Program to count digits in an integer (4 Different Methods)

Count the number of digits in a long integer entered by a user.

Method 1: 

Simple Iterative Solution
The integer entered by the user is stored in the variable n. Then the while loop is iterated until the test expression n != 0 is evaluated to 0 (false). 

1. After the first iteration, the value of n will be 345 and the count is incremented to 1.
2. After the second iteration, the value of n will be 34 and the count is incremented to 2.
3. After the third iteration, the value of n will be 3 and the count is incremented to 3.
4. At the start of the fourth iteration, the value of n will be 0 and the loop is terminated.

Then the test expression is evaluated for false and the loop terminates. 

C++

// Iterative C++ program to count
// number of digits in a number
#include <bits/stdc++.h>
using namespace std;

int countDigit(long long n)
{
   if(n == 0)
     return 1;
    int count = 0;
    while (n != 0)
    {
        n = n / 10;
        ++count;
    }
    return count;
}

// Driver code
int main(void)
{
    long long n = 345289467;
    cout << "Number of digits : " << countDigit(n);
    return 0;
}

// This code is contributed
// by Akanksha Rai

C

// Iterative C program to count number of
// digits in a number
#include <stdio.h>

int countDigit(long long n)
{
    if(n == 0)
     return 1;
    int count = 0;
    while (n != 0)
    {
        n = n / 10;
        ++count;
    }
    return count;
}

// Driver code
int main(void)
{
    long long n = 345289467;
    printf("Number of digits : %d", countDigit(n));
    return 0;
}

Java

// JAVA Code to count number of
// digits in an integer
class GFG {

    static int countDigit(long n)
    {
        int count = 0;
        while (n != 0) {
            n = n / 10;
            ++count;
        }
        return count;
    }

    /* Driver code */
    public static void main(String[] args)
    {
        long n = 345289467;
        System.out.print("Number of digits : "
                         + countDigit(n));
    }
}
// This code is contributed by Arnav Kr. Mandal.

Python3

# Iterative Python program to count
# number of digits in a number


def countDigit(n):
    count = 0
    while n != 0:
        n //= 10
        count += 1
    return count


# Driver Code
n = 345289467
print("Number of digits : % d" % (countDigit(n)))

# This code is contributed by Shreyanshi Arun

C#

// C# Code to count number of
// digits in an integer
using System;

class GFG {

    static int countDigit(long n)
    {
        int count = 0;
        while (n != 0)
        {
            n = n / 10;
            ++count;
        }
        return count;
    }

    /* Driver code */
    public static void Main()
    {
        long n = 345289467;
        Console.WriteLine("Number of"
                          + " digits : " + countDigit(n));
    }
}

// This code is contributed by anuj_67.

PHP

<?php
// Iterative PHP program to count
// number of digits in a number

function countDigit($n)
{
    $count = 0;
    while ($n != 0)
    {
        $n = round($n / 10);
        ++$count;
    }
    return $count;
}

// Driver code
$n = 345289467;
echo "Number of digits : "
        . countDigit($n);

//This code is contributed by mits
?>

Javascript

<script>

// Iterative Javascript program to count
// number of digits in a number

function countDigit(n)
{
    let count = 0;
    while (n != 0)
    {
        n = Math.floor(n / 10);
        ++count;
    }
    return count;
}

// Driver code

    n = 345289467;
    document.write("Number of digits : "+ countDigit(n));

// This code is contributed by Mayank Tyagi

</script>

Output

Number of digits : 9

• Time Complexity : O(n) or linear 
• Auxiliary Space : O(1) or constant

Method 2: 

Recursive Solution:

C++

// Recursive C++ program to count number of
// digits in a number
#include <bits/stdc++.h>
using namespace std;

int countDigit(long long n)
{
    if (n/10 == 0)
        return 1;
    return 1 + countDigit(n / 10);
}

// Driver code
int main(void)
{
    long long n = 345289467;
    cout << "Number of digits :" << countDigit(n);
    return 0;
}
// This code is contributed by Mukul Singh.

C

// Recursive C program to count number of
// digits in a number
#include <stdio.h>

int countDigit(long long n)
{
    if (n/10 == 0)
        return 1;
    return 1 + countDigit(n / 10);
}

// Driver code
int main(void)
{
    long long n = 345289467;
    printf("Number of digits : %d", countDigit(n));
    return 0;
}

Java

// JAVA Code to count number of
// digits in an integer
import java.util.*;

class GFG {

    static int countDigit(long n)
    {
        if (n/10 == 0)
            return 1;
        return 1 + countDigit(n / 10);
    }

    /* Driver code */
    public static void main(String[] args)
    {
        long n = 345289467;
        System.out.print("Number of digits : "
                         + countDigit(n));
    }
}

// This code is contributed by Arnav Kr. Mandal.

Python3

# Recursive Python program to count
# number of digits in a number


def countDigit(n):
    if n//10 == 0:
        return 1
    return 1 + countDigit(n // 10)


# Driver Code
n = 345289467
print("Number of digits : % d" % (countDigit(n)))

# This code is contributed by Shreyanshi Arun

C#

// C# Code to count number of
// digits in an integer
using System;

class GFG {

    static int countDigit(long n)
    {
        if (n/10 == 0)
            return 1;
        return 1 + countDigit(n / 10);
    }

    /* Driver Code */
    public static void Main()
    {
        long n = 345289467;
        Console.WriteLine("Number of "
                          + "digits : "
                          + countDigit(n));
    }
}

// This code is contributed by anuj_67.

