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Count total bits in a number

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Given a positive number n, count total bit in it.
Examples: 
 

Input : 13
Output : 4
Binary representation of 13 is 1101

Input  : 183
Output : 8

Input  : 4096
Output : 13


 


 

Method 1 (Using Log)


The log2(n) logarithm in base 2 of n, which is the exponent to which 2 is raised to get n only integer and we add 1 find total bit in a number in log(n) time.
 

C++

// C++ program to find total bit in given number
#include <iostream>    
#include <cmath>
  
unsigned countBits(unsigned int number)
{    
    
    // log function in base 2
    // take only integer part
    return (int)log2(number)+1;
}
  
// Driven program    
int main()
{
    unsigned int num = 65;
    std::cout<<countBits(num)<<'\n';
    return 0;
}
  
// This code is contributed by thedev05.

                    

C

// C program to find total bit in given number
#include <stdio.h>      
#include <math.h>
  
unsigned countBits(unsigned int number)
{      
      // log function in base 2 
      // take only integer part
      return (int)log2(number)+1;
}
  
// Driven program       
int main()
{
    unsigned int num = 65;
    printf("%d\n", countBits(num));
    return 0;

                    

Java

// Java program to 
// find total bit
// in given number
import java.io.*;
  
class GFG 
{
    static int countBits(int number)
    
          
        // log function in base 2 
        // take only integer part
        return (int)(Math.log(number) / 
                     Math.log(2) + 1);
    }
      
    // Driver code
    public static void main (String[] args) 
    {
        int num = 65;
          
        System.out.println(countBits(num));
                                  
    }
}
  
// This code is contributed by vij

                    

Python3

# Python3 program to find 
# total bit in given number
import math
def countBits(number):
      
    # log function in base 2 
    # take only integer part
    return int((math.log(number) / 
                math.log(2)) + 1);
  
# Driver Code
num = 65;
print(countBits(num));
  
# This code is contributed by mits

                    

C#

// C# program to find total bit
// in given number
using System;
  
class GFG {
      
    static uint countBits(uint number)
    {     
          
        // log function in base 2 
        // take only integer part
        return (uint)Math.Log(number , 2.0) + 1;
    }
      
    // Driver code
    public static void Main() 
    {
        uint num = 65;
          
        Console.WriteLine(countBits(num));
                                  
    }
}
  
// This code is contributed by Sam007.

                    

PHP

<?php
// PHP program to find total
// bit in given number
  
function countBits($number)
      
    // log function in base 2 
    // take only integer part
    return (int)(log($number) / 
                   log(2)) + 1;
}
  
// Driver Code
$num = 65;
echo(countBits($num));
  
// This code is contributed by Ajit.
?>

                    

Javascript

<script>
// JavaScript program to find total bit in given number 
  
    function countBits(number) {       
      // log function in base 2  
      // take only integer part 
      return Math.floor(Math.log2(number)+1); 
    
    
    // Driven program        
  
    let num = 65; 
    document.write(countBits(num)); 
   
// This code is contributed by Surbhi Tyagi 
</script>

                    

Output
7


 Time Complexity : O(logn)

Auxiliary Space : O(1)

Method 2 (Using Bit Traversal)


 

C

/* Function to get no of bits in binary
   representation of positive integer */
#include <stdio.h>         
     
unsigned int countBits(unsigned int n)
{
   unsigned int count = 0;
   while (n)
   {
        count++;
        n >>= 1;
    }
    return count;
}
   
/* Driver program*/
int main()
{
    int i = 65;
    printf("%d", countBits(i));
    return 0;
}

                    

Java

/* Function to get no of bits in binary
representation of positive integer */
class GFG {
  
    static int countBits(int n)
    {
        int count = 0;
        while (n != 0)
        {
            count++;
            n >>= 1;
        }
          
        return count;
    }
      
    /* Driver program*/
    public static void main(String[] arg)
    {
        int i = 65;
        System.out.print(countBits(i));
    }
}
  
// This code is contributed by Smitha.

                    

Python3


                    

C#


                    

PHP


                    

Javascript


                    

Output
7

Time Complexity : O(logn)

Auxiliary Space : O(1)

Method 3 ( Using conversion from binary to string)

C++

// C++ program to implement the approach
#include <bits/stdc++.h>
using namespace std;
  
// function to count the number of bits in a number n
int count_bits(int n)
{
  
  // to_string() returns the binary string 
  // representation of the number n
  string binary = bitset< 64 >(n).to_string();
    
  // returning the length of the binary string
  return 64 - binary.find('1');
}
  
int main()
{
    int a = 65;
    int b = 183;
  
    cout << "Total bits in " << a << " : " << count_bits(a) << endl;
    cout << "Total bits in " << b << " : " << count_bits(b) << endl;
}
  
// This code is contributed by phasing17

                    

Java

// Java code to implement the approach
class GFG {
  
    // function to count the number of bits in a number n
    static int count_bits(int n)
    {
        // return the length of the binary string
        return Integer.toBinaryString(n).length();
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        int a = 65;
        int b = 183;
  
        // function call
        System.out.printf("Total bits in %d: %d\n", a,
                          count_bits(a));
        System.out.printf("Total bits in %d: %d\n", b,
                          count_bits(b));
    }
}
  
// this code is contributed by phasing17

                    

Python3


                    

C#


                    

Javascript

// JavaScript program to implement the approach
  
// function to count the number of bits in a number n
function count_bits(n)
{
  
  // toString(2) returns the binary string 
  // representation of the number n
  let binary = n.toString(2);
    
  // returning the length of the binary string
  return binary.length;
}
  
let a = 65;
let b = 183;
  
console.log("Total bits in", a, ":", count_bits(a));
console.log("Total bits in", b, ":", count_bits(b));
  
// This code is contributed by phasing17

                    

Output
Total bits : 7
Total bits : 8

Time Complexity : O(logn)

Auxiliary Space : O(1)




 



Last Updated : 15 Jul, 2022
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