Determine whether the given integer N is a Peculiar Number or not

Given an integer N, our task is the determine if the integer N is Peculiar Number. If it is then print "yes" otherwise output "no".
The peculiar number is the number which is three times the sum of digits of the number. 
Examples:
 

Input: N = 27 
Output: Yes 
Explanation: 
Digit sum for 27 is 9 and 3 * 9 = 27 which is equal to N. Hence the output is yes.
Input: N = 36 
Output: No 
Explanation: 
Digit sum for 36 is 9 and 3 * 9 = 27 which is not equal to N. Hence the output is no. 
 

Approach:
To solve the problem mentioned above we have to first find the sum of the digits of a number N. Then check if the sum of digits of the number multiplied by 3 is actually the number N itself. If it is then print Yes otherwise the output is no.
Below is the implementation of the above approach: 
 

// C++ implementation to check if the
// number is peculiar

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

// Function to find sum of digits
// of a number
int sumDig(int n)
{
    int s = 0;

    while (n != 0) {
        s = s + (n % 10);

        n = n / 10;
    }

    return s;
}

// Function to check if the 
// number is peculiar
bool Pec(int n)
{
    // Store a duplicate of n
    int dup = n;

    int dig = sumDig(n);

    if (dig * 3 == dup)
        return true;

    else
        return false;
}

// Driver code
int main()
{
    int n = 36;

    if (Pec(n) == true)
        cout << "Yes" << endl;

    else
        cout << "No" << endl;

    return 0;
}

// Java implementation to check if the
// number is peculiar
import java.io.*;

class GFG{

// Function to find sum of digits
// of a number
static int sumDig(int n)
{
    int s = 0;

    while (n != 0)
    {
        s = s + (n % 10);
        n = n / 10;
    }
    return s;
}

// Function to check if number is peculiar
static boolean Pec(int n)
{
    
    // Store a duplicate of n
    int dup = n;
    int dig = sumDig(n);

    if (dig * 3 == dup)
        return true;
    else
        return false;
}

// Driver code
public static void main (String[] args) 
{
    int n = 36;

    if (Pec(n) == true)
        System.out.println("Yes");
    else
        System.out.println("No");
}
}

// This code is contributed by shubhamsingh10

# Python3 implementation to check if the
# number is peculiar

# Function to get sum of digits
# of a number 
def sumDig(n):
    
    s = 0
    while (n != 0): 
        s = s + int(n % 10) 
        n = int(n / 10) 
    
    return s
    
# Function to check if the  
# number is peculiar     
def Pec(n):
    
    dup = n
    dig = sumDig(n)

    if(dig * 3 == dup):
        return "Yes"
    else :
        return "No"
        
# Driver code 
n = 36

if Pec(n) == True:
    print("Yes")
else:
    print("No")

# This code is contributed by grand_master

// C# implementation to check if the
// number is peculiar
using System;

class GFG{

// Function to find sum of digits
// of a number
static int sumDig(int n)
{
    int s = 0;

    while (n != 0)
    {
        s = s + (n % 10);
        n = n / 10;
    }
    return s;
}

// Function to check if the number is peculiar
static bool Pec(int n)
{
    
    // Store a duplicate of n
    int dup = n;
    int dig = sumDig(n);

    if (dig * 3 == dup)
        return true;
    else
        return false;
}

// Driver code
public static void Main() 
{
    int n = 36;

    if (Pec(n) == true)
        Console.Write("Yes");
    else
        Console.Write("No");
}
}

// This code is contributed by Akanksha_Rai

<script>

// Javascript implementation to check if the
// number is peculiar

// Function to find sum of digits
// of a number
function sumDig(n)
{
    var s = 0;

    while (n != 0) {
        s = s + (n % 10);

        n = parseInt(n / 10);
    }

    return s;
}

// Function to check if the 
// number is peculiar
function Pec(n)
{
    // Store a duplicate of n
    var dup = n;

    var dig = sumDig(n);

    if (dig * 3 == dup)
        return true;

    else
        return false;
}

// Driver code
var n = 36;
if (Pec(n) == true)
    document.write( "Yes");
else
    document.write( "No" );

// This code is contributed by noob2000.
</script>

No

Time Complexity: O(log₁₀n), time used to find the sum of digits of a number
Auxiliary Space: O(1), as no extra space is required