Find all factorial numbers less than or equal to n

A number N is called a factorial number if it is the factorial of a positive integer. For example, the first few factorial numbers are
1, 2, 6, 24, 120, ...
Given a number n, print all factorial numbers smaller than or equal to n.
Examples :

Input: n = 100
Output: 1 2 6 24
Input: n = 1500
Output: 1 2 6 24 120 720

Try It Yourself

A simple solution is to generate all factorials one by one until the generated factorial is greater than n.
An efficient solution is to find next factorial using previous factorial.

C++

// CPP program to find all factorial numbers
// smaller than or equal to n.
#include <iostream>
using namespace std;

void printFactorialNums(int n)
{
    int fact = 1;
    int x = 2;
    while (fact <= n) {
        cout << fact << " ";

        // Compute next factorial 
        // using previous
        fact = fact * x;

        x++;
    }
}

// Driver code
int main()
{
    int n = 100;
    printFactorialNums(n);
    return 0;
}

Java

// Java program to find all factorial numbers
// smaller than or equal to n.

class GFG
{
    static void printFactorialNums(int n)
    {
        int fact = 1;
        int x = 2;
        while (fact <= n)
        {
            System.out.print(fact + " ");
    
            // Compute next factorial 
            // using previous
            fact = fact * x;
    
            x++;
        }
    }
    
    // Driver code
    public static void main (String[] args) 
    {
        int n = 100;
        printFactorialNums(n);
    }
}

// This code is contributed by Anant Agarwal.

Python3

# Python3 program to find all factorial 
# numbers smaller than or equal to n.

def printFactorialNums( n):
    fact = 1
    x = 2
    while fact <= n:
        print(fact, end = " ")
        
        # Compute next factorial
        # using previous
        fact = fact * x
        
        x += 1

# Driver code
n = 100
printFactorialNums(n)

# This code is contributed by "Abhishek Sharma 44"

// C# program to find all factorial numbers
// smaller than or equal to n.
using System;

class GFG
{
    static void printFactorialNums(int n)
    {
        int fact = 1;
        int x = 2;
        while (fact <= n)
        {
            Console.Write(fact + " ");
    
            // Compute next factorial 
            // using previous
            fact = fact * x;
    
            x++;
        }
    }
    
    // Driver code
    public static void Main () 
    {
        int n = 100;
        printFactorialNums(n);
    }
}

// This code is contributed by vt_m.

JavaScript

<script>

// Javascript program to find all factorial numbers 
// smaller than or equal to n. 

function printFactorialNums(n) 
{ 
    let fact = 1; 
    let x = 2; 
    while (fact <= n) { 
        document.write(fact + " "); 

        // Compute next factorial 
        // using previous 
        fact = fact * x; 

        x++; 
    } 
} 

// Driver code 

    let n = 100; 
    printFactorialNums(n); 

// This code is contributed by Mayank Tyagi
    
</script>

PHP

<?php
// PHP program to find all
// factorial numbers smaller
// than or equal to n.

function printFactorialNums($n)
{
    $fact = 1;
    $x = 2;
    while ($fact <= $n) 
    {
        echo $fact, " ";

        // Compute next factorial 
        // using previous
        $fact = $fact * $x;
        $x++;
    }
}

    // Driver code
    $n = 100;
    echo printFactorialNums($n);

// This code is contributed by ajit.
?>

Output

1 2 6 24

Time Complexity: O(x)
Auxiliary Space: O(1)

Another solution : we can print factorial of a number such that factorial <=n by recursion.

C++

// C++ implementation of the above approach

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

// Function to Print factorial using Recursion
void PrintFactorialNums(int n, int fac, int i)
{
    i++;
    if (fac > n) // if fac>n return because
    {
        return;
    } // we have to find in [1,N]

    cout << fac << " "; // print Factors

    // recursive call
    PrintFactorialNums(n, fac * i, i);
}

// Drive Code
int main()
{
    int n = 100;

    // Function call
    PrintFactorialNums(n, 1, 1);

    return 0;
}

// This code is contributed by nikhilsainiofficial546

Java

import java.util.*;

class Main {
    // Function to Print factorial using Recursion
    static void PrintFactorialNums(int n, int fac, int i) {
        i++;
        if (fac > n) { // if fac>n return because
            return;
        } // we have to find in [1,N]

        System.out.print(fac + " "); // print Factors

        // recursive call
        PrintFactorialNums(n, fac * i, i);
    }

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

        // Function call
        PrintFactorialNums(n, 1, 1);
    }
}

Python

def print_factorial_nums(n, fac, i):
    # Recursive function to print factors of n!

    i += 1
    if fac > n:
        # If fac becomes greater than n, return from the function
        # because we only need to find the factors up to n
        return

    print(fac, end=' ')  # print the current factor

    # Recursive call to print the next factor
    print_factorial_nums(n, fac * i, i)


# Driver code
if __name__ == '__main__':
    n = 100
    print_factorial_nums(n, 1, 1)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;

// C# code implementation of the above approach. 
class HelloWorld {

  // Function to Print factorial using Recursion
  public static void PrintFactorialNums(int n, int fac, int i) {
    i++;
    if (fac > n) { // if fac>n return because
      return;
    } // we have to find in [1,N]

    Console.Write(fac + " "); // print Factors

    // recursive call
    PrintFactorialNums(n, fac * i, i);
  }

  static void Main() {

    int n = 100;

    // Function call
    PrintFactorialNums(n, 1, 1);
  }
}

// The code is contributed by Nidhi goel.

JavaScript

temp=""; 
function print_factorial_nums(n, fac, i) {

    // Recursive function to print factors of n!
    i += 1;
    if (fac > n) {
        // If fac becomes greater than n, return from the function
        // because we only need to find the factors up to n
        return;
    }

    temp = temp + fac + " "; // print the current factor

    // Recursive call to print the next factor
    print_factorial_nums(n, fac * i, i);
}

// Driver code
let n = 100;
print_factorial_nums(n, 1, 1);
console.log(temp);

Output

1 2 6 24

Time Complexity: O(n)
Auxiliary Space: O(n)
If there are multiple queries, then we can cache all previously computed factorial numbers to avoid re-computations.

Find all factorial numbers less than or equal to n

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