Given an integer N, the task is to check if the concatenation of first N natural numbers is divisible by 3 or not. Print Yes if divisible and No if not.
Examples:
Input: N = 3
Output: Yes
Explanation:
The concatenated number = 123
Since it is divisible by 3, the output is YesInput: N = 7
Output: No
Explanation: The concatenated number = 1234567
Since it is not divisible by 3, the output is No.
Brute Force Approach:
A brute force approach to solve this problem would be to generate the concatenation of the first N natural numbers and then check if it is divisible by 3 or not. We can use a string to store the concatenation and then convert it into an integer for checking the divisibility.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function that returns True if // concatenation of first N natural // numbers is divisible by 3 bool isDivisible( int N)
{ string concat = "" ;
for ( int i = 1; i <= N; i++){
concat += to_string(i);
}
int num = stoi(concat);
return (num % 3 == 0);
} // Driver Code int main()
{ // Given Number
int N = 6;
// Function Call
if (isDivisible(N))
cout << ( "Yes" );
else
cout << ( "No" );
return 0;
} |
import java.util.*;
public class Main {
// Function that returns True if
// concatenation of first N natural
// numbers is divisible by 3
public static boolean isDivisible( int N) {
String concat = "" ;
for ( int i = 1 ; i <= N; i++) {
concat += Integer.toString(i);
}
int num = Integer.parseInt(concat);
return (num % 3 == 0 );
}
// Driver Code
public static void main(String[] args) {
// Given Number
int N = 6 ;
// Function Call
if (isDivisible(N)) {
System.out.println( "Yes" );
} else {
System.out.println( "No" );
}
}
} |
# Python code of the above approac # Defining the Function that returns True if # concatenation of first N natural # numbers is divisible by 3 def isDivisible(n : int ) - > bool :
concat = ""
for i in range ( 1 ,n + 1 ):
concat + = str (i)
num = int (concat)
return num % 3 = = 0
# Driver Code n = 6
# Storing the result res = isDivisible(n)
# Checking if the value # stored in variable res # is either False # or True if res = = False :
print ( "No" )
else :
print ( "Yes" )
|
using System;
public class MainClass
{ // Function that returns True if
// concatenation of first N natural
// numbers is divisible by 3
public static bool isDivisible( int N)
{
string concat = "" ;
for ( int i = 1; i <= N; i++){
concat += i.ToString();
}
int num = int .Parse(concat);
return (num % 3 == 0);
}
// Driver Code
public static void Main()
{
// Given Number
int N = 6;
// Function Call
if (isDivisible(N))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
} |
function isDivisible(N) {
let concat = "" ;
for (let i = 1; i <= N; i++) {
concat += i.toString(); } let num = parseInt(concat); return (num % 3 === 0);
} // Driver Code let N = 6; if (isDivisible(N))
console.log( "Yes" );
else console.log( "No" );
|
Yes
Time Complexity: O(N)
Auxiliary Space:: O(1)
Efficient Approach:
To optimize the above approach, we can observe a pattern. The concatenation of first N natural numbers is not divisible by 3 for the following series 1, 4, 7, 10, 13, 16, 19, and so on. The Nth term of the series is given by the formula 3×n +1. Hence, if (N – 1) is not divisible by 3, then the resultant number is divisible by 3, so print Yes. Otherwise, print No.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function that returns True if // concatenation of first N natural // numbers is divisible by 3 bool isDivisible( int N)
{ // Check using the formula
return (N - 1) % 3 != 0;
} // Driver Code int main()
{ // Given Number
int N = 6;
// Function Call
if (isDivisible(N))
cout << ( "Yes" );
else
cout << ( "No" );
return 0;
} // This code is contributed by Mohit Kumar |
// Java program for the above approach class GFG{
// Function that returns True if // concatenation of first N natural // numbers is divisible by 3 static boolean isDivisible( int N)
{ // Check using the formula
return (N - 1 ) % 3 != 0 ;
} // Driver Code public static void main(String[] args)
{ // Given Number
int N = 6 ;
// Function Call
if (isDivisible(N))
System.out.println( "Yes" );
else
System.out.println( "No" );
} } // This code is contributed by Ritik Bansal |
# Python program for the above approach # Function that returns True if # concatenation of first N natural # numbers is divisible by 3 def isDivisible(N):
# Check using the formula
return (N - 1 ) % 3 ! = 0
# Driver Code if __name__ = = "__main__" :
# Given Number
N = 6
# Function Call
if (isDivisible(N)):
print ( "Yes" )
else :
print ( "No" )
|
// C# program for the above approach using System;
class GFG{
// Function that returns True if // concatenation of first N natural // numbers is divisible by 3 static bool isDivisible( int N)
{ // Check using the formula
return (N – 1) % 3 != 0;
} // Driver Code public static void Main()
{ // Given Number
int N = 6;
// Function Call
if (isDivisible(N))
Console.Write( "Yes" );
else
Console.Write( "No" );
} } // This code is contributed by Code_Mech |
<script> // javascript program for the above approach // Function that returns True if // concatenation of first N natural // numbers is divisible by 3 function isDivisible(N)
{ // Check using the formula
return (N - 1) % 3 != 0;
} // Driver Code // Given Number var N = 6;
// Function Call if (isDivisible(N))
document.write( "Yes" );
else document.write( "No" );
// This code is contributed by Princi Singh. </script> |
Yes
Time Complexity: O(1)
Auxiliary Space: O(1)