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:
- Initially all the doors have status closed.
- 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
The person with id 1 opens all doors : open open open
id 2 closes the second door : open closed open
id 3 closes the third door : open closed closedInput 4
Output open closed closed openclosed closed closed
Explanation
The person with id 1 opens all doors : open open open open
id 2 closes the 2nd and 4th doors : open closed open closed
id 3 closes the third door : open closed closed closed
id 4 opens the fourth door : open closed closed open
[Naive Approach] Using Nested Loop – O(n2) time and O(1) space
Each door is toggled once for every divisor of its number. Doors with an odd number of divisors remain open, and doors with an even number of divisors remain closed. Only perfect square doors (1, 4, 9, 16, 25, ...) have an odd number of divisors, so they remain open. Please note that, for every number, divisors appear in pairs. For example, 12 has divisors (1, 12), (3, 4) and (2, 6). For a perfect square, there is a pair that has same values and this makes the number of different divisors odd. For example for 16, divisors are (1, 16), (2, 8) and (4, 4)
#include <iostream>
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;
}
public class GfG{
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);
}
}
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)
using System;
class GfG{
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);
}
}
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
[Expected Approach] Finding Square Root – O(n log(n)) time and O(1) space
We can use Check if count of divisors is even or odd logic to solve the problem effectively.
#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;
}
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);
}
}
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)
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);
}
}
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))
console.log("closed" +
" ");
// else odd number of factors
// final status is open
else
console.log("open" +
" ");
}
}
// Driver Code
let n = 5;
printStatusOfDoors(n);
Output
open closed closed open closed
