🚀 Math vs. Strings: Two Ways to Manipulate Digits in C++

#include <iostream>
#include <string>

using namespace std;

int main()
{
    int number;
    int number_copy;
    int nt = 0; // Stores the result for the first method

    cout << "Enter your number: ";
    cin >> number;

    // --- EDGE CASE HANDLING ---
    // Rule 1: We only accept positive numbers for this algorithm
    if (number < 0) {
        cout << "Please enter a positive number!" << endl;
        return 0;
    }

    // Rule 2: Handing zero separately because it's an even number 
    // and mathematical division loops won't process it.
    if (number == 0) {
        cout << "The changed number for the first method is: 1" << endl;
        cout << "The changed number for the second method is: 1" << endl;
        return 0;
    }
    else {
        // CRITICAL STEP: Back up the original number.
        // The while loop below will destroy the 'number' variable (reducing it to 0).
        number_copy = number;

        long long p = 1; // Base 10 multiplier to reconstruct the number in the correct order

        // --- METHOD 1: THE MATHEMATICAL APPROACH ---
        while (number != 0) {
            int l = number % 10; // Extract the last digit

            // Check if the digit is odd
            if (l % 2 != 0) {
                nt = nt + l * p; // Keep the odd digit as it is
            }
            else {
                nt = nt + 1 * p; // Replace the even digit with 1
            }

            p = p * 10;       // Move to the next decimal place (units, tens, hundreds...)
            number = number / 10; // Remove the last digit from the number
        }
    }
    cout << "The changed number for the first method is: " << nt << '\n';

    // --- METHOD 2: THE STRING MANIPULATION APPROACH ---
    // We use 'number_copy' because 'number' is now 0.
    string numberStr = to_string(number_copy); // Convert the integer to a string

    // Loop through each character of the string
    for (int i = 0; i < numberStr.length(); i++) {
        // Check if the current character represents an even digit
        if (numberStr[i] == '0' || numberStr[i] == '2' || numberStr[i] == '4' || numberStr[i] == '6' || numberStr[i] == '8') {
            numberStr[i] = '1'; // Replace the character with '1'
        }
    }

    // Convert the modified string back into a 64-bit integer (long long) to prevent overflow
    long long nt2 = stoll(numberStr);
    cout << "The changed number for the second method is: " << nt2 << endl;

    return 0;
}