Find all Factors of Large Perfect Square Natural Number in O(sqrt(sqrt(N))

Given a perfect square natural number N. The task is to find all the factors of N.

Examples

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

Input: N = 900 
Output: 1 2 4 3 6 12 9 18 36 5 10 20 15 30 60 45 90 180 25 50 100 75 150 300 225 450 900 

Approach:

1. Find the square root of N in temp.
2. Find all the prime factors of temp in O(sqrt(temp)) using the approach discussed in this article.
3. Initialise an array factor[] with element 1 in it.
4. Store all the prime factors of temp obtained in above step twice in an array factor[].
5. Initialise a matrix M such that for every element in factor[] starting from index 1:
   • If factor[i] is equals to factor[i-1], then store factor[i]*factor[i-1] in matrix M in row i – 1 . 
     
   • Else factor[i] is not equals to factor[i-1], then store factor[i]*factor[i-1] in matrix M in row i.
6. Initialise two arrays arr1[] and arr2[] with the element 1 in both the array.
7. Iterate over every row of matrix M such that the product of every element in arr1[] with every element of current row must be stored in arr2[].
8. After above step, copy every element of arr2[] in arr1[].
9. Repeat above two steps, till all the element of matrixM is traverse.
10. The array arr2[] contains all the factors of number N.

Below is the implementation of the above approach:  

C++

// C++ program to find the factors
// of large perfect square number
// in O(sqrt(sqrt(N))) time
#include "bits/stdc++.h"
using namespace std;

int MAX = 100000;

// Function that find all the prime
// factors of N
void findFactors(int N)
{
    // Store the sqrt(N) in temp
    int temp = sqrt(N);

    // Initialise factor array with
    // 1 as a factor in it
    int factor[MAX] = { 1 };
    int i, j, k;
    int len1 = 1;

    // Check divisibility by 2
    while (temp % 2 == 0) {

        // Store the factors twice
        factor[len1++] = 2;
        factor[len1++] = 2;

        temp /= 2;
    }

    // Check for other prime
    // factors other than 2
    for (j = 3; j < sqrt(temp); j += 2) {

        // If j is a prime factor
        while (temp % j == 0) {

            // Store the prime
            // factor twice
            factor[len1++] = j;
            factor[len1++] = j;
            temp /= j;
        }
    }

    // If j is prime number left
    // other than 2
    if (temp > 2) {

        // Store j twice
        factor[len1++] = temp;
        factor[len1++] = temp;
    }

    // Initialise Matrix M to
    // to store all the factors
    int M[len1][MAX] = { 0 };

    // tpc for rows
    // tpr for column
    int tpc = 0, tpr = 0;

    // Initialise M[0][0] = 1 as
    // it also factor of N
    M[0][0] = 1;
    j = 1;

    // Traversing factor array
    while (j < len1) {

        // If current and previous
        // factors are not same then
        // move to next row and
        // insert the current factor
        if (factor[j] != factor[j - 1]) {
            tpr++;
            M[tpr][0] = factor[j];
            j++;
            tpc = 1;
        }

        // If current and previous
        // factors are same then,
        // Insert the factor with
        // previous factor inserted
        // in matrix M
        else {
            M[tpr][tpc]
                = M[tpr][tpc - 1] * factor[j];
            j++;
            tpc++;
        }
    }

    // The arr1[] and arr2[] used to
    // store all the factors of N
    int arr1[MAX], arr2[MAX];
    int l1, l2;
    l1 = l2 = 1;

    // Initialise arrays as 1
    arr1[0] = arr2[0] = 1;

    // Traversing the matrix M
    for (i = 1; i < tpr + 1; i++) {

        // Traversing till column
        // element doesn't become 0
        for (j = 0; M[i][j] != 0; j++) {

            // Store the product of
            // every element of current
            // row with every element
            // in arr1[]
            for (k = 0; k < l1; k++) {
                arr2[l2++]
                    = arr1[k] * M[i][j];
            }
        }

        // Copying every element of
        // arr2[] in arr1[]
        for (j = l1; j < l2; j++) {
            arr1[j] = arr2[j];
        }

        // length of arr2[] and arr1[]
        // are equal after copying
        l1 = l2;
    }

    // Print all the factors
    for (i = 0; i < l2; i++) {
        cout << arr2[i] << ' ';
    }
}

// Drivers Code
int main()
{
    int N = 900;
    findFactors(N);
    return 0;
}

