Check if the door is open or closed

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:

Approach to solve this problem is to observe that a door is toggled once for every divisor it has. If the number of divisors is odd, the door will end up open, otherwise it will be closed. Using this observation, we can iterate through each door and count its divisors. If the count is odd, the door is open, otherwise it’s closed.

• Initialize a variable divisors to 0 to count the number of divisors of the door number.
• Loop through the divisors of the door number from 1 to the door number itself and for each divisor.
• Check if the divisor is a factor of the door number by checking if the remainder of dividing the door number by the divisor is zero.
• If the divisor is a factor, increment the divisors variable.
• Check if the number of divisors is even by checking if the remainder of dividing the divisors variable by 2 is zero.
• If the number of divisors is even, print “closed” for the door.
• If the number of divisors is odd, print “open” for the door.

Below is the implication of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;

void printStatusOfDoors(int n) {
    for (int i = 1; i <= n; i++) {
        int divisors = 0;
        for (int j = 1; j <= i; j++) {
            if (i % j == 0) {
                divisors++;
            }
        }
        if (divisors % 2 == 0) {
            cout << "closed ";
        } else {
            cout << "open ";
        }
    }
}

int main() {
    int n = 5;
    printStatusOfDoors(n);
    return 0;
}

Java

public class Main {
    public static void printStatusOfDoors(int n) {
        for (int i = 1; i <= n; i++) {
            int divisors = 0;
            for (int j = 1; j <= i; j++) {
                if (i % j == 0) {
                    divisors++;
                }
            }
            if (divisors % 2 == 0) {
                System.out.print("closed ");
            } else {
                System.out.print("open ");
            }
        }
    }

    public static void main(String[] args) {
        int n = 5;
        printStatusOfDoors(n);
    }
}

Python3

def print_status_of_doors(n):
    for i in range(1, n + 1):
        divisors = 0
        for j in range(1, i + 1):
            if i % j == 0:
                divisors += 1
        if divisors % 2 == 0:
            print("closed", end=" ")
        else:
            print("open", end=" ")

n = 5
print_status_of_doors(n)

C#

using System;

class Program {
    static void PrintStatusOfDoors(int n) {
        for (int i = 1; i <= n; i++) {
            int divisors = 0;
            for (int j = 1; j <= i; j++) {
                if (i % j == 0) {
                    divisors++;
                }
            }
            if (divisors % 2 == 0) {
                Console.Write("closed ");
            } else {
                Console.Write("open ");
            }
        }
    }

    static void Main(string[] args) {
        int n = 5;
        PrintStatusOfDoors(n);
    }
}

Javascript

function printStatusOfDoors(n) {
    for (let i = 1; i <= n; i++) {
        let divisors = 0;
        for (let j = 1; j <= i; j++) {
            if (i % j === 0) {
                divisors++;
            }
        }
        if (divisors % 2 === 0) {
            process.stdout.write("closed ");
        } else {
            process.stdout.write("open ");
        }
    }
}

const n = 5;
printStatusOfDoors(n);

Output:

     open closed closed open closed 

Time Complexity: O(n^2), as there are two nested loops, one iterating over n elements and the other iterating over i elements.

Space Complexity: O(1), as we are not using any extra space.

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

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>

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@
?>

Output

open closed closed open closed 

Time complexity : O(nlogn)
Auxiliary Space: O(1)

References: Asked in an interview of TCS