Data Structure / Maths

Count digits in a number

Problem Statement: Given an integer N, write a program to count the number of digits in N.

Examples 
Example 1:
Input: N = 12345
Output: 5
Explanation: N has 5 digits

Example 2:
Input: N = 8394
Output: 4
Explanation: N has 4 digits
Practice:
Solve Problem code-studio

Disclaimer: Don’t jump directly to the solution, try it out yourself first.

Solution 1 Solution 2 Solution 3

Expand any one approach by clicking the given options in the bar. Clicking one approach on bar, closes all other expands. You can manually expand more than one approach at a time

Solution 1:
Algorithm / Intuition

Approach: 

  1. Store the integer in a variable X and initialize a counter variable to count the number of digits.
  2. We know that in programming languages when we divide X by Y it will result in an integer (given both the variables are integers). For example, 133/10 will result in 13 similarly 1/10 will result in 0.
  3. Using a for loop and above observation keep on dividing X by 10 and increment the count in every iteration when X equals 0 terminate the loop and the count will have the number of digits in N.
Code 



#include<bits/stdc++.h>

using namespace std;

int count_digits( int n )
{
   int x = n; int count =0;
   while( x !=0 ) 
   {
       x = x / 10;
       count++;
   }
   return count;
}


int main()
{
   int n = 12345;
   cout<< "Number of digits in "<<n<<" is "<<count_digits(n);
   return 0;
}

Output: Number of digits in 12345 is 5 




public class tUf { 
    static int count_digits(int n)
    {
       int x = n; int count =0;
       while( x!=0 ) 
      {
          x = x / 10;
          count++;
      }
       return count;
    } 
    public static void main(String args[]) 
    { 
         int n = 12345;
        System.out.println("Number of digits in "+n+" is "+count_digits(n));
    } 
}

Output: Number of digits in 12345 is 5 




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

n = 12345
print("Number of digits in ",n," is ",count_digits(n))

Output: Number of digits in 12345 is 5 




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

let n = 12345;
console.log("Number of digits in " + n + " is " + countDigits(n));

Output: Number of digits in 12345 is 5

Complexity Analysis

Time Complexity: O (n) where n is the number of digits in the given integer

Space Complexity: O(1)

Solution 2:
Algorithm / Intuition

Solution 2 : 

  1. Convert the integer into a string.
  2. Find the length of the string      
Code 



#include<bits/stdc++.h>

using namespace std;

int count_digits( int n )
{
  string x = to_string(n);
  return x.length(); 
}

int main()
{
   int n = 12345;
   cout<< "Number of digits in "<<n<<" is "<<count_digits(n);
   return 0;
}

Number of digits in 12345 is 5 




import java.util.*;
public class tUf {
  static int count_digits(int n) {
    String n2 = Integer.toString(n);
    return n2.length();
  }

  public static void main(String args[]) {
    int n = 12345;
    System.out.println("Number of digits in " + n + " is " + count_digits(n));
  }
}

Number of digits in 12345 is 5 




def count_digits(n):
  x = str(n)
return len(x)

n = 12345
print("Number of digits in ", n, " is ", count_digits(n))

Number of digits in 12345 is 5 




function countDigits(n) {
  let x = n.toString();
  return x.length;
}

let n = 12345;
console.log("Number of digits in " + n + " is " + countDigits(n));

Number of digits in 12345 is 5

Complexity Analysis

Time Complexity: O (1) 

Space Complexity: O(1)

Solution 3:
Algorithm / Intuition

Use logarithm base 10 to count the number of digits. As

The number of digits in an integer = the upper bound of log10(n).

Example :

n = 12345
log10(12345) = 4.091
Integral part of 4.091 is 4 and 4 + 1 =  5 which is also the number of digits in 12345
Code 



#include<bits/stdc++.h>

using namespace std;

int count_digits( int n )
{
  int digits = floor(log10(n) + 1);
  return digits;
}

int main()
{
   int n = 12345;
   cout<< "Number of digits in "<<n<<" is "<<count_digits(n);
   return 0;
}

Output: Number of digits in 12345 is 5 




import java.util.*;
public class tUf { 
    static int count_digits(int n)
    {
       int digits = (int) Math.floor(Math.log10(n) + 1);
        return digits;
    } 

    public static void main(String args[]) 
    { 
         int n = 12345;
        System.out.println("Number of digits in "+n+" is "+count_digits(n));
    } 
}

Output: Number of digits in 12345 is 5 




import math
def count_digits(n):
  digits = math.floor(math.log10(n) + 1)
return digits
n = 12345
print("Number of digits in ", n, " is ", count_digits(n))

Output: Number of digits in 12345 is 5 




function countDigits(n) {
  let digits = Math.floor(Math.log10(n) + 1);
  return digits;
}

let n = 12345;
console.log("Number of digits in " + n + " is " + countDigits(n));

Output: Number of digits in 12345 is 5

Complexity Analysis

Time Complexity: O (1) 

Space Complexity: O(1)

Video Explanation

Special thanks to Harsh Verma for contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam, please check out this article