Find all divisors of N2 using N

Given a number N, the task is to print all distinct divisors of N2.

Examples:

Input: N = 4
Output: 1 2 4 8 16
Explanation:
N = 4, N2 = 16
Divisors of 16 are: 1 2 4 8 16

Input: N = 8
Output: 1 2 4 8 16 32 64

Naive Approach:
Find all divisors of a natural number using sqrt(N) approach. But this solution is not efficient as the time complexity would be O(N).



Efficient Approach:

  • We try to generate divisors of N2 from divisors of N using the sqrt(N) approach. As,
  • Divisors of N2 = All distinct numbers obtained from pairwise multiplication of divisors of N.

For example: If N = 4, to generate divisors of 42 = 16, we would first calculate the divisors of 4 = 1, 2, 4. Now we will iterate over this generated divisors to calculate divisors of 42 that are 1, 2, 4, 8, and 16.

Below is the implementation of the above approach.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ code to print all
// divisors of N*N using N
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find Divisor of N
void DivisorOfN(vector<int>& v,
                map<int, bool>& marked,
                int n)
{
    // sqrt(N) approach
    // to find divisors of N
    for (int i = 1; i <= sqrt(n); i++) {
  
        if (n % i == 0) {
            if (n / i == i) {
                v.push_back(i);
                marked[i] = true;
            }
            else {
                v.push_back(i);
                v.push_back(n / i);
                marked[i] = true;
                marked[n / i] = true;
            }
        }
    }
}
  
// Function to print all divisor of N*N
void PrintDivisors(int n)
{
    // Vector v to store divisors of n
    vector<int> v;
  
    // Map to avoid repeated divisors
    map<int, bool> marked;
  
    // Store all divisor of n
    DivisorOfN(v, marked, n);
  
    int size = v.size();
  
    // Iterating over vector v
    // to generate divisors of N*N
    for (int i = 0; i < size; i++) {
        for (int j = i; j < size; j++) {
            int check = v[i] * v[j];
  
            // Checking if element is
            // already present
            if (marked[check] != true) {
                v.push_back(v[i] * v[j]);
  
                // marking element true
                // after adding in vector
                marked[v[i] * v[j]] = true;
            }
        }
    }
  
    sort(v.begin(), v.end());
  
    printf("Divisors of %d are: ", n * n);
    for (int i = 0; i < v.size(); i++) {
        printf("%d ", v[i]);
    }
    printf("\n");
}
  
// Driver Code
int main()
{
    PrintDivisors(4);
    PrintDivisors(8);
    PrintDivisors(10);
  
    return 0;
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 code to print all
# divisors of N*N using 
from math import sqrt
  
# Function to find Divisor of N
def DivisorOfN(v, marked, n):
    # sqrt(N) approach
    # to find divisors of N
    for i in range(1,int(sqrt(n)) + 1, 1):
        if (n % i == 0):
            if (n // i == i):
                v.append(i)
                marked[i] = True
            else:
                v.append(i)
                v.append(n // i)
                marked[i] = True
                marked[n // i] = True
  
# Function to print all divisor of N*N
def PrintDivisors(n):
    # Vector v to store divisors of n
    v = []
  
    # Map to avoid repeated divisors
    marked = {i:False for i in range(1000)}
  
    # Store all divisor of n
    DivisorOfN(v, marked, n)
  
    size = len(v)
  
    # Iterating over vector v
    # to generate divisors of N*N
    for i in range(size):
        for j in range(i,size,1):
            check = v[i] * v[j]
  
            # Checking if element is
            # already present
            if (marked[check] != True):
                v.append(v[i] * v[j])
  
                # marking element true
                # after adding in vector
                marked[v[i] * v[j]] = True
  
    v.sort(reverse = False)
  
    print("Divisors of",n * n,"are: ",end = "")
    for i in range(len(v)):
        print(v[i],end = " ")
      
    print("\n",end = "")
  
# Driver Code
if __name__ == '__main__':
    PrintDivisors(4)
    PrintDivisors(8)
    PrintDivisors(10)
  
# This code is contributed by Bhupendra_Singh

chevron_right


Output:

Divisors of 16 are: 1 2 4 8 16 
Divisors of 64 are: 1 2 4 8 16 32 64 
Divisors of 100 are: 1 2 4 5 10 20 25 50 100

Time Complexity: O(sqrt(N) + a2) where, a is number of divisor of N.

Note: How this approach is different from Find all divisors of a natural number?

Let N = 5, therefore we need to find all divisors of 25.

  • Using approach used in Find all divisors of a natural number: we will iterate using i from 1 to sqrt(25) = 5 and check for i and n/i.
    Time Complexity: O(sqrt(25))
  • Using approach used in this article: we will find divisor of 5 by using the above-mentioned articles approach which will be done in sqrt(5) time complexity. Now for all divisor of 5 i.e. 1, 5 we will store this in an array and multiply it pairwise with the help of 2 loops { (1*1, 1*5, 5*1, 5*5) } and choose the unique ones i.e. 1, 5, 25. This will take a^2 time (where a is the number of the divisor of 5, which is 2 here)
    Time Complexity: O(sqrt(5) + 2^2)

This article only works better than the above-mentioned article when the number of divisors of the number is less.

competitive-programming-img




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : bgangwar59, Akanksha_Rai