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

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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)

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