# 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 ` `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 s = ``new` `HashSet(); `   `    ``// 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

 ``

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 ` `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 s = ``new` `HashSet(); ` `    ``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

 ``

Output:

`0 1 2 5`

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

Auxiliary Space: O(n)

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