Count Digits and Reverse Number

import sys

def solve_digit_manipulation():
    try:
        N = int(sys.stdin.readline().strip())
    except:
        return

    original_n = N
    
    # Handle negative numbers by converting to positive for digit logic
    num = abs(N) 
    
    # --- 1. Count Digits ---
    temp_count = num
    count = 0
    while temp_count > 0:
        temp_count //= 10
        count += 1
    print(f"Count: {count}")
    
    # --- 2. Reverse Number ---
    temp_reverse = num
    reversed_num = 0
    while temp_reverse > 0:
        last_digit = temp_reverse % 10
        reversed_num = reversed_num * 10 + last_digit
        temp_reverse //= 10
        
    # Apply original sign
    if original_n < 0:
        reversed_num *= -1
        
    print(f"Reversed: {reversed_num}")

if __name__ == "__main__":
    solve_digit_manipulation()

import java.util.Scanner;

class Solution {

    // O(log N) Time, O(1) Space
    public static void solve_digit_manipulation() {
        Scanner scanner = new Scanner(System.in);
        if (!scanner.hasNextLong()) return;
        long N = scanner.nextLong();
        
        long originalN = N;
        long num = Math.abs(N);

        // --- 1. Count Digits ---
        long tempCount = num;
        int count = 0;
        
        if (num == 0) {
            count = 1;
        } else {
            while (tempCount > 0) {
                tempCount /= 10;
                count++;
            }
        }
        System.out.println("Count: " + count);

        // --- 2. Reverse Number ---
        long tempReverse = num;
        long reversedNum = 0;
        
        while (tempReverse > 0) {
            int lastDigit = (int)(tempReverse % 10);
            reversedNum = reversedNum * 10 + lastDigit;
            tempReverse /= 10;
        }
        
        if (originalN < 0) {
            reversedNum *= -1;
        }
        
        System.out.println("Reversed: " + reversedNum);
        scanner.close();
    }

    public static void main(String[] args) {
        solve_digit_manipulation();
    }
}

#include <iostream>
#include <algorithm>

using namespace std;

// O(log N) Time, O(1) Space
void solve_digit_manipulation() {
    long long N;
    if (!(cin >> N)) return;

    long long original_n = N;
    long long num = abs(N); 

    // --- 1. Count Digits ---
    long long temp_count = num;
    int count = 0;
    
    if (num == 0) { // Edge case: count is 1 for number 0
        count = 1;
    } else {
        while (temp_count > 0) {
            temp_count /= 10;
            count++;
        }
    }
    cout << "Count: " << count << endl;

    // --- 2. Reverse Number ---
    long long temp_reverse = num;
    long long reversed_num = 0;
    
    while (temp_reverse > 0) {
        int last_digit = temp_reverse % 10;
        // Important: check for overflow here in production code
        reversed_num = reversed_num * 10 + last_digit;
        temp_reverse /= 10;
    }
    
    if (original_n < 0) {
        reversed_num *= -1;
    }
        
    cout << "Reversed: " << reversed_num << endl;
}

int main() {
    solve_digit_manipulation();
    return 0;
}