Java

// Java program to find the factors
// of large perfect square number
// in O(Math.sqrt(Math.sqrt(N))) time
import java.util.*;

class GFG{

static int MAX = 100000;

// Function that find all the prime
// factors of N
static void findFactors(int N)
{
    // Store the Math.sqrt(N) in temp
    int temp = (int) Math.sqrt(N);

    // Initialise factor array with
    // 1 as a factor in it
    int []factor = new int[MAX];
    Arrays.fill(factor, 1);
    int i, j, k;
    int len1 = 1;

    // Check divisibility by 2
    while (temp % 2 == 0) {

        // Store the factors twice
        factor[len1++] = 2;
        factor[len1++] = 2;

        temp /= 2;
    }

    // Check for other prime
    // factors other than 2
    for (j = 3; j < Math.sqrt(temp); j += 2) {

        // If j is a prime factor
        while (temp % j == 0) {

            // Store the prime
            // factor twice
            factor[len1++] = j;
            factor[len1++] = j;
            temp /= j;
        }
    }

    // If j is prime number left
    // other than 2
    if (temp > 2) {

        // Store j twice
        factor[len1++] = temp;
        factor[len1++] = temp;
    }

    // Initialise Matrix M to
    // to store all the factors
    int [][]M = new int[len1][MAX];

    // tpc for rows
    // tpr for column
    int tpc = 0, tpr = 0;

    // Initialise M[0][0] = 1 as
    // it also factor of N
    M[0][0] = 1;
    j = 1;

    // Traversing factor array
    while (j < len1) {

        // If current and previous
        // factors are not same then
        // move to next row and
        // insert the current factor
        if (factor[j] != factor[j - 1]) {
            tpr++;
            M[tpr][0] = factor[j];
            j++;
            tpc = 1;
        }

        // If current and previous
        // factors are same then,
        // Insert the factor with
        // previous factor inserted
        // in matrix M
        else {
            M[tpr][tpc]
                = M[tpr][tpc - 1] * factor[j];
            j++;
            tpc++;
        }
    }

    // The arr1[] and arr2[] used to
    // store all the factors of N
    int []arr1 = new int[MAX];
    int []arr2 = new int[MAX];
    int l1, l2;
    l1 = l2 = 1;

    // Initialise arrays as 1
    arr1[0] = arr2[0] = 1;

    // Traversing the matrix M
    for (i = 1; i < tpr + 1; i++) {

        // Traversing till column
        // element doesn't become 0
        for (j = 0; M[i][j] != 0; j++) {

            // Store the product of
            // every element of current
            // row with every element
            // in arr1[]
            for (k = 0; k < l1; k++) {
                arr2[l2++]
                    = arr1[k] * M[i][j];
            }
        }

        // Copying every element of
        // arr2[] in arr1[]
        for (j = l1; j < l2; j++) {
            arr1[j] = arr2[j];
        }

        // length of arr2[] and arr1[]
        // are equal after copying
        l1 = l2;
    }

    // Print all the factors
    for (i = 0; i < l2; i++) {
        System.out.print(arr2[i] + " ");
    }
}

// Drivers Code
public static void main(String[] args)
{
    int N = 900;
    findFactors(N);
}
}

// This code is contributed by sapnasingh4991

Python3

# Python 3 program to find the factors
# of large perfect square number
# in O(sqrt(sqrt(N))) time

import math

MAX = 100000

# Function that find all the prime
# factors of N
def findFactors( N):

    # Store the sqrt(N) in temp
    temp = int(math.sqrt(N))

    # Initialise factor array with
    # 1 as a factor in it
    factor = [1]*MAX
    len1 = 1

    # Check divisibility by 2
    while (temp % 2 == 0) :

        # Store the factors twice
        factor[len1] = 2
        len1 += 1
        factor[len1] = 2
        len1 += 1
        temp //= 2

    # Check for other prime
    # factors other than 2
    sqt = math.sqrt(temp)

    for j in range(3, math.ceil(sqt), 2):

        # If j is a prime factor
        while (temp % j == 0):

            # Store the prime
            # factor twice
            factor[len1] = j
            len1 += 1
            factor[len1] = j
            len1 += 1
            temp //= j

