Related Articles
Check if the door is open or closed
• Difficulty Level : Medium
• Last Updated : 23 Mar, 2021

Given n doors and n persons. The doors are numbered 1 to n and persons are given id’s numbered 1 to n. Each door can have only 2 status open and closed. Initially all the doors have status closed. Find the final status of all the doors if a person changes the current status of all the doors, i.e. if status open then change to status closed and vice versa, for which he is authorized. A person with id ‘i’ is authorized to change the status of door numbered ‘j’ if ‘j’ is a multiple of ‘i’.
Note:
– A person has to change the current status of all the doors for which he is authorized exactly once.
– There can be a situation that before a person changes the status of the door, another person who is also authorized for the same door changes the status of the door.
Example :

```Input : 3
Output : open closed closed```

Explanation : As n = 3, therefore there are
3 doors {1, 2, 3} and
3 persons with ids {1, 2, 3}
person with id = 1 can change the status of door 1, 2, 3
person with id = 2 can change the status of door 2
person with id = 3 can change the status of door 3
Current status of all doors: closed closed closed
Consider a sequence of events,

1. Person with id = 1 changes status of door 2
Current status of all doors: closed open closed
2. Person with id = 3 changes status of door 3
Current status of all doors: closed open open
3. Person with id = 1 changes status of door 1, 3
Current status of all doors: open open closed
4. Person with id = 2 changes status of door 2
Current status of all doors: open closed closed

Another Example :

```Input : 5
Output : open closed closed open closed

Note: Sequence of open/closed is displayed in
increasing door number ```

Approach: It is mathematical and logical approach. If we observe it properly, then we find that the final status of a door numbered i is open if ‘i’ has odd number of factors and status is closed if ‘i’ has even number of factors. It does not depend in which sequence the status of doors are changed. To find whether count of divisors of number is even or odd, we can see Check if count of divisors is even or odd post.

C++

 `// C++ implementation of``// doors open or closed``#include ``using` `namespace` `std;` `// Function to check whether 'n'``// has even number of factors or not``bool` `hasEvenNumberOfFactors(``int` `n)``{``    ``int` `root_n = ``sqrt``(n);` `    ``// if 'n' is a perfect square``    ``// it has odd number of factors``    ``if` `((root_n*root_n) == n)``        ``return` `false``;` `    ``// else 'n' has even``    ``// number of factors``    ``return` `true``;``}` `// Function to find and print``// status of each door``void` `printStatusOfDoors(``int` `n)``{``    ``for` `(``int` `i=1; i<=n; i++)``    ``{``        ``// If even number of factors``        ``// final status is closed``        ``if` `(hasEvenNumberOfFactors(i))``            ``cout << ``"closed"` `<< ``" "``;` `        ``// else odd number of factors``        ``// final status is open``        ``else``            ``cout << ``"open"` `<< ``" "``;``    ``}``}` `// Driver program``int` `main()``{``    ``int` `n = 5;``    ``printStatusOfDoors(n);``    ``return` `0;``}`

Java

 `// java implementation of``// doors open or closed``import` `java.io.*;` `class` `GFG {``    ` `    ``// Function to check whether 'n'``    ``// has even number of factors or not``    ``static` `boolean` `hasEvenNumberOfFactors(``int` `n)``    ``{``        ``double` `root_n = Math.sqrt(n);``    ` `        ``// if 'n' is a perfect square``        ``// it has odd number of factors``        ``if` `((root_n*root_n) == n)``            ``return` `false``;``    ` `        ``// else 'n' has even``        ``// number of factors``        ``return` `true``;``    ``}``    ` `    ``// Function to find and print``    ``// status of each door``    ``static` `void` `printStatusOfDoors(``int` `n)``    ``{``        ``for` `(``int` `i = ``1` `; i <= n; i++)``        ``{``            ``// If even number of factors``            ``// final status is closed``            ``if` `(hasEvenNumberOfFactors(i))``                ``System .out.print( ``"closed"` `+ ``" "``);``    ` `            ``// else odd number of factors``            ``// final status is open``            ``else``                ``System.out.print( ``"open"` `+ ``" "``);``        ``}``    ``}``    ` `    ``// Driver program``    ``public` `static` `void` `main (String[] args) {``        ``int` `n = ``5``;``        ``printStatusOfDoors(n);``        ` `    ``}``}` `// This article is contributed by vt_m`

Python3

 `# Python 3 implementation of``# doors open or closed``import` `math` `# Function to check whether``# 'n' has even number of``# factors or not``def` `hasEvenNumberOfFactors(n):` `    ``root_n ``=` `math.sqrt(n)` `    ``# if 'n' is a perfect square``    ``# it has odd number of factors``    ``if` `((root_n ``*` `root_n) ``=``=` `n):``        ``return` `False` `    ``# else 'n' has even``    ``# number of factors``    ``return` `True` `# Function to find and print``# status of each door``def` `printStatusOfDoors(n):` `    ``for` `i ``in` `range``(``1``, n ``+` `1``):``    ` `        ``# If even number of factors``        ``# final status is closed``        ``if` `(hasEvenNumberOfFactors(i) ``=``=` `True``):``            ``print``(``"closed"``, end ``=``" "``)` `        ``# else odd number of factors``        ``# final status is open``        ``else``:``            ``print``(``"open"``, end ``=``" "``)``    ` `# Driver program``n ``=` `5` `printStatusOfDoors(n)` `# This code is contributed by Smitha Dinesh Semwal`

C#

 `// C# implementation of``// doors open or closed``using` `System;` `class` `GFG``{` `// Function to check whether``// 'n' has even number of``// factors or not``static` `bool` `hasEvenNumberOfFactors(``int` `n)``{``    ``double` `root_n = Math.Sqrt(n);` `    ``// if 'n' is a perfect square``    ``// it has odd number of factors``    ``if` `((root_n * root_n) == n)``        ``return` `false``;` `    ``// else 'n' has even``    ``// number of factors``    ``return` `true``;``}` `// Function to find and print``// status of each door``static` `void` `printStatusOfDoors(``int` `n)``{``    ``for` `(``int` `i = 1 ; i <= n; i++)``    ``{``        ``// If even number of factors``        ``// final status is closed``        ``if` `(hasEvenNumberOfFactors(i))``            ``Console.Write(``"closed"` `+ ``" "``);` `        ``// else odd number of factors``        ``// final status is open``        ``else``            ``Console.Write(``"open"` `+ ``" "``);``    ``}``}` `// Driver Code``static` `public` `void` `Main ()``{``    ``int` `n = 5;``    ``printStatusOfDoors(n);``}``}` `// This Code is contributed by ajit`

PHP

 ``

Javascript

 ``

Output :

`open closed closed open closed`

Time complexity : O(n)
References: Asked in an interview of TCS
This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.