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:
 

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// 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; 
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// 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 
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# 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
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// 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
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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.
 

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// 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; 
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// 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 
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# 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
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// 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
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Output: 
0 1 2 5

 

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

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