C Exercises: Check if a number is Keith or not

#include <stdio.h>
#include <stdlib.h>

// Function to count the number of digits in a given number
int lenCount(int nm)
{
    int ctr = 0;
    while (nm > 0)
    {
        nm = nm / 10;
        ctr++;
    }
    return ctr;
}

int main()
{
    int num1 = 0, arr1[10], num2 = 0, flg = 0, i = 0, sum = 0;

    // Input section and description of Keith numbers
    printf("\n\n Check whether a number is Keith or not: \n");
    printf(" Sample Keith numbers: 197, 742, 1104, 1537, 2208, 2580, 3684, 4788, 7385..\n");
    printf(" --------------------------------------------------------------------------\n");
    printf(" Input a number : ");
    scanf("%d", &num1);
    num2 = num1;

    // Breaking the input number into an array of its digits
    for (i = lenCount(num2) - 1; i >= 0; i--)
    {
        arr1[i] = num1 % 10;
        num1 /= 10;
    }

    // Algorithm to check if the number is a Keith number
    while (flg == 0)
    {
        for (i = 0; i < lenCount(num2); i++)
            sum += arr1[i]; // Summing up the digits

        // Checking if the sum matches the original number or if it exceeds it
        if (sum == num2)
        {
            printf(" The given number is a Keith Number.\n");
            flg = 1;
        }
        if (sum > num2)
        {
            printf(" The given number is not a Keith number.\n");
            flg = 1;
        }

        // Rearranging the array by shifting elements
        for (i = 0; i < lenCount(num2); i++)
        {
            if (i != lenCount(num2) - 1)
                arr1[i] = arr1[i + 1];
            else
                arr1[i] = sum;
        }
        sum = 0;
    }
}