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

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).
- After the first iteration, the value of n will be 345 and the count is incremented to 1.
- After the second iteration, the value of n will be 34 and the count is incremented to 2.
- After the third iteration, the value of n will be 3 and the count is incremented to 3.
- 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.
Below is the implementation of the above approach:
C++
#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;
}
int main( void )
{
long long n = 345289467;
cout << "Number of digits : " << countDigit(n);
return 0;
}
|
C
#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;
}
int main( void )
{
long long n = 345289467;
printf ( "Number of digits : %d" , countDigit(n));
return 0;
}
|
Java
class GFG {
static int countDigit( long n)
{
int count = 0 ;
while (n != 0 ) {
n = n / 10 ;
++count;
}
return count;
}
public static void main(String[] args)
{
long n = 345289467 ;
System.out.print( "Number of digits : "
+ countDigit(n));
}
}
|
Python3
def countDigit(n):
count = 0
while n ! = 0 :
n / / = 10
count + = 1
return count
n = 345289467
print ( "Number of digits : % d" % (countDigit(n)))
|
C#
using System;
class GFG {
static int countDigit( long n)
{
int count = 0;
while (n != 0) {
n = n / 10;
++count;
}
return count;
}
public static void Main()
{
long n = 345289467;
Console.WriteLine( "Number of"
+ " digits : " + countDigit(n));
}
}
|
PHP
<?php
function countDigit( $n )
{
$count = 0;
while ( $n != 0)
{
$n = round ( $n / 10);
++ $count ;
}
return $count ;
}
$n = 345289467;
echo "Number of digits : "
. countDigit( $n );
?>
|
Javascript
<script>
function countDigit(n)
{
let count = 0;
while (n != 0)
{
n = Math.floor(n / 10);
++count;
}
return count;
}
n = 345289467;
document.write( "Number of digits : " + countDigit(n));
</script>
|
OutputNumber 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++
#include <bits/stdc++.h>
using namespace std;
int countDigit( long long n)
{
if (n/10 == 0)
return 1;
return 1 + countDigit(n / 10);
}
int main( void )
{
long long n = 345289467;
cout << "Number of digits :" << countDigit(n);
return 0;
}
|
C
#include <stdio.h>
int countDigit( long long n)
{
if (n/10 == 0)
return 1;
return 1 + countDigit(n / 10);
}
int main( void )
{
long long n = 345289467;
printf ( "Number of digits : %d" , countDigit(n));
return 0;
}
|
Java
import java.util.*;
class GFG {
static int countDigit( long n)
{
if (n/ 10 == 0 )
return 1 ;
return 1 + countDigit(n / 10 );
}
public static void main(String[] args)
{
long n = 345289467 ;
System.out.print( "Number of digits : "
+ countDigit(n));
}
}
|
Python3
def countDigit(n):
if n / / 10 = = 0 :
return 1
return 1 + countDigit(n / / 10 )
n = 345289467
print ( "Number of digits : % d" % (countDigit(n)))
|
C#
using System;
class GFG {
static int countDigit( long n)
{
if (n/10 == 0)
return 1;
return 1 + countDigit(n / 10);
}
public static void Main()
{
long n = 345289467;
Console.WriteLine( "Number of "
+ "digits : "
+ countDigit(n));
}
}
|
PHP
<?php
function countDigit( $n )
{
if ( $n /10 == 0)
return 1;
return 1 + countDigit((int)( $n / 10));
}
$n = 345289467;
print ( "Number of digits : " .
(countDigit( $n )));
?>
|
Javascript
<script>
function countDigit(n)
{
if (n/10 == 0)
return 1;
return 1 + countDigit(parseInt(n / 10));
}
var n = 345289467;
document.write( "Number of digits :" + countDigit(n));
</script>
|
OutputNumber 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++
#include <bits/stdc++.h>
using namespace std;
int countDigit( long long n) {
return floor ( log10 (n) + 1);
}
int main( void )
{
long long n = 345289467;
cout << "Number of digits : "
<< countDigit(n);
return 0;
}
|
C
#include <math.h>
#include <stdio.h>
int countDigit( long long n) {
return floor ( log10 (n) + 1);
}
int main( void )
{
long long n = 345289467;
printf ( "Number of digits : %d" , countDigit(n));
return 0;
}
|
Java
import java.util.*;
class GFG {
static int countDigit( long n)
{
return ( int )Math.floor(Math.log10(n) + 1 );
}
public static void main(String[] args)
{
long n = 345289467 ;
System.out.print( "Number of digits : "
+ countDigit(n));
}
}
|
Python3
import math
def countDigit(n):
return math.floor(math.log10(n) + 1 )
n = 345289467
print ( "Number of digits : % d" % (countDigit(n)))
|
C#
using System;
class GFG {
static int countDigit( long n)
{
return ( int )Math.Floor(Math.Log10(n) + 1);
}
public static void Main()
{
long n = 345289467;
Console.WriteLine( "Number of digits : "
+ countDigit(n));
}
}
|
PHP
<?php
function countDigit( $n )
{
return floor (log10( $n )+1);
}
$n = 345289467;
echo "Number of digits : " ,
countDigit( $n );
?>
|
Javascript
<script>
function countDigit(n)
{
return Math.floor(Math.log10(n) + 1);
}
var n = 345289467;
document.write( "Number of digits : " +
countDigit(n));
</script>
|
OutputNumber 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;
void count_digits( int n)
{
string num = to_string(n);
cout << num.size() << endl;
}
int main()
{
int n = 345;
count_digits(n);
return 0;
}
|
Java
import java.util.*;
public class GFG {
static void count_digits( int n)
{
String num = Integer.toString(n);
System.out.println(+num.length());
}
public static void main(String args[])
{
int n = 345 ;
count_digits(n);
}
}
|
Python3
def count_digits(n):
n = str (n)
return len (n)
n = 456533457776
print (count_digits(n))
|
C#
using System;
using System.Collections.Generic;
class GFG {
static void count_digits( int n)
{
string num = Convert.ToString(n);
Console.WriteLine(+num.Length);
}
public static void Main( string [] args)
{
int n = 345;
count_digits(n);
}
}
|
Javascript
<script>
function count_digits(n)
{
let num = n.toString();
document.write(num.length);
}
let n = 345;
count_digits(n);
</script>
|
Time Complexity: O(1) or constant
Auxiliary Space: O(Number of digits in an integer)
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