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

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 an 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) {
 
            // check if divisors are equal
            if (n / i == i) 
                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) {
 
                // check if divisors are equal
                if (n / i == i) 
                    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))

A O(sqrt(n)) Time and O(1) Space Solution :

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)
{
    for (int i = 1; i*i < n; i++) {
        if (n % i == 0)
            printf("%d ", i);
    }
    for (int i = sqrt(n); i >= 1; i--) {
        if (n % i == 0)
            printf("%d ", n / 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)) program that prints all
// divisors in sorted order 
import java.lang.Math; 
 
class GFG{
     
// Function to print the divisors 
public static void printDivisors(int n) 
{ 
    for(int i = 1; i * i < n; i++)
    { 
        if (n % i == 0)
            System.out.print(i + " ");
    } 
    for(int i = (int)Math.sqrt(n);
            i >= 1; i--)
    { 
        if (n % i == 0) 
            System.out.print(n / i + " ");
    } 
} 
 
// Driver code
public static void main(String[] args) 
{
    System.out.println("The divisors of 100 are: ");
     
    printDivisors(100); 
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# A O(sqrt(n)) program that prints all divisors 
# in sorted order
from math import *
 
# Function to print the divisors
def printDivisors (n):
 
    i = 1
    while (i * i < n):
        if (n % i == 0):
            print(i, end = " ")
 
        i += 1
 
    for i in range(int(sqrt(n)), 0, -1):
        if (n % i == 0):
            print(n // i, end = " ")
 
# Driver Code
print("The divisors of 100 are: ")
 
printDivisors(100)
 
# This code is contributed by himanshu77

C#

// A O(sqrt(n)) program that prints
// all divisors in sorted order
using System;
using System.Collections; 
using System.Collections.Generic; 
 
class GFG{
 
// Function to print the divisors 
static void printDivisors(int n) 
{ 
    for(int i = 1; i * i < n; i++)
    { 
        if (n % i == 0) 
            Console.Write(i + " "); 
    } 
    for(int i = (int)Math.Sqrt(n); i >= 1; i--)
    { 
        if (n % i == 0) 
            Console.Write(n / i + " "); 
    } 
}  
  
// Driver code   
public static void Main(string []arg)
{
    Console.Write("The divisors of 100 are: \n"); 
     
    printDivisors(100); 
}
}
 
// This code is contributed by rutvik_56

Output:

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

Thanks to Mysterious Mind for suggesting above solution.

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.

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My Personal Notes

Find all divisors of a natural number | Set 2

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

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