Skip to content
Related Articles

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 <bits/stdc++.h>
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




<?php
// PHP implementation of
// doors open or closed
 
// Function to check whether
// 'n' has even number of
// factors or not
 
function hasEvenNumberOfFactors($n)
{
    $root_n = sqrt($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
function printStatusOfDoors($n)
{
    for ($i = 1; $i <= $n; $i++)
    {
        // If even number of factors
        // final status is closed
        if (hasEvenNumberOfFactors($i))
            echo "closed" ," ";
 
        // else odd number of factors
        // final status is open
        else
            echo "open" ," ";
    }
}
 
// Driver Code
$n = 5;
printStatusOfDoors($n);
 
// This code is contributed by ajit@
?>

Javascript




<script>
 
// JavaScript program for the above approach
 
    // Function to check whether 'n'
    // has even number of factors or not
    function hasEvenNumberOfFactors(n)
    {
        let 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
    function printStatusOfDoors(n)
    {
        for (let i = 1 ; i <= n; i++)
        {
            // If even number of factors
            // final status is closed
            if (hasEvenNumberOfFactors(i))
                document.write( "closed" + " ");
     
            // else odd number of factors
            // final status is open
            else
                document.write( "open" + " ");
        }
    }
     
// Driver Code
     
    let n = 5;
    printStatusOfDoors(n);
 
// This code is contributed by susmitakundugoaldanga.
</script>

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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

My Personal Notes arrow_drop_up
Recommended Articles
Page :