    # If j is prime number left
    # other than 2
    if (temp > 2) :

        # Store j twice
        factor[len1] = temp
        len1 += 1
        factor[len1] = temp
        len1 += 1

    # Initialise Matrix M to
    # to store all the factors
    M = [ [ 0 for x in range(MAX)] for y in range(len1)]

    # tpc for rows
    # tpr for column
    tpc , tpr = 0 , 0

    # Initialise M[0][0] = 1 as
    # it also factor of N
    M[0][0] = 1
    j = 1

    # Traversing factor array
    while (j < len1):

        # If current and previous
        # factors are not same then
        # move to next row and
        # insert the current factor
        if (factor[j] != factor[j - 1]):
            tpr+=1
            M[tpr][0] = factor[j]
            j += 1
            tpc = 1

        # If current and previous
        # factors are same then,
        # Insert the factor with
        # previous factor inserted
        # in matrix M
        else :
            M[tpr][tpc]= M[tpr][tpc - 1] * factor[j]
            j += 1
            tpc += 1

    # The arr1[] and arr2[] used to
    # store all the factors of N
    arr1 = [0]*MAX
    arr2 = [0]*MAX
    l1 = l2 = 1

    # Initialise arrays as 1
    arr1[0] = 1
    arr2[0] = 1

    # Traversing the matrix M
    # print("tpr ",tpr)
    for i in range(1 , tpr + 1) :

        # Traversing till column
        # element doesn't become 0
        j = 0
        while M[i][j] != 0:

            # Store the product of
            # every element of current
            # row with every element
            # in arr1[]
            for k in range(l1):
                arr2[l2]= arr1[k] * M[i][j]
                l2 += 1

            j += 1

        # Copying every element of
        # arr2[] in arr1[]
        for j in range(l1, l2):
            arr1[j] = arr2[j]

        # length of arr2[] and arr1[]
        # are equal after copying
        l1 = l2

    # Print all the factors
    for i in range(l2):
        print(arr2[i] ,end= " ")

# Drivers Code
if __name__ == "__main__":

    N = 900
    findFactors(N)

# This code is contributed by chitranayal

C#

// C# program to find the factors
// of large perfect square number
// in O(Math.Sqrt(Math.Sqrt(N))) time
using System;

class GFG{

static int MAX = 100000;

// Function that find all the prime
// factors of N
static void findFactors(int N)
{
    // Store the Math.Sqrt(N) in temp
    int temp = (int) Math.Sqrt(N);

    // Initialise factor array with
    // 1 as a factor in it
    int []factor = new int[MAX];
    for(int l= 0; l < MAX; l++)
        factor[l] = 1;
    int i, j, k;
    int len1 = 1;

    // Check divisibility by 2
    while (temp % 2 == 0) {

        // Store the factors twice
        factor[len1++] = 2;
        factor[len1++] = 2;

        temp /= 2;
    }

    // Check for other prime
    // factors other than 2
    for (j = 3; j < Math.Sqrt(temp); j += 2) {

        // If j is a prime factor
        while (temp % j == 0) {

            // Store the prime
            // factor twice
            factor[len1++] = j;
            factor[len1++] = j;
            temp /= j;
        }
    }

    // If j is prime number left
    // other than 2
    if (temp > 2) {

        // Store j twice
        factor[len1++] = temp;
        factor[len1++] = temp;
    }

    // Initialise Matrix M to
    // to store all the factors
    int [,]M = new int[len1, MAX];

    // tpc for rows
    // tpr for column
    int tpc = 0, tpr = 0;

    // Initialise M[0,0] = 1 as
    // it also factor of N
    M[0, 0] = 1;
    j = 1;

    // Traversing factor array
    while (j < len1) {

        // If current and previous
        // factors are not same then
        // move to next row and
        // insert the current factor
        if (factor[j] != factor[j - 1]) {
            tpr++;
            M[tpr, 0] = factor[j];
            j++;
            tpc = 1;
        }

        // If current and previous
        // factors are same then,
        // Insert the factor with
        // previous factor inserted
        // in matrix M
        else {
            M[tpr,tpc]
                = M[tpr,tpc - 1] * factor[j];
            j++;
            tpc++;
        }
    }

    // The arr1[] and arr2[] used to
    // store all the factors of N
    int []arr1 = new int[MAX];
    int []arr2 = new int[MAX];
    int l1, l2;
    l1 = l2 = 1;

    // Initialise arrays as 1
    arr1[0] = arr2[0] = 1;

    // Traversing the matrix M
    for (i = 1; i < tpr + 1; i++) {

        // Traversing till column
        // element doesn't become 0
        for (j = 0; M[i, j] != 0; j++) {

