# Count the number of digits in an integer

## Table of Contents

It’s easy to count the characters in a char array or a string but it’s a little tricky to count the number of digits \((k)\) in an integer \((n)\).

The simple \(O(n)\) solution in C:

```
#include<stdio.h>
int count\_digits(int n){
int k=0;
if(n==0){
return 1;
}
while(n!=0){
n=n/10;
k++;
}
return k;
}
int main(void){
int n;
printf("Enter a no.: ");
scanf("%d", &n);
printf("Number of digits: %d", count\_digits(n));
return 0;
}
```

We can do it in logarithmic time.

## It’s time to go mathematical #

We know that a binary number has base 2 and uses digits 0 and 1, and a decimal number has base 10 and uses digits from 0 to 9.

Now, let’s take a decimal number, say, \(n = 4591\).

We can represent it as, \[4591 = 4 * 10^3 + 5 * 10^2 + 9 * 10^1 + 1 * 10^0\]

Here, the maximum exponent of 10 is 3 to which if we add 1, we get the total no. of digits \((k)\) in integer \(n\) i.e. 4.

We know a logarithm is an inverse function of exponentiation. That means the logarithm of a given no. \(n\) is the exponent to which *another fixed number*, the base \(b\), must be raised, to produce that no. \(n\).

\[b^{\log_b n} = n\\ or, b^y = n\\ where, y = {\log_b n}\] For example, if \(b = 2, y = 5\ \&\ n = 32\), \[b^y = n\\ 2^5 = 32\\ and, \log_{2} 32 = 5\]

To put it simply, logarithm counts repeated multiplication of the *same factor.* Logarithm answers the question ‘How many of one number do we multiply to get another number?’. The answer here is we multiplied the no. 2 by itself 5 times to get another no. 32.

Take another example, when \(b = 10, y = 4\ \&\ n = 10,000\), \[b^y = n\\ 10^4 = 10,000\\ and, {\log_{10} 10,000} = 4\]

The interesting thing is we use \(\log_{10}\) for counting the no. of digits \((k)\) in an integer \((n)\). \(\log_{10}\) tries to find the no. of times \(n\) can be divided by 10 until the value of n is less than 10. The value of \({\log_{10} n}\) always turns out to be less than \(k\).

The reason for the value of \(({\log_{10} n}) < k\) can be explained using a simple number line representation. We can place \(n = 4591\) between 1000 and 10,000 since 1000 < 4591 < 10,000 , and the corresponding value of \({\log_{10} 4591}\) turns out to be between 3 and 4 i.e 3.661.

The points in this number line are multiples of 10.

Now, if we consider the floor value i.e 3 and add 1 to it, we get 4. Count the number of digits in 4591 yourself, it’s 4.

```
#include <stdio.h>
#include<math.h>
int count\_digits(int n){
if(n==0)
return 1;
else
return floor(log10(n)) + 1;
}
int main(void){
int n;
printf("Enter a no.: ");
scanf("%d", &n);
printf("Number of digits: %d", count\_digits(n));
return 0;
}
```

### References: #

- http://mathforum.org/library/drmath/view/70510.html
- https://youtu.be/kWndpHMpUlY
- https://www.youtube.com/watch?v=Xe9aq1WLpjU
- https://www.mathsisfun.com/algebra/logarithms.html
- https://stackoverflow.com/questions/41416329/how-to-write-a-log-base-10-function-in-c
- https://en.wikipedia.org/wiki/Logarithm