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 100Input: 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:
- Find the square root of N in temp.
- Find all the prime factors of temp in O(sqrt(temp)) using the approach discussed in this article.
- Initialise an array factor[] with element 1 in it.
- Store all the prime factors of temp obtained in above step twice in an array factor[].
- 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.
- If factor[i] is equals to factor[i-1], then store factor[i]*factor[i-1] in matrix M in row i – 1 .
- Initialise two arrays arr1[] and arr2[] with the element 1 in both the array.
- 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[].
- After above step, copy every element of arr2[] in arr1[].
- Repeat above two steps, till all the element of matrixM is traverse.
- 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”
- Initialize an empty vector called factors to store the factors of N.
- 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.
- 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.
- 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.
- 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))
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