Find N numbers such that a number and its reverse are divisible by sum of its digits

Given a number N, the task is to print the first N numbers such that every number and the reverse of the number is divisible by its sum of digits.
Example:

Input: N = 4 
Output: 1 2 3 4 
Explanation:
The reverse of every single digit number is the same number. And, every number is divisible by itself. 
Input: N = 12 
Output: 1 2 3 4 5 6 7 8 9 10 12 18 

Approach: The idea is to iterate through every number from 1 and compute the sum of the digits. For every such number, check if the number and the reverse of the number are divisible by the sum or not. Therefore, the following steps are followed to compute the answer: 

1. Initialize the counter to one and iterate through all the numbers one by one.
2. For every number, find the reverse of the number.
3. While finding the reverse of the number, compute the sum of the digits of the number.
4. Now, check if the number and the reverse of the number are divisible by the sum of its digits.
5. If it is, then increment the counter. Repeat the above steps until this counter is equal to N.

Below is the implementation of the above approach:

C++

// C++ program to print the first N numbers
// such that every number and the reverse
// of the number is divisible by its
// sum of digits

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

// Function to calculate the sum of digits
int digit_sum(int n)
{
    int sum = 0, m;

    // Loop to iterate through every
    // digit of the number
    while (n > 0) {
        m = n % 10;
        sum = sum + m;
        n = n / 10;
    }

    // Returning the sum of digits
    return (sum);
}

// Function to calculate the reverse
// of a number
int reverse(int n)
{
    int r = 0;

    // Loop to calculate the reverse
    // of the number
    while (n != 0) {
        r = r * 10;
        r = r + n % 10;
        n = n / 10;
    }

    // Return the reverse of the
    // number
    return (r);
}

// Function to print the first N numbers
// such that every number and the reverse
// of the number is divisible by its
// sum of digits
void operation(int n)
{
    int i = 1, a, count = 0, r;

    // Loop to continuously check and
    // generate number until there
    // are n outputs
    while (count < n) {

        // Variable to hold the sum of
        // the digit of the number
        a = digit_sum(i);

        // Computing the reverse of the
        // number
        r = reverse(i);

        // Checking if the condition satisfies.
        // Increment the count and print the
        // number if it satisfies.
        if (i % a == 0 && r % a == 0) {
            cout << i << " ";
            count++;
            i++;
        }
        else
            i++;
    }
}

// Driver code
int main()
{
    int n = 10;

    operation(n);
}

Java

// Java program to print the first N numbers
// such that every number and the reverse
// of the number is divisible by its
// sum of digits
import java.util.*;

class GFG{

// Function to calculate the sum of digits
static int digit_sum(int n)
{
    int sum = 0, m;

    // Loop to iterate through
    // every digit of the number
    while (n > 0)
    {
        m = n % 10;
        sum = sum + m;
        n = n / 10;
    }

    // Returning the sum of digits
    return (sum);
}

// Function to calculate the
// reverse of a number
static int reverse(int n)
{
    int r = 0;

    // Loop to calculate the
    // reverse of the number
    while (n != 0)
    {
        r = r * 10;
        r = r + n % 10;
        n = n / 10;
    }

    // Return the reverse
    // of the number
    return (r);
}

// Function to print the first N numbers
// such that every number and the reverse
// of the number is divisible by its
// sum of digits
static void operation(int n)
{
    int i = 1, a, count = 0, r;

    // Loop to continuously check and
    // generate number until there
    // are n outputs
    while (count < n)
    {

        // Variable to hold the sum
        // of the digit of the number
        a = digit_sum(i);

        // Computing the reverse of the
        // number
        r = reverse(i);

        // Checking if the condition satisfies.
        // Increment the count and print the
        // number if it satisfies.
        if (i % a == 0 && r % a == 0)
        {
            System.out.print(i + " ");
            count++;
            i++;
        }
        else
            i++;
    }
}

// Driver code
public static void main(String args[])
{
    int n = 10;

    operation(n);
}
}

// This code is contributed by shivanisinghss2110

Python3

# Python3 program to print the first N numbers
# such that every number and the reverse
# of the number is divisible by its
# sum of digits

# Function to calculate the sum of digits
def digit_sum(n):
    sum = 0

    # Loop to iterate through every
    # digit of the number
    while (n > 0):
        m = n % 10;
        sum = sum + m;
        n = n //10

