All possible values of floor(N/K) for all values of K

Given a function f(K) = floor(N/K) (N>0 and K>0), the task is to find all possible values of f(K) for a given N where K takes all values in the range [1, Inf].
Examples:

Input: N = 5
Output: 0 1 2 5
Explanation:
5 divide 1 = 5
5 divide 2 = 2
5 divide 3 = 1
5 divide 4 = 1
5 divide 5 = 1
5 divide 6 = 0
5 divide 7 = 0
So all possible distinct values of f(k) are {0, 1, 2, 5}.
Input: N = 11
Output: 0 1 2 3 5 11
Explanation:
11 divide 1 = 11
11 divide 2 = 5
11 divide 3 = 3
11 divide 4 = 2
11 divide 5 = 2
11 divide 6 = 1
11 divide 7 = 1
…
…
11 divided 11 = 1
11 divides 12 = 0
So all possible distinct values of f(k) are {0, 1, 2, 3, 5, 11}.

Naive Approach:
The simplest approach to the iterate over [1, N+1] and store in a set, all values of (N/i) ( 1 ? i ? N + 1) to avoid duplication.
Below is the implementation of the above approach:

C++

// C++ Program for the  
// above approach  
#include <bits/stdc++.h>  
using namespace std;  
  
// Function to print all  
// possible values of  
// floor(N/K)  
void allQuotients(int N)  
{  
    set<int> s;  
  
    // loop from 1 to N+1  
    for (int k = 1; k <= N + 1; k++) {  
        s.insert(N / k);  
    }  
  
    for (auto it : s)  
        cout << it << " ";  
}  
  
int main()  
{  
    int N = 5;  
    allQuotients(N);  
  
    return 0;  
}  

Java

// Java program for the above approach  
import java.util.*;  
  
class GFG{  
  
// Function to print all  
// possible values of  
// Math.floor(N/K)  
static void allQuotients(int N)  
{  
    HashSet<Integer> s = new HashSet<Integer>();  
  
    // loop from 1 to N+1  
    for(int k = 1; k <= N + 1; k++)  
    {  
        s.add(N / k);  
    }  
      
    for(int it : s)  
        System.out.print(it + " ");  
}  
  
// Driver code  
public static void main(String[] args)  
{  
    int N = 5;  
      
    allQuotients(N);  
}  
}  
  
// This code is contributed by Rajput-Ji  

Python3

# Python3 program for the above approach 
  
# Function to print all possible  
# values of floor(N/K) 
def allQuotients(N): 
  
    s = set() 
  
    # Iterate from 1 to N+1 
    for k in range(1, N + 2): 
        s.add(N // k) 
  
    for it in s: 
        print(it, end = ' ') 
  
# Driver code 
if __name__ == '__main__': 
  
    N = 5
      
    allQuotients(N) 
  
# This code is contributed by himanshu77 

C#

// C# program for the above approach 
using System; 
using System.Collections.Generic; 
  
class GFG{ 
      
// Function to print all possible 
// values of Math.floor(N/K)  
static void allQuotients(int N)  
{  
    SortedSet<int> s = new SortedSet<int>();  
      
    // Loop from 1 to N+1  
    for(int k = 1; k <= N + 1; k++)  
    {  
        s.Add(N / k);  
    }  
      
    foreach(int it in s)  
    {  
        Console.Write(it + " ");  
    } 
}  
  
// Driver code 
static void Main() 
{ 
    int N = 5;  
      
    allQuotients(N); 
} 
} 
  
// This code is contributed by divyeshrabadiya07 

Output:

0 1 2 5

Time Complexity: O(N)
Auxiliary Space: O(N)
Efficient Approach:
An optimized solution is to iterate over [1, ?N] and insert values K and (N/K) into the set.

C++

// C++ Program for the  
// above approach  
#include <bits/stdc++.h>  
using namespace std;  
  
// Function to print all  
// possible values of  
// floor(N/K)  
void allQuotients(int N)  
{  
    set<int> s;  
    s.insert(0);  
  
    for (int k = 1; k <= sqrt(N); k++) {  
        s.insert(k);  
        s.insert(N / k);  
    }  
  
    for (auto it : s)  
        cout << it << " ";  
}  
  
int main()  
{  
    int N = 5;  
    allQuotients(N);  
  
    return 0;  
}  

Java

// Java program for the above approach  
import java.util.*;  
class GFG{  
  
// Function to print all  
// possible values of  
// Math.floor(N/K)  
static void allQuotients(int N)  
{  
    HashSet<Integer> s = new HashSet<Integer>();  
    s.add(0);  
      
    // loop from 1 to N+1  
    for(int k = 1; k <= Math.sqrt(N); k++)  
    {  
        s.add(k);  
        s.add(N / k);  
    }  
      
    for(int it : s)  
        System.out.print(it + " ");  
}  
  
// Driver code  
public static void main(String[] args)  
{  
    int N = 5;  
      
    allQuotients(N);  
}  
}  
  
// This code is contributed by rock_cool  

Python3

# Python3 program for the above approach 
from math import *
  
# Function to print all possible  
# values of floor(N/K) 
def allQuotients(N): 
  
    s = set() 
    s.add(0) 
  
    for k in range(1, int(sqrt(N)) + 1): 
        s.add(k) 
        s.add(N // k) 
  
    for it in s: 
        print(it, end = ' ') 
  
# Driver code 
if __name__ == '__main__': 
  
    N = 5
      
    allQuotients(N) 
  
# This code is contributed by himanshu77 

C#

// C# program for the above approach 
using System; 
using System.Collections.Generic; 
  
class GFG{ 
      
// Function to print all possible 
// values of Math.floor(N/K)  
static void allQuotients(int N)  
{  
    SortedSet<int> s = new SortedSet<int>();  
    s.Add(0);  
      
    // loop from 1 to N+1  
    for(int k = 1; k <= Math.Sqrt(N); k++)  
    {  
        s.Add(k); 
        s.Add(N / k);  
    }  
      
    foreach(int it in s)  
    {  
        Console.Write(it + " ");  
    } 
}  
  
// Driver code     
static void Main()  
{ 
    int N = 5;  
      
    allQuotients(N); 
} 
} 
  
// This code is contributed by divyeshrabadiya07 

Output:

0 1 2 5

Time Complexity: O(?N)
Auxiliary Space: O(N)

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