            // Store the product of
            // every element of current
            // row with every element
            // in arr1[]
            for (k = 0; k < l1; k++) {
                arr2[l2++]
                    = arr1[k] * M[i, j];
            }
        }

        // Copying every element of
        // arr2[] in arr1[]
        for (j = l1; j < l2; j++) {
            arr1[j] = arr2[j];
        }

        // length of arr2[] and arr1[]
        // are equal after copying
        l1 = l2;
    }

    // Print all the factors
    for (i = 0; i < l2; i++) {
        Console.Write(arr2[i] + " ");
    }
}

// Drivers Code
public static void Main(String[] args)
{
    int N = 900;
    findFactors(N);
}
}

// This code is contributed by sapnasingh4991

Javascript

<script>

// Javascript program to find the factors
// of large perfect square number
// in O(Math.sqrt(Math.sqrt(N))) time
let MAX = 100000;

// Function that find all the prime
// factors of N
function findFactors(N)
{

    // Store the Math.sqrt(N) in temp
    let temp = Math.floor(Math.sqrt(N));

    // Initialise factor array with
    // 1 as a factor in it
    let factor = new Array(MAX);
    for(let i = 0; i < MAX; i++)
    {
        factor[i] = 1;
    }
    let i, j, k;
    let len1 = 1;

    // Check divisibility by 2
    while (temp % 2 == 0)
    {

        // Store the factors twice
        factor[len1++] = 2;
        factor[len1++] = 2;

        temp = Math.floor(temp / 2);
    }

    // Check for other prime
    // factors other than 2
    for(j = 3; j < Math.sqrt(temp); j += 2)
    {

        // If j is a prime factor
        while (temp % j == 0)
        {

            // Store the prime
            // factor twice
            factor[len1++] = j;
            factor[len1++] = j;
            temp = Math.floor(temp / j);
        }
    }

    // If j is prime number left
    // other than 2
    if (temp > 2)
    {

        // Store j twice
        factor[len1++] = temp;
        factor[len1++] = temp;
    }

    // Initialise Matrix M to
    // to store all the factors
    let M = new Array(len1);
    for(let i = 0; i < len1; i++)
    {
        M[i] = new Array(MAX);
        for(let j = 0; j < MAX; j++)
        {
            M[i][j] = 0;
        }
    }

    // tpc for rows
    // tpr for column
    let tpc = 0, tpr = 0;

    // Initialise M[0][0] = 1 as
    // it also factor of N
    M[0][0] = 1;
    j = 1;

    // Traversing factor array
    while (j < len1)
    {

        // If current and previous
        // factors are not same then
        // move to next row and
        // insert the current factor
        if (factor[j] != factor[j - 1])
        {
            tpr++;
            M[tpr][0] = factor[j];
            j++;
            tpc = 1;
        }

        // If current and previous
        // factors are same then,
        // Insert the factor with
        // previous factor inserted
        // in matrix M
        else
        {
            M[tpr][tpc] = M[tpr][tpc - 1] * factor[j];
            j++;
            tpc++;
        }
    }

    // The arr1[] and arr2[] used to
    // store all the factors of N
    let arr1 = new Array(MAX);
    let arr2 = new Array(MAX);
    for(let i = 0; i < MAX; i++)
    {
        arr1[i] = 0;
        arr2[i] = 0;
    }
    let l1, l2;
    l1 = l2 = 1;

    // Initialise arrays as 1
    arr1[0] = arr2[0] = 1;

    // Traversing the matrix M
    for(i = 1; i < tpr + 1; i++)
    {

        // Traversing till column
        // element doesn't become 0
        for(j = 0; M[i][j] != 0; j++)
        {

            // Store the product of
            // every element of current
            // row with every element
            // in arr1[]
            for(k = 0; k < l1; k++)
            {
                arr2[l2++] = arr1[k] * M[i][j];
            }
        }

        // Copying every element of
        // arr2[] in arr1[]
        for(j = l1; j < l2; j++)
        {
            arr1[j] = arr2[j];
        }

        // length of arr2[] and arr1[]
        // are equal after copying
        l1 = l2;
    }

    // Print all the factors
    for(i = 0; i < l2; i++)
    {
        document.write(arr2[i] + " ");
    }
}

// Driver Code
let N = 900;

findFactors(N);

// This code is contributed by avanitrachhadiya2155

</script>

Output

1 2 4 3 6 12 9 18 36 5 10 20 15 30 60 45 90 180 25 50 100 75 150 300 225 450 900 

Time Complexity: O(sqrt(sqrt(N)))

