Print prime numbers from 1 to N in reverse order

Given a number N, print all prime number smaller than or equal to N in reverse order .
For example, if N is 9, the output should be “7, 5, 3, 2”.

Examples:

Input :  N = 5
Output : 5 3 2

Input : N = 20
Output : 19 17 13 11 7 5 3 2

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

A simple solution is to traverse from N to 1. For every number, check if it is a prime using school method to check for prime. Print the number if it is prime.

An efficient solution is to use Sieve of Eratosthenes. We efficiently find all number from 1 to N, then print them.

C++

// C++ program to print all primes between 1 
// to N in reverse order using Sieve of 
// Eratosthenes 
#include <bits/stdc++.h> 
using namespace std; 
  
void Reverseorder(int n) 
{ 
    // Create a boolean array "prime[0..n]" and 
    // initialize all entries it as true. A value 
    // in prime[i] will finally be false if i 
    // is Not a prime, else true. 
    bool prime[n + 1]; 
    memset(prime, true, sizeof(prime)); 
  
    for (int p = 2; p * p <= n; 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 <= n; i += p) 
                prime[i] = false; 
        } 
    } 
  
    // Print all prime numbers in reverse order 
    for (int p = n; p >= 2; p--) 
        if (prime[p]) 
            cout << p << " "; 
} 
  
// Driver Program 
int main() 
{ 
    // static input 
    int N = 25; 
  
    // to display 
    cout << "Prime number in reverse order" << endl; 
  
    if (N == 1) 
        cout << "No prime no exist in this range"; 
    else
        Reverseorder(N); // calling the function 
  
    return 0; 
} 

Java

// Java program to print all primes between 1 
// to N in reverse order using Sieve of 
// Eratosthenes 
import java.io.*; 
import java.util.*; 
  
class GFG { 
    static void reverseorder(int n) 
    { 
  
        // Create a boolean array "prime[0..n]" and 
        // initialize all entries it as true. A value 
        // in prime[i] will finally be false if i 
        // is Not a prime, else true. 
        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] is not changed, then 
            // it is a prime 
            if (prime[p] == true) { 
  
                // Update all multiples of p 
                for (int i = p * 2; i <= n; i += p) 
                    prime[i] = false; 
            } 
        } 
  
        // Print all prime numbers 
        for (int i = n; i >= 2; i--) 
            if (prime[i] == true) 
                System.out.print(i + " "); 
    } 
  
    // Driver Program to test above function 
    public static void main(String args[]) 
    { 
        // static input 
        int N = 25; 
        // To display 
        System.out.println("Prime number in reverse order"); 
  
        if (N == 1) 
            System.out.println("No prime no exist in this range"); 
        else
            reverseorder(N); // calling the function 
    } 
} 

Python3

# Python3 program to print all primes  
# between 1 to N in reverse order  
# using Sieve of Eratosthenes 
def Reverseorder(n): 
  
    # Create a boolean array "prime[0..n]"  
    # and initialize all entries it as true.  
    # A value in prime[i] will finally be  
    # false if i is Not a prime, else true. 
    prime = [True] * (n + 1); 
  
    p = 2;  
    while(p * p <= n): 
  
        # 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), (n + 1), p): 
                prime[i] = False; 
        p += 1; 
  
    # Print all prime numbers in  
    # reverse order 
    for p in range(n, 1, -1): 
        if (prime[p]): 
            print(p, end = " "); 
  
# Driver Code 
  
# static input 
N = 25; 
  
# to display 
print("Prime number in reverse order"); 
  
if (N == 1): 
    print("No prime no exist in this range"); 
else: 
    Reverseorder(N); # calling the function 
  
# This code is contributed by mits 

C#

// C# program to print all primes between 1 
// to N in reverse order using Sieve of 
// Eratosthenes 
using System; 
  
class GFG { 
  
    static void reverseorder(int n) 
    { 
  
        // Create a boolean array "prime[0..n]" 
        // and initialize all entries it as 
        // true. A value in prime[i] will  
        // finally be false if i is Not a 
        // prime, else true. 
        bool []prime = new bool[n + 1]; 
          
        for (int i = 0; i < n; i++) 
            prime[i] = true; 
  
        for (int p = 2; p * p <= n; 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 <= n; 
                                     i += p) 
                    prime[i] = false; 
            } 
        } 
  
        // Print all prime numbers 
        for (int i = n; i >= 2; i--) 
            if (prime[i] == true) 
                Console.Write(i + " "); 
    } 
      
    // Driver code 
    public static void Main() 
    { 
          
        // static input 
        int N = 25; 
          
        // To display 
        Console.WriteLine("Prime number in"
                       + " reverse order"); 
  
        if (N == 1) 
            Console.WriteLine("No prime no"
                + " exist in this range"); 
        else
          
            // calling the function 
            reverseorder(N); 
    } 
} 
  
// This code is contributed by Sam007. 

PHP

<?php 
// PHP program to print all primes  
// between 1 to N in reverse order  
// using Sieve of Eratosthenes 
  
function Reverseorder($n) 
{ 
    // Create a boolean array "prime[0..n]"  
    // and initialize all entries it as true.  
    // A value in prime[i] will finally be  
    // false if i is Not a prime, else true. 
    $prime = array_fill(0, $n + 1, true); 
  
    for ($p = 2; $p * $p <= $n; $p++) 
    { 
  
        // If prime[p] is not changed,  
        // then it is a prime 
        if ($prime[$p] == true)  
        { 
  
            // Update all multiples of p 
            for ($i = $p * 2; $i <= $n; $i += $p) 
                $prime[$i] = false; 
        } 
    } 
  
    // Print all prime numbers in  
    // reverse order 
    for ($p = $n; $p >= 2; $p--) 
        if ($prime[$p]) 
            echo $p." "; 
} 
  
// Driver Code 
  
// static input 
$N = 25; 
  
// to display 
echo "Prime number in reverse order\n"; 
  
if ($N == 1) 
    echo "No prime no exist in this range"; 
else
    Reverseorder($N); // calling the function 
  
// This code is contributed by mits 
?> 

Output:

Prime number in reverse order
23 19 17 13 11 7 5 3 2

My Personal Notes

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

C++

Java

Python3

C#

PHP

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