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

Given a number N, the task is to return the count of digits in this number.

Example:

Program to count digits in an integer

Simple Iterative Solution to count digits in an integer

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

Below is the implementation of the above approach:

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(log₁₀(n)) or O(num digits)
Auxiliary Space: O(1) or constant

Recursive Solution to count digits in an integer

Keep dividing the number by 10 this reduces the input number size by 1 and keeps track of the number of sizes reduced.

Algorithm:

• The base condition of this recursive approach is when we divide the number by 10 and the number gets reduced to 0, so return 1 for this operation.
• Make a function call by dividing the number by 10, reducing the input size of the given number by 1, and adding 1 for this operation.

Below is the implementation of the above approach:

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))

Log-based Solution to count digits in an integer

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). 

Below is the implementation of the above idea:

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

Converting given number to string solution to count digits in an integer

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)

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