Sieve of Eratosthenes

Last Updated : 23 Jun, 2025

Given a number n, find all prime numbers less than or equal to n.

Examples:

Input: n = 10
Output: [2, 3, 5, 7]
Explanation: The prime numbers up to 10 obtained by Sieve of Eratosthenes are [2, 3, 5, 7].
Input: n = 35
Output: [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31]
Explanation: The prime numbers up to 35 obtained by Sieve of Eratosthenes are [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31].

Table of Content

[Naive Approach] - Using Loop O(n*sqrt(n)) Time and O(1) Space

The Naive Approach for finding all prime numbers from 1 to n involves checking each number individually to determine whether it is prime.

Step By Step Implementations:

Loop through all numbers i from 2 to n.
For each i, check if it is divisible by any number from 2 to i - 1.
If it is divisible, then i is not prime.
If it is not divisible by any number in that range, then i is prime.

[Efficient Approach] - Sieve of Eratosthenes

The Sieve of Eratosthenes efficiently finds all primes up to n by repeatedly marking multiples of each prime as non-prime, starting from 2. This avoids redundant checks and quickly filters out all composite numbers.

Step By Step Implementations:

Initialize a Boolean array prime[0..n] and set all entries to true, except for 0 and 1 (which are not primes).
Start from 2, the smallest prime number.
For each number p from 2 up to √n:
- If p is marked as prime(true):
- Mark all multiples of p as not prime(false), starting from p * p (since smaller multiples have already been marked by smaller primes).
After the loop ends, all the remaining true entries in prime represent prime numbers.

C++

#include <iostream> #include <vector> using namespace std;  vector<int> sieve(int n) {          // creation of boolean array     vector<bool> prime(n + 1, true);     for (int p = 2; p * p <= n; p++) {         if (prime[p] == true) {                          // marking as false             for (int i = p * p; i <= n; i += p)                 prime[i] = false;         }     }          vector<int> res;     for (int p = 2; p <= n; p++){         if (prime[p]){              res.push_back(p);         }     }     return res; }  int main(){     int n = 35;     vector<int> res = sieve(n);     for(auto ele : res){         cout << ele << ' ';     }     return 0; }

Java

class GfG {     static int[] sieve(int n) {                  // creation of boolean array         boolean[] prime = new boolean[n + 1];         for (int i = 0; i <= n; i++) {             prime[i] = true;         }           for (int p = 2; p * p <= n; p++) {             if (prime[p]) {                 // marking as false                 for (int i = p * p; i <= n; i += p)                     prime[i] = false;             }         }           // Count number of primes         int count = 0;         for (int p = 2; p <= n; p++) {             if (prime[p])                 count++;         }           // Store primes in an array         int[] res = new int[count];         int index = 0;         for (int p = 2; p <= n; p++) {             if (prime[p])                 res[index++] = p;         }           return res;     }       public static void main(String[] args) {         int n = 35;         int[] res = sieve(n);         for (int ele : res) {             System.out.print(ele + " ");         }     } }

Python

def sieve(n):         #Create a boolean list to track prime status of numbers     prime = [True] * (n + 1)     p = 2      # Sieve of Eratosthenes algorithm     while p * p <= n:         if prime[p]:                          # Mark all multiples of p as non-prime             for i in range(p * p, n + 1, p):                 prime[i] = False         p += 1      # Collect all prime numbers     res = []     for p in range(2, n + 1):         if prime[p]:             res.append(p)          return res  if __name__ == "__main__":     n = 35     res = sieve(n)     for ele in res:         print(ele, end=' ')

using System; using System.Collections.Generic;  class GfG {          // Function to return all prime numbers up to n     static List<int> sieve(int n)  {                  // Boolean array to mark primes         bool[] prime = new bool[n + 1];         for (int i = 0; i <= n; i++) {             prime[i] = true;         }          // Sieve of Eratosthenes         for (int p = 2; p * p <= n; p++)         {             if (prime[p])             {                 for (int i = p * p; i <= n; i += p)                 {                     prime[i] = false;                 }             }         }          // Store primes in list         List<int> res = new List<int>();         for (int i = 2; i <= n; i++)         {             if (prime[i])             {                 res.Add(i);             }         }          return res;     }      static void Main()     {         int n = 35;         List<int> res = sieve(n);          foreach (int ele in res)         {             Console.Write(ele + " ");         }     } }

JavaScript

function sieve(n) {          // Create a boolean array to mark primes     let prime = new Array(n + 1).fill(true);          // 0 and 1 are not prime     prime[0] = prime[1] = false;       // Apply Sieve of Eratosthenes     for (let p = 2; p * p <= n; p++) {         if (prime[p]) {                        // Mark all multiples of p as not prime             for (let i = p * p; i <= n; i += p) {                 prime[i] = false;             }         }     }      // Collect all primes into result array     let res = [];     for (let p = 2; p <= n; p++) {         if (prime[p]) {             res.push(p);         }     }      return res; }  // Driver code let n = 35; let res = sieve(n); console.log(res.join(' '));

Output

2 3 5 7 11 13 17 19 23 29 31

Time Complexity: O(n*log(log(n))). For each prime number, we mark its multiples, which takes around n/p steps. The total time is proportional to n*(1/2 + 1/3 + 1/5 + ....).
This sum over primes grows slowly and is approximately O(n*log(log(n))) making the algorithm very efficient.
Auxiliary Space: O(n)

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How is the time complexity of Sieve of Eratosthenes is n*log(log(n))?

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Sieve of Eratosthenes

[Naive Approach] - Using Loop O(n*sqrt(n)) Time and O(1) Space

[Efficient Approach] - Sieve of Eratosthenes

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