Find all divisors of a natural number | Set 2

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 referring to the 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 them in 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
?>

Javascript

<script>

// 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
    let v = [];
    let t = 0;
    for (let i = 1;
         i <= parseInt(Math.sqrt(n)); i++)
    {
        if (n % i == 0)
        {
            // check if divisors are equal
            if (parseInt(n / i) == i)
                document.write(i + " ");
            else
            {
                document.write(i + " ");

                // push the second
                // divisor in the vector
                v[t++] = parseInt(n / i);
            }
        }
    }

    // The vector will be
    // printed in reverse
    for (let i = v.length - 1;
         i >= 0; i--){
         document.write(v[i] + " ");
    }
}

// Driver code
document.write("The divisors of 100 are: \n");
printDivisors(100);

// This code is contributed by _saurabh_jaiswal

</script>

Output

The divisors of 100 are: n1 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 <iostream>
#include <math.h>
using namespace std;

// Function to print the divisors
void printDivisors(int n)
{
    int i;
    for (i = 1; i * i < n; i++) {
        if (n % i == 0)
            cout<<i<<" ";
    }
    if (i - (n / i) == 1) {
        i--;
    }
    for (; i >= 1; i--) {
        if (n % i == 0)
            cout<<n / i<<" ";
    }
}

// Driver code
int main()
{
    cout << "The divisors of 100 are: \n";

    printDivisors(100);

    return 0;
}

// This code is contributed by siteshbiswal

C

// A O(sqrt(n)) program that prints all divisors
// in sorted order
#include <stdio.h>
#include <math.h>

// function to print the divisors
void printDivisors(int n)
{ int i;
    for ( i = 1; i*i < n; i++) {
        if (n % i == 0)
            printf("%d ", i);
    }
   if(i-(n/i)==1)
    {
      i--;
    }
    for (; 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)
{ int i;
    for( i = 1; i * i < n; i++)
    {
        if (n % i == 0)
            System.out.print(i + " ");
    }
    if(i-(n/i)==1)
    {
      i--;
    }
    for(; 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 Prakash Veer Singh Tomar.

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

Javascript

<script>

// A O(sqrt(n)) program that prints all divisors
// in sorted order

// function to print the divisors
function printDivisors(n)
{
    for (var i = 1; i*i < n; i++) {
        if (n % i == 0)
            document.write(i + " ");
    }
    for (var i = Math.sqrt(n); i >= 1; i--) {
        if (n % i == 0)
            document.write(" " + n / i);
    }
}

// Driver program to test above function

    document.write("The divisors of 100 are: \n");
    printDivisors(100);

// This code is contributed by simranarora5sos

</script>

Output

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

Thanks to Mysterious Mind for suggesting the above solution.

The if condition between the two loops is used when corner factors in loops condition have a difference of 1(example- factors of 30 (5,6)here,5 will be printed two times; to resolve that issue this step is required.

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.