# Given a number N in decimal base, find number of its digits in any base (base b)

Given A Number n in a base 10, find the number of digits in its base b representation.
Constraints : Whole
Examples :

Input : Number = 48
Base = 4
Output: 3
Explanation : (48)10 = (300)4

Input : Number = 1446
Base = 7
Output: 4
Explanation : (446)10 = (4134)7

A simple approach: convert the decimal number into the given base r and then count number of digits.
An efficient approach : It resides on the relationship between the base of the number and number of digits of that number.
Typically : Let n be a positive integer. The base representation of has digits if , which is the case if or .The number of digits in the base b representation of n is therefore

In above equation the base changing logarithmic property has been used. So we calculate the logarithm of the number in that base which we want to calculate the number of digits. And take its floor value and then add 1.
This idea can be further used to find the number of digits of a given number n of base b in base r. All have to be done is to convert the number in base 10 and then apply the above formula of finding digits. It would be easier to calculate log of any base when number is in base 10.

## C++

 // C++ program to Find Number of digits  // in base b. #include  #include  using namespace std;   // function to print number of // digits void findNumberOfDigits(long n, int base) {     // Calculating log using base     // changing property and then     // taking it floor and then      // adding 1.     int dig = (int)(floor( log(n) /                           log(base)) + 1);            // printing output     cout << "The Number of digits of "          << "Number " << n << " in base "          << base << " is " << dig; }   // Driver method  int main() {     // taking inputs     long n = 1446;     int base = 7;           // calling the method     findNumberOfDigits(n, base);     return 0; }   // This code is contributed by Manish Shaw  // (manishshaw1)

## Java

 // Java program to Find Number // of digits in base b. import java.io.*; public class GFG {           // function to print number of digits     static void findNumberOfDigits(long n, int base)     {                   // Calculating log using base changing         // property and then taking it          // floor and then adding 1.         int dig = (int)(Math.floor(                         Math.log(n) / Math.log(base))                         + 1);                              // printing output         System.out.println("The Number of digits of Number "                             + n + " in base " + base                              + " is " + dig);     }       // Driver method         public static void main(String[] args)     {         // taking inputs         long n = 1446;         int base = 7;                   // calling the method         findNumberOfDigits(n, base);     } }

## Python3

 # Python3 program to Find Number of digits  # in base b.   import math   # function to print number of # digits def findNumberOfDigits(n, base):           # Calculating log using base     # changing property and then     # taking it floor and then      # adding 1.     dig = (math.floor(math.log(n) /                  math.log(base)) + 1)           # printing output     print ("The Number of digits of"       " Number {} in base {} is {}"             . format(n, base, dig))   # Driver method    # taking inputs n = 1446 base = 7   # calling the method findNumberOfDigits(n, base)   # This code is contributed by  # Manish Shaw (manishshaw1)

## C#

 // C# program to Find Number of digits  // in base b. using System;   class GFG {           // function to print number of     // digits     static void findNumberOfDigits(long n,                                     int b)     {         // Calculating log using base         // changing property and then         // taking it floor and then          // adding 1.         int dig = (int)(Math.Floor(           Math.Log(n) / Math.Log(b)) + 1);                    // printing output         Console.Write("The Number of digits"            + " of Number " + n + " in base "                         + b + " is " + dig);     }       // Driver method      public static void Main()     {         // taking inputs         long n = 1446;         int b = 7;                   // calling the method         findNumberOfDigits(n, b);     } }   // This code is contributed by Manish Shaw  // (manishshaw1)

## PHP

 

## Javascript

 

Output :

The Number of digits of Number 1446 in base 7 is 4

Complexity Analysis:
Time complexity : O(logN)
Space complexity : O(1)

Optimized Approach:

Here are few optimizations that can be made to the given program:

• Avoid using math.h library: The log function provided by the math.h library is a costly operation. Instead of using it, we can use the base changing property of logarithm and use log10 function to calculate the logarithm of n in base b. This will reduce the overhead of including the math.h library.
• Replace floor function with casting: Instead of using the floor function, we can cast the result of the logarithmic operation to an integer type. This is faster and simpler than calling the floor function.

Here’s the optimized version of the program with the above optimizations:

## C++

 #include  using namespace std;   void findNumberOfDigits(long n, int base) {     int dig = 0;     while (n > 0) {         n /= base;         dig++;     }     cout << "The Number of digits of "          << "Number " << n << " in base "          << base << " is " << dig; }   int main() {     long n = 1446;     int base = 7;     findNumberOfDigits(n, base);     return 0; }

## Java

 import java.util.*;   public class Main {     public static void findNumberOfDigits(long n, int base) {         int dig = 0;         while (n > 0) {             n /= base;             dig++;         }         System.out.println("The Number of digits of Number " + n + " in base " + base + " is " + dig);     }       public static void main(String[] args) {         long n = 1446;         int base = 7;         findNumberOfDigits(n, base);     } }

## Python3

 def findNumberOfDigits(n, base):     dig = 0     while n > 0:         n //= base         dig += 1     print(f"The Number of digits of Number {n} in base {base} is {dig}")   n = 1446 base = 7 findNumberOfDigits(n, base)

## C#

 using System;   public class Program {     public static void FindNumberOfDigits(long n, int baseVal)     {         int dig = 0;         while (n > 0)         {             n /= baseVal;             dig++;         }         Console.WriteLine("The Number of digits of Number " + n + " in base " + baseVal + " is " + dig);     }       public static void Main()     {         long n = 1446;         int baseVal = 7;         FindNumberOfDigits(n, baseVal);     } }

## Javascript

 function findNumberOfDigits(n, base) {     let dig = 0;     while (n > 0) {         n = Math.floor(n / base);         dig++;     }     console.log(The Number of digits of Number ${n} in base${base} is \${dig}); }   let n = 1446; let base = 7; findNumberOfDigits(n, base);

Output :
The Number of digits of Number 1446 in base 7 is 4

Complexity Analysis:
Time complexity : O(logN)
Auxiliary Space : O(1)

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Previous
Next