Find all divisors of a natural number

Given a natural number n, print all distinct divisors of it.

Examples:

 Input : n = 10       
 Output: 1 2 5 10

 Input:  n = 100
 Output: 1 2 4 5 10 20 25 50 100

 Input:  n = 125
 Output: 1 5 25 125

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

We strongly recommend to refer below article as a prerequisite.
Find all divisors of a natural number | Set 1

In the above post, we had found a way to find all the divisors in O(sqrt(n)) .

However there is still a minor problem in the solution, can you guess?
Yes! the output is not in a sorted fashion which we had got using brute-force technique.

How to print the output in sorted order?
If we observe the output which we had got, we can analyze that the divisors are printed in a zig-zag fashion (small, large pairs). Hence if we store half of them then we can print then in a sorted order.

Below is a implementation for the same:

C++

// A O(sqrt(n)) program that prints all divisors 
// in sorted order 
#include <bits/stdc++.h> 
using namespace std; 
  
// function to print the divisors 
void printDivisors(int n) 
{ 
    // Vector to store half of the divisors 
    vector<int> v; 
    for (int i=1; i<=sqrt(n); i++) 
    { 
        if (n%i==0) 
        { 
            if (n/i == i) // check if divisors are equal 
                printf("%d ", i); 
            else
            { 
                printf("%d ", i); 
  
                // push the second divisor in the vector 
                v.push_back(n/i); 
            } 
        } 
    } 
  
    // The vector will be printed in reverse 
    for (int i=v.size()-1; i>=0; i--) 
        printf("%d ", v[i]); 
} 
  
/* Driver program to test above function */
int main() 
{ 
    printf("The divisors of 100 are: n"); 
    printDivisors(100); 
    return 0; 
} 

Java

// A O(sqrt(n)) java program that prints all divisors 
// in sorted order 
  
import java.util.Vector; 
  
class Test 
{ 
    // method to print the divisors 
    static void printDivisors(int n) 
    { 
        // Vector to store half of the divisors 
        Vector<Integer> v = new Vector<>(); 
        for (int i=1; i<=Math.sqrt(n); i++) 
        { 
            if (n%i==0) 
            { 
                if (n/i == i) // check if divisors are equal 
                    System.out.printf("%d ", i); 
                else
                { 
                     System.out.printf("%d ", i); 
       
                    // push the second divisor in the vector 
                    v.add(n/i); 
                } 
            } 
        } 
       
        // The vector will be printed in reverse 
        for (int i=v.size()-1; i>=0; i--) 
             System.out.printf("%d ", v.get(i)); 
    } 
  
    // Driver method 
    public static void main(String args[]) 
    { 
        System.out.println("The divisors of 100 are: "); 
        printDivisors(100); 
    } 
} 

Python3

# A O(sqrt(n)) java program that prints  
# all divisors in sorted order 
import math  
  
# Method to print the divisors 
def printDivisors(n) : 
    list = []  
      
    # List to store half of the divisors 
    for i in range(1, int(math.sqrt(n) + 1)) : 
          
        if (n % i == 0) : 
              
            # Check if divisors are equal 
            if (n / i == i) : 
                print (i, end=" ") 
            else : 
                # Otherwise print both 
                print (i, end=" ") 
                list.append(int(n / i)) 
                  
    # The list will be printed in reverse     
    for i in list[::-1] : 
        print (i, end=" ") 
          
# Driver method 
print ("The divisors of 100 are: ") 
printDivisors(100) 
  
#This code is contributed by Gitanjali 

C#

// A O(sqrt(n)) C# program that  
// prints all divisors in sorted order 
using System; 
  
class GFG 
{ 
// method to print the divisors 
static void printDivisors(int n) 
{ 
    // Vector to store half 
    // of the divisors 
    int[] v = new int[n]; 
    int t = 0; 
    for (int i = 1;  
             i <= Math.Sqrt(n); i++) 
    { 
        if (n % i == 0) 
        { 
            // check if divisors are equal 
            if (n / i == i)  
                Console.Write(i + " "); 
            else
            { 
                Console.Write(i + " "); 
  
                // push the second divisor 
                // in the vector 
                v[t++] = n / i; 
            } 
        } 
    } 
  
    // The vector will be 
    // printed in reverse 
    for (int i = t - 1; i >= 0; i--) 
        Console.Write(v[i] + " "); 
} 
  
// Driver Code 
public static void Main() 
{ 
    Console.Write("The divisors of 100 are: \n"); 
    printDivisors(100); 
} 
} 
  
// This code is contributed 
// by ChitraNayal 

PHP

<?php 
// A O(sqrt(n)) program that  
// prints all divisors in 
// sorted order 
  
// function to print 
// the divisors 
function printDivisors($n) 
{ 
    // Vector to store half 
    // of the divisors 
    $v; 
    $t = 0; 
    for ($i = 1;  
         $i <= (int)sqrt($n); $i++) 
    { 
        if ($n % $i == 0) 
        { 
            // check if divisors are equal 
            if ((int)$n / $i == $i)  
                echo $i . " "; 
            else
            { 
                echo $i . " "; 
  
                // push the second  
                // divisor in the vector 
                $v[$t++] = (int)$n / $i; 
            } 
        } 
    } 
  
    // The vector will be 
    // printed in reverse 
    for ($i = count($v) - 1;  
         $i >= 0; $i--) 
        echo $v[$i] . " "; 
} 
  
// Driver code 
echo "The divisors of 100 are: \n"; 
printDivisors(100); 
  
// This code is contributed by mits 
?> 

Output :

The divisors of 100 are: 
1 2 4 5 10 20 25 50 100

Time Complexity : O(sqrt(n))
Auxiliary Space : O(sqrt(n))

This article is contributed by Ashutosh Kumar. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details

My Personal Notes

Find all divisors of a natural number | Set 2

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

Recommended Posts: