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Find all factorial numbers less than or equal to n

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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


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#

// 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.

                    

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.
?>

                    

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>

                    

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);
    }
}

                    

Python3

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)

                    

C#

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.



Last Updated : 17 Mar, 2023
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