Auxiliary Space: O(MAX)

Method: “Naive Factorization Method”

1. Initialize an empty vector called factors to store the factors of N.
2. Loop through all the numbers from 1 to sqrt(N). For each number i, check if N is divisible by i (i.e., N % i == 0). If it is, then i is a factor of N.
3. Add i to the vector factors. If i is not equal to the quotient N/i (i.e., i != N/i), then add N/i to the vector factors as well. This is done to avoid adding duplicate factors to the vector.
4. Once all the factors have been added to the vector factors, print all the elements in the vector to get all the factors of N.
5. Since we loop only till sqrt(N), the time complexity of this algorithm is O(sqrt(sqrt(N))).
   

C++

#include <iostream>
#include <vector>
using namespace std;

void findFactors(int n) {
    vector<int> factors; // to store factors of n
    for (int i = 1; i*i <= n; i++) {
        if (n % i == 0) {
            factors.push_back(i);
            if (i != n/i) { // to avoid duplicates
                factors.push_back(n/i);
            }
        }
    }
    // print factors
    for (int i = 0; i < factors.size(); i++) {
        cout << factors[i] << " ";
    }
    cout << endl;
}

int main() {
    int N1 = 100, N2 = 900;
    cout << "Factors of " << N1 << " are: ";
    findFactors(N1);
    cout << "Factors of " << N2 << " are: ";
    findFactors(N2);
    return 0;
}

Java

import java.util.ArrayList;
import java.util.List;

class Main {
    static void findFactors(int n) {
        List<Integer> factors = new ArrayList<>(); // to store factors of n
        for (int i = 1; i * i <= n; i++) {
            if (n % i == 0) {
                factors.add(i);
                if (i != n / i) { // to avoid duplicates
                    factors.add(n / i);
                }
            }
        }
        // print factors
        for (int i = 0; i < factors.size(); i++) {
            System.out.print(factors.get(i) + " ");
        }
        System.out.println();
    }

    public static void main(String[] args) {
        int N1 = 100, N2 = 900;
        System.out.print("Factors of " + N1 + " are: ");
        findFactors(N1);
        System.out.print("Factors of " + N2 + " are: ");
        findFactors(N2);
    }
}

Python3

def find_factors(n):
    factors = []  # to store factors of n
    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            factors.append(i)
            if i != n // i:  # to avoid duplicates
                factors.append(n // i)
    # print factors
    for factor in factors:
        print(factor, end=' ')
    print()

N1, N2 = 100, 900
print(f"Factors of {N1} are: ", end='')
find_factors(N1)
print(f"Factors of {N2} are: ", end='')
find_factors(N2)

C#

using System;
using System.Collections.Generic;

class Program {
    static void Main()
    {
        int N1 = 100, N2 = 900;
        Console.Write("Factors of {0} are: ", N1);
        FindFactors(N1);
        Console.Write("Factors of {0} are: ", N2);
        FindFactors(N2);
    }

    static void FindFactors(int n)
    {
        List<int> factors
            = new List<int>(); // to store factors of n
        for (int i = 1; i * i <= n; i++) {
            if (n % i == 0) {
                factors.Add(i);
                if (i != n / i) // to avoid duplicates
                {
                    factors.Add(n / i);
                }
            }
        }
        // print factors
        foreach(int factor in factors)
        {
            Console.Write(factor + " ");
        }
        Console.WriteLine();
    }
}

Javascript

var findFactors = function(n) {
var factors = []; // to store factors of n
for (var i = 1; i * i <= n; i++) {
if (n % i == 0) {
factors.push(i);
if (i != n / i) { // to avoid duplicates
factors.push(n / i);
}
}
}
// print factors
for (var i = 0; i < factors.length; i++) {
console.log(factors[i] + " ");
}
console.log();
}

var N1 = 100, N2 = 900;
console.log("Factors of " + N1 + " are: ");
findFactors(N1);
console.log("Factors of " + N2 + " are: ");
findFactors(N2);
// This code is contributed by sarojmcy2e

Output

Factors of 100 are: 1 100 2 50 4 25 5 20 10 
Factors of 900 are: 1 900 2 450 3 300 4 225 5 180 6 150 9 100 10 90 12 75 15 60 18 50 20 45 25 36 30 

The time complexity of the given approach is O(sqrt(sqrt(N)))

The auxiliary space used by the algorithm is O(sqrt(sqrt(N))