    # Returning the sum of digits
    return (sum)

# Function to calculate the reverse
# of a number
def reverse(n):
    r = 0

    # Loop to calculate the reverse
    # of the number
    while (n != 0):
        r = r * 10
        r = r + n % 10
        n = n // 10

    # Return the reverse of the
    # number
    return (r)

# Function to print the first N numbers
# such that every number and the reverse
# of the number is divisible by its
# sum of digits
def operation(n):
    i = 1
    count = 0

    # Loop to continuously check and
    # generate number until there
    # are n outputs
    while (count < n):

        # Variable to hold the sum of
        # the digit of the number
        a = digit_sum(i)

        # Computing the reverse of the
        # number
        r = reverse(i)

        # Checking if the condition satisfies.
        # Increment the count and print the
        # number if it satisfies.
        if (i % a == 0 and r % a == 0):

            print(i, end = " ")
            count += 1
            i += 1
        else:
            i += 1

# Driver code
if __name__ == '__main__':

    n = 10
    operation(n)

# This code is contributed by Samarth

C#

// C# program to print the first N numbers
// such that every number and the reverse
// of the number is divisible by its
// sum of digits
using System;
class GFG{

// Function to calculate the sum of digits
static int digit_sum(int n)
{
    int sum = 0, m;

    // Loop to iterate through
    // every digit of the number
    while (n > 0)
    {
        m = n % 10;
        sum = sum + m;
        n = n / 10;
    }

    // Returning the sum of digits
    return (sum);
}

// Function to calculate the
// reverse of a number
static int reverse(int n)
{
    int r = 0;

    // Loop to calculate the
    // reverse of the number
    while (n != 0)
    {
        r = r * 10;
        r = r + n % 10;
        n = n / 10;
    }

    // Return the reverse
    // of the number
    return (r);
}

// Function to print the first N numbers
// such that every number and the reverse
// of the number is divisible by its
// sum of digits
static void operation(int n)
{
    int i = 1, a, count = 0, r;

    // Loop to continuously check and
    // generate number until there
    // are n outputs
    while (count < n)
    {

        // Variable to hold the sum
        // of the digit of the number
        a = digit_sum(i);

        // Computing the reverse of the
        // number
        r = reverse(i);

        // Checking if the condition satisfies.
        // Increment the count and print the
        // number if it satisfies.
        if (i % a == 0 && r % a == 0)
        {
            Console.Write(i + " ");
            count++;
            i++;
        }
        else
            i++;
    }
}

// Driver code
public static void Main()
{
    int n = 10;

    operation(n);
}
}

// This code is contributed by Code_Mech

Javascript

<script>

    // Javascript program to print the first N numbers
    // such that every number and the reverse
    // of the number is divisible by its
    // sum of digits

    // Function to calculate the sum of digits
    function digit_sum(n)
    {
        let sum = 0, m;

        // Loop to iterate through every
        // digit of the number
        while (n > 0) {
            m = n % 10;
            sum = sum + m;
            n = parseInt(n / 10, 10);
        }

        // Returning the sum of digits
        return (sum);
    }

    // Function to calculate the reverse
    // of a number
    function reverse(n)
    {
        let r = 0;

        // Loop to calculate the reverse
        // of the number
        while (n != 0) {
            r = r * 10;
            r = r + n % 10;
            n = parseInt(n / 10, 10);
        }

        // Return the reverse of the
        // number
        return (r);
    }

    // Function to print the first N numbers
    // such that every number and the reverse
    // of the number is divisible by its
    // sum of digits
    function operation(n)
    {
        let i = 1, a, count = 0, r;

        // Loop to continuously check and
        // generate number until there
        // are n outputs
        while (count < n) {

            // Variable to hold the sum of
            // the digit of the number
            a = digit_sum(i);

            // Computing the reverse of the
            // number
            r = reverse(i);

            // Checking if the condition satisfies.
            // Increment the count and print the
            // number if it satisfies.
            if (i % a == 0 && r % a == 0) {
                document.write(i + " ");
                count++;
                i++;
            }
            else
                i++;
        }
    }

    let n = 10;

    operation(n);

</script>

Output:

1 2 3 4 5 6 7 8 9 10

Time Complexity: O(n * log₁₀n)

Auxiliary Space: O(1)