Find the k-th smallest divisor of a natural number N

You’re given a number N and a number K. Our task is to find the kth smallest divisor of N.

Examples:

Input : N = 12, K = 5
Output : 6
The divisors of 12 after sorting are 1, 2, 3, 4, 6 and 12. 
Where the value of 5th divisor is equal to 6.

Input : N = 16, K 2
Output : 2

Simple Approach: A simple approach is to run a loop from 1 to √N and find all factors of N and push them into a vector. Finally, sort the vector and print the K-th value from the vector.



Note: Elements in the vector will not be sorted initially as we are pushing both factors (i) and (n/i). That is why it is needed to sort the vector before printing the K-th factor.

Below is the implementation of above approach :

C++

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// C++ program to find K-th smallest factor
  
#include <bits/stdc++.h>
using namespace std;
  
// function to find the k'th divisor
void findkth(int n, int k)
{
    // initialize a vector v
    vector<long long> v;
  
    // store all the divisors
    // so the loop will needs to run till sqrt ( n )
    for (int i = 1; i <= sqrt(n); i++) {
        if (n % i == 0) {
            v.push_back(i);
            if (i != sqrt(n))
                v.push_back(n / i);
        }
    }
  
    // sort the vector in an increasing order
    sort(v.begin(), v.end());
  
    // if k is greater than the size of vector
    // then no divisor can be possible
    if (k > v.size())
        cout << "Doesn't Exist";
    // else print the ( k - 1 )th value of vector
    else
        cout << v[k - 1];
}
  
// Driver code
int main()
{
    int n = 15, k = 2;
  
    findkth(n, k);
  
    return 0;
}

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Python3

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# Python3 program to find K-th smallest factor
from math import sqrt
  
# function to find the k'th divisor
def findkth(n, k):
      
    # initialize a vector v
    v = []
  
    # store all the divisors so the loop 
    # will needs to run till sqrt ( n )
    p = int(sqrt(n)) + 1
    for i in range(1, p, 1):
        if (n % i == 0):
            v.append(i)
            if (i != sqrt(n)):
                v.append(n / i);
          
    # sort the vector in an increasing order
    v.sort(reverse = False)
  
    # if k is greater than the size of vector
    # then no divisor can be possible
    if (k > len(v)):
        print("Doesn't Exist")
          
    # else print the (k - 1)th 
    # value of vector
    else:
        print(v[k - 1])
  
# Driver code
if __name__ == '__main__':
    n = 15
    k = 2
  
    findkth(n, k)
  
# This code is contributed by
# Surendra_Gangwar

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Output:

3

Time Complexity: √N log( √N )

Efficient Approach: An efficient approach will be to store the factors in two separate vectors. That is, factors i will be stored in a separate vector and N/i will be stored in a separate vector for all i from 1 to √N.
Now, if observed carefully, it can be seen that the first vector is already sorted in increasing order and the second vector is sorted in decreasing order. So, reverse the second vector and print the K-th element from either of the vectors in which it is lying.

Below is the implementation of the above approach:

C++

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// C++ program to find the K-th smallest factor
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the k'th divisor
void findkth ( int n, int k)
{
    // initialize vectors v1 and v2
    vector <int> v1;
    vector <int> v2;
      
    // store all the divisors in the two vectors
    // accordingly
    for( int i = 1 ; i <= sqrt( n ); i++ )
    {
        if ( n % i == 0 ) 
        {
            v1.push_back ( i );
              
            if ( i != sqrt ( n ) )
                v2.push_back ( n / i );
        }
    }
      
    // reverse the vector v2 to sort it
    // in increasing order
    reverse(v2.begin(), v2.end());
      
    // if k is greater than the size of vectors 
    // then no divisor can be possible
    if ( k > (v1.size() + v2.size()))
        cout << "Doesn't Exist" ;
    // else print the ( k - 1 )th value of vector 
    else
    {
        // If K is lying in first vector
        if(k <= v1.size())
            cout<<v1[k-1];
        // If K is lying in second vector
        else
            cout<<v2[k-v1.size()-1];
    }
}
  
// Driver code
int main()
{
    int n = 15, k = 2;
      
    findkth ( n, k) ;
  
    return 0;

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Python3

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# Python3 program to find the K-th 
# smallest factor
import math as mt
  
# Function to find the k'th divisor
def findkth (n, k):
  
    # initialize vectors v1 and v2
    v1 = list()
    v2 = list()
      
    # store all the divisors in the 
    # two vectors accordingly
    for i in range(1, mt.ceil(n**(.5))):
      
        if (n % i == 0):
          
            v1.append(i)
              
            if (i != mt.ceil(mt.sqrt(n))):
                v2.append(n // i)
          
    # reverse the vector v2 to sort it
    # in increasing order
    v2[::-1]
      
    # if k is greater than the size of vectors 
    # then no divisor can be possible
    if ( k > (len(v1) + len(v2))):
        print("Doesn't Exist", end = "")
          
    # else print the ( k - 1 )th value of vector 
    else:
      
        # If K is lying in first vector
        if(k <= len(v1)):
            print(v1[k - 1])
              
        # If K is lying in second vector
        else:
            print(v2[k - len(v1) - 1])
      
# Driver code
n = 15
k = 2
findkth (n, k) 
  
# This code is contributed by Mohit kumar

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Output:

3

Time Complexity: √N



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