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 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

Javascript

<script>

// Javascript Program for the
// above approach

// Function to print all
// possible values of
// floor(N/K)
function allQuotients(N)
{
    var s = new Set();

    // loop from 1 to N+1
    for (var k = 1; k <= N + 1; k++) {
        s.add(parseInt(N / k));
    }

    var ls = Array.from(s).reverse();
    ls.forEach(v => document.write(v+ " "))
}

var N = 5;
allQuotients(N);


</script>

Output:

0 1 2 5

Time Complexity: O(nlogn)

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

Javascript

<script>


// Javascript Program for the
// above approach

// Function to print all
// possible values of
// floor(N/K)
function allQuotients(N)
{
    var s = new Set();
    s.add(0);

    for (var k = 1; k <= parseInt(Math.sqrt(N)); k++) {
        s.add(k);
        s.add(parseInt(N / k));
    }
    var tmp = [...s];
    tmp.sort((a,b)=>a-b)

    tmp.forEach(it => {
        document.write( it + " ");
    });
}

var N = 5;
allQuotients(N);

// This code is contributed by noob2000.
</script>

Output:

0 1 2 5

Time Complexity: O(sqrt(n)*logn)

Auxiliary Space: O(n)