)
{
if (vis[i])
sum += i;
}
return sum;
}
int main()
{
int n = 10;
int result = primeSum(n);
cout << result << endl;
return 0;
}
import java.util.*;
class GFG {
static int primeSum(int n)
{
boolean[] vis = new boolean[n + 1];
Arrays.fill(vis, true);
int sum = 0;
// 0 and 1 are not prime numbers
if (n >= 0)
vis[0] = false;
if (n >= 1)
vis[1] = false;
// Find all prime numbers using Sieve of
// Eratosthenes
for (int i = 2; i * i <= n; i++) {
if (vis[i]) {
// Mark all multiples of i as non-prime
for (int j = i * i; j <= n; j += i) {
vis[j] = false;
}
}
}
// Calculate the sum of prime numbers
for (int i = 2; i <= n; i++) {
if (vis[i])
sum += i;
}
return sum;
}
public static void main(String[] args)
{
int n = 10;
int result = primeSum(n);
System.out.println(result);
}
}
def primeSum(n):
vis = [True] * (n + 1)
sum = 0
# 0 and 1 are not prime numbers
if n >= 0:
vis[0] = False
if n >= 1:
vis[1] = False
# Find all prime numbers using Sieve of Eratosthenes
i = 2
while i * i <= n:
if vis[i]:
# Mark all multiples of i as non-prime
for j in range(i * i, n + 1, i):
vis[j] = False
i += 1
# Calculate the sum of prime numbers
for i in range(2, n + 1):
if vis[i]:
sum += i
return sum
# Driver Code
if __name__ == "__main__":
n = 10
result = primeSum(n)
print(result)
using System;
class GFG {
static int primeSum(int n)
{
bool[] vis = new bool[n + 1];
for (int i = 0; i <= n; i++)
vis[i] = true;
int sum = 0;
// 0 and 1 are not prime numbers
if (n >= 0)
vis[0] = false;
if (n >= 1)
vis[1] = false;
// Find all prime numbers using Sieve of
// Eratosthenes
for (int i = 2; i * i <= n; i++) {
if (vis[i]) {
// Mark all multiples of i as non-prime
for (int j = i * i; j <= n; j += i) {
vis[j] = false;
}
}
}
// Calculate the sum of prime numbers
for (int i = 2; i <= n; i++) {
if (vis[i])
sum += i;
}
return sum;
}
public static void Main()
{
int n = 10;
int result = primeSum(n);
Console.WriteLine(result);
}
}
function primeSum(n)
{
let vis = new Array(n + 1).fill(true);
let sum = 0;
// 0 and 1 are not prime numbers
if (n >= 0)
vis[0] = false;
if (n >= 1)
vis[1] = false;
// Find all prime numbers using Sieve of Eratosthenes
for (let i = 2; i * i <= n; i++) {
if (vis[i]) {
// Mark all multiples of i as non-prime
for (let j = i * i; j <= n; j += i) {
vis[j] = false;
}
}
}
// Calculate the sum of prime numbers
for (let i = 2; i <= n; i++) {
if (vis[i])
sum += i;
}
return sum;
}
// Driver Code
const n = 10;
const result = primeSum(n);
console.log(result);
Output
17
