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).
- 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.
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(n) or linear
- Auxiliary Space : O(1) or constant
Method 2:
Recursive Solution:
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))
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++
#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
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;
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)
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.
This article is contributed by Suruchi Kumari . If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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