⚽ Live Match: dsa/mersenne prime - Jadwal Siaran Langsung

A Mersenne Prime is a prime number that is one less than a power of two. In other words, it is a prime number of the form 2^k - 1, where k ≥ 2.

Given a positive integer n, return all Mersenne Primes less than or equal to n in increasing order.

Examples:

Input: n = 10
Output: [3, 7]
Explanation: 3 and 7 are the only Mersenne Primes less than or equal to 10.
Input: n = 100
Output: [3, 7, 31]
Explanation: 3, 7, and 31 are the only Mersenne Primes less than or equal to 10.

Try It Yourself

Table of Content

[Naive Approach] One by One Check Mersenne - O(sqrt(n) * log n) Time and O(1) Space
[Optimal Approach] Sieve of Eratosthenes - O(n log log n) Time and O(n) Space

[Naive Approach] One by One Check Mersenne - O(sqrt(n) * log n) Time and O(1) Space

Generate numbers of the form 2^k - 1 up to n and check each candidate for primality using trial division.

Working of the Approach:

Start with k = 2.
Generate the Mersenne number 2^k - 1.
If it exceeds n, stop generating further numbers.
Check whether the generated number is prime by testing divisibility up to its square root.
If it is prime, add it to the result.
Increment k and repeat the process.
Return all the Mersenne Primes in increasing order.

C++

#include <bits/stdc++.h>
using namespace std;

bool isPrime(int n) {
    if (n < 2)
        return false;

    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0)
            return false;
    }

    return true;
}

vector<int> allMersennePrimeNo(int n) {
    vector<int> ans;

    for (int k = 2; ; k++) {
        int num = (1 << k) - 1;

        if (num > n)
            break;

        if (isPrime(num))
            ans.push_back(num);
    }

    return ans;
}

int main() {
    int n = 100;

    vector<int> ans = allMersennePrimeNo(n);

    cout << "[";
    for (int i = 0; i < ans.size(); i++) {
        if (i > 0)
            cout << ", ";
        cout << ans[i];
    }
    cout << "]";

    return 0;
}

Java

import java.util.List;
import java.util.ArrayList;

class GFG {
    static boolean isPrime(int n) {
        if (n < 2)
            return false;

        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0)
                return false;
        }

        return true;
    }

    static List<Integer> allMersennePrimeNo(int n) {
        List<Integer> ans = new ArrayList<>();

        for (int k = 2; ; k++) {
            int num = (1 << k) - 1;

            if (num > n)
                break;

            if (isPrime(num))
                ans.add(num);
        }

        return ans;
    }

    public static void main(String[] args) {
        int n = 100;

        List<Integer> ans = allMersennePrimeNo(n);

        System.out.println(ans);
    }
}

Python

def is_prime(n):
    if n < 2:
        return False

    i = 2
    while i * i <= n:
        if n % i == 0:
            return False
        i += 1

    return True


def allMersennePrimeNo(n):
    ans = []

    k = 2
    while True:
        num = (1 << k) - 1

        if num > n:
            break

        if is_prime(num):
            ans.append(num)

        k += 1

    return ans


if __name__ == "__main__":
    n = 100

    ans = allMersennePrimeNo(n)

    print(ans)

using System;
using System.Collections.Generic;

class GFG {
    static bool isPrime(int n) {
        if (n < 2)
            return false;

        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0)
                return false;
        }

        return true;
    }

    static List<int> allMersennePrimeNo(int n) {
        List<int> ans = new List<int>();

        for (int k = 2; ; k++) {
            int num = (1 << k) - 1;

            if (num > n)
                break;

            if (isPrime(num))
                ans.Add(num);
        }

        return ans;
    }

    static void Main() {
        int n = 100;

        List<int> ans = allMersennePrimeNo(n);

        Console.Write("[");
        for (int i = 0; i < ans.Count; i++) {
            if (i > 0)
                Console.Write(", ");
            Console.Write(ans[i]);
        }
        Console.Write("]");
    }
}

JavaScript

function isPrime(n) {
    if (n < 2)
        return false;

    for (let i = 2; i * i <= n; i++) {
        if (n % i === 0)
            return false;
    }

    return true;
}

function allMersennePrimeNo(n) {
    const ans = [];

    for (let k = 2; ; k++) {
        const num = (2 ** k) - 1;

        if (num > n)
            break;

        if (isPrime(num))
            ans.push(num);
    }

    return ans;
}

// Driver Code
const n = 100;
const ans = allMersennePrimeNo(n);

console.log("[" + ans.join(", ") + "]");

Output

[3, 7, 31]

[Optimal Approach] Sieve of Eratosthenes - O(n log log n) Time and O(n) Space

The idea is to use the Sieve of Eratosthenes to precompute all prime numbers up to n. Then generate numbers of the form 2^k - 1 and use the sieve to check whether each candidate is prime.