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Find all divisors of N2 using N

  • Difficulty Level : Medium
  • Last Updated : 03 Jun, 2021

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

Examples: 

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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 the 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
     

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 these generated divisors to calculate divisors of 42 that are 1, 2, 4, 8, and 16.
Below is the implementation of the above approach. 

C++




// 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;
}

Java




// Java code to print all
// divisors of N*N using N
import java.util.*;
class GFG
{
 
  // Vector v to store divisors of n
  static  List<Integer> v = new ArrayList<>();
 
  // Map to avoid repeated divisors
  static HashMap<Integer, Boolean> marked = new HashMap<>();
 
  // Function to find Divisor of N
  static void DivisorOfN(int n)
  {
 
    // sqrt(N) approach
    // to find divisors of N
    for (int i = 1; i <= Math.sqrt(n); i++)
    {
 
      if (n % i == 0)
      {
        if (n / i == i)
        {
          v.add(i);
          marked.put(i, true);
        }
        else
        {
          v.add(i);
          v.add(n / i);
          marked.put(i, true); 
          marked.put(n / i, true);
        }
      }
    }
  }
 
  // Function to print all divisor of N*N
  static void PrintDivisors(int n)
  {
 
    // Store all divisor of n
    DivisorOfN(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.get(i) * v.get(j);
 
        // Checking if element is
        // already present
        if (!marked.containsKey(check))
        {
          v.add(v.get(i) * v.get(j));
 
          // marking element true
          // after adding in vector
          marked.put(v.get(i) * v.get(j), true);
        }
      }
    }
 
    Collections.sort(v);      
    System.out.print("Divisors of " + n * n + " are: ");
    for (int i = 0; i < v.size(); i++)
    {
      System.out.print(v.get(i) + " ");
    }
    System.out.println();
    v.clear();
    marked.clear();
  }
 
  // Driver code
  public static void main(String[] args)
  {
    PrintDivisors(4);
    PrintDivisors(8);
    PrintDivisors(10);
  }
}
 
// This code is contributed by divyeshrabadiya07

Python3




# 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

C#




// C# code to print all
// divisors of N*N using N
using System;
using System.Collections.Generic;
class GFG {
     
    // Vector v to store divisors of n
    static List<int> v = new List<int>();
   
    // Map to avoid repeated divisors
    static Dictionary<int, bool> marked = new Dictionary<int, bool>();
     
    // Function to find Divisor of N
    static void DivisorOfN(int n)
    {
        // sqrt(N) approach
        // to find divisors of N
        for (int i = 1; i <= Math.Sqrt(n); i++)
        {
       
            if (n % i == 0)
            {
                if (n / i == i)
                {
                    v.Add(i);
                    marked[i] = true;
                }
                else
                {
                    v.Add(i);
                    v.Add(n / i);
                    marked[i] = true;
                    marked[n / i] = true;
                }
            }
        }
    }
 
    // Function to print all divisor of N*N
    static void PrintDivisors(int n)
    {
         
        // Store all divisor of n
        DivisorOfN(n);
       
        int size = v.Count;
       
        // 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.ContainsKey(check))
                {
                    v.Add(v[i] * v[j]);
       
                    // marking element true
                    // after adding in vector
                    marked[v[i] * v[j]] = true;
                }
            }
        }
       
        v.Sort();
       
        Console.Write("Divisors of " + n * n + " are: ");
        for (int i = 0; i < v.Count; i++)
        {
            Console.Write(v[i] + " ");
        }
        Console.WriteLine();
         
        v.Clear();
        marked.Clear();
    }
 
  // Driver code
  static void Main()
  {
    PrintDivisors(4);
    PrintDivisors(8);
    PrintDivisors(10);
  }
}
 
// This code is contributed by divyesh072019

Javascript




<script>
 
// Javascript code to print all
// divisors of N*N using N
 
// Function to find Divisor of N
function DivisorOfN(v, marked, n)
{
    // sqrt(N) approach
    // to find divisors of N
    for (var i = 1; i <= Math.sqrt(n); i++) {
 
        if (n % i == 0) {
            if (n / i == i) {
                v.push(i);
                marked[i] = true;
            }
            else {
                v.push(i);
                v.push(n / i);
                marked[i] = true;
                marked[n / i] = true;
            }
        }
    }
}
 
// Function to print all divisor of N*N
function PrintDivisors(n)
{
    // Vector v to store divisors of n
    var v = [];
 
    // Map to avoid repeated divisors
    var marked = new Map();
 
    // Store all divisor of n
    DivisorOfN(v, marked, n);
 
    var size = v.length;
 
    // Iterating over vector v
    // to generate divisors of N*N
    for (var i = 0; i < size; i++) {
        for (var j = i; j < size; j++) {
            var check = v[i] * v[j];
 
            // Checking if element is
            // already present
            if (marked[check] != true) {
                v.push(v[i] * v[j]);
 
                // marking element true
                // after adding in vector
                marked[v[i] * v[j]] = true;
            }
        }
    }
    v.sort((a,b)=>a-b)
 
    document.write("Divisors of "+n*n+" are: ");
    for (var i = 0; i < v.length; i++) {
        document.write(v[i]+ " ");
    }
    document.write("<br>");
}
 
// Driver Code
PrintDivisors(4);
PrintDivisors(8);
PrintDivisors(10);
 
</script>
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 a number of divisors of N.
Note: How is this approach different from Find all divisors of a natural number?
Let N = 5, therefore we need to find all divisors of 25.
 

  • Using the 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 the approach used in this article: we will find a 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.
     

 




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