PHP

<?php
// Recursive PHP program to count
// number of digits in a number

function countDigit($n)
{
    if ($n/10 == 0)
        return 1;
    return 1 + countDigit((int)($n / 10));
}

// Driver Code
$n = 345289467;
print ("Number of digits : " .
            (countDigit($n)));

// This code is contributed by mits
?>

Javascript

<script>

// Recursive Javascript program to count number of
// digits in a number

function countDigit(n)
{
    if (n/10 == 0)
        return 1;
    return 1 + countDigit(parseInt(n / 10));
}

// Driver code
var n = 345289467;
document.write("Number of digits :" + countDigit(n));

</script>

Output

Number of digits :9

• Time Complexity : O(log(n)) 
• Auxiliary Space : O(log(n))

Method 3: 

Log-based Solution:
We can use log10(logarithm of base 10) to count the number of digits of positive numbers (logarithm is not defined for negative numbers).
Digit count of N = upper bound of log10(N). 

C++

// Log based C++ program to count number of
// digits in a number
#include <bits/stdc++.h>
using namespace std;

int countDigit(long long n) {
  return floor(log10(n) + 1);
}

// Driver code
int main(void)
{
    long long n = 345289467;
    cout << "Number of digits : "
         << countDigit(n);
    return 0;
}

// This code is contributed by shivanisinghss2110

C

// Log based C program to count number of
// digits in a number
#include <math.h>
#include <stdio.h>

int countDigit(long long n) {
  return floor(log10(n) + 1);
}

// Driver code
int main(void)
{
    long long n = 345289467;
    printf("Number of digits : %d", countDigit(n));
    return 0;
}

Java

// JAVA Code to count number of
// digits in an integer
import java.util.*;

class GFG {

    static int countDigit(long n)
    {
        return (int)Math.floor(Math.log10(n) + 1);
    }

    /* Driver code */
    public static void main(String[] args)
    {
        long n = 345289467;
        System.out.print("Number of digits : "
                         + countDigit(n));
    }
}
// This code is contributed by Arnav Kr. Mandal.

Python3

# Log based Python program to count number of
# digits in a number

# function to import ceil and log
import math


def countDigit(n):
    return math.floor(math.log10(n)+1)


# Driver Code
n = 345289467
print("Number of digits : % d" % (countDigit(n)))

# This code is contributed by Shreyanshi Arun

C#

// C# Code to count number of
// digits in an integer
using System;

class GFG {

    static int countDigit(long n)
    {
        return (int)Math.Floor(Math.Log10(n) + 1);
    }

    /* Driver code */
    public static void Main()
    {
        long n = 345289467;
        Console.WriteLine("Number of digits : "
                          + countDigit(n));
    }
}

// This code is contributed by anuj_67..

PHP

<?php
// Log based php program to
// count number of digits
// in a number

function countDigit($n)
{
    return floor(log10($n)+1);
}

// Driver code
$n = 345289467;
echo "Number of digits : ",
    countDigit($n);

// This code is contributed by ajit
?>

Javascript

<script>

// Log based Javascript program to count number of
// digits in a number
function countDigit(n)
{
    return Math.floor(Math.log10(n) + 1);
}

// Driver code
var n = 345289467;
document.write("Number of digits : "  +
               countDigit(n));

// This code is contributed by noob2000

</script>

Output

Number of digits : 9

• Time Complexity : O(1) or constant
•  Auxiliary Space : O(1) or constant

Method 4:
We can convert the number into a string and then find the length of the string to get the number of digits in the original number. 

C++

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

// To count the no. of digits in a number
void count_digits(int n)
{
    // converting number to string using
    // to_string in C++
    string num = to_string(n);

    // calculate the size of string
    cout << num.size() << endl;
}
//Driver Code
int main()
{
    // number
    int n = 345;
    count_digits(n);
    return 0;
}

// This code is contributed by Shashank Pathak

Java

import java.util.*;
public class GFG {

    // To count the no. of digits in a number
    static void count_digits(int n)
    {
        // converting number to string using
        // to_string in C++
        String num = Integer.toString(n);

        // calculate the size of string

        System.out.println(+num.length());
    }
    // Driver code
    public static void main(String args[])
    {
        // number
        int n = 345;
        count_digits(n);
    }
}
// Code is contributed by shivanisinghss2110

Python3

# Python3 implementation of the approach
def count_digits(n):
    n = str(n)
    return len(n)


# Driver code
n = 456533457776
print(count_digits(n))

C#

// C# implementation of the above approach
using System;
using System.Collections.Generic;

class GFG {

    // To count the no. of digits in a number
    static void count_digits(int n)
    {
        // converting number to string using
        // to_string in C#

        string num = Convert.ToString(n);

        // calculate the size of string
        Console.WriteLine(+num.Length);
    }

    // Driver Code
    public static void Main(string[] args)
    {
        // number
        int n = 345;
        count_digits(n);
    }
}

// This code is contributed by shivanisinghss2110

Javascript

<script>
    // Javascript implementation of the above approach

    // To count the no. of digits in a number
    function count_digits(n)
    {
        // converting number to string using
        // to_string in javascript

        let num = n.toString();

        // calculate the size of string
        document.write(num.length);
    }

    // number
    let n = 345;
    count_digits(n);

</script>

Output

3

• Time Complexity : O(1) or constant
• Auxiliary Space :  O(Number of digits in an integer)

Here in all these four different methods to count the number of digits in an integer , the best approach will be the Method : 3 as the time complexity as well as auxiliary space both are constant , so our program will run in constant time and takes constant space in terms of input size.

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