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Count total bits in a number

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  • Difficulty Level : Basic
  • Last Updated : 28 Jun, 2022

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




# Function to get no of bits
# in binary representation
# of positive integer
 
def countBits(n):
 
    count = 0
    while (n):
        count += 1
        n >>= 1
         
    return count
 
# Driver program
i = 65
print(countBits(i))
 
# This code is contributed
# by Smitha

C#




/* Function to get no of bits
in binary representation of
positive integer */
using System;
 
class GFG
{
    static int countBits(int n)
    {
        int count = 0;
        while (n != 0)
        {
            count++;
            n >>= 1;
        }
         
        return count;
    }
     
    // Driver Code
    static public void Main ()
    {
        int i = 65;
        Console.Write(countBits(i));
    }
}
 
// This code is contributed
// by akt_mit.

PHP




<?php
// PHP Code to get no of bits in binary
// representation of positive integer
 
// Function to get no of bits in binary
// representation of positive integer
function countBits($n)
{
    $count = 0;
    while ($n)
    {
        $count++;
        $n >>= 1;
    }
    return $count;
}
 
// Driver Code
$i = 65;
echo(countBits($i));
 
// This code is contributed by Ajit.
?>

Javascript




<script>
 
/* Function to get no of bits
in binary representation of
positive integer */
function countBits(n)
{
    var count = 0;
    while (n != 0)
    {
        count++;
        n >>= 1;
    }
     
    return count;
}
 
// Driver Code
var i = 65;
document.write(countBits(i));
 
</script>

Output

7

Time Complexity : O(logn)

Auxiliary Space : O(1)

Method 3 ( Using conversion from binary to string)

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




# function to count the number of bits in a number n
def count_bits(n):
  # bin(n) returns a binary string representation of n preceded by '0b' in python
  binary = bin(n)
   
  # we did -2 from length of binary string to ignore '0b'
  return len(binary)-2
 
a = 65
b = 183
 
print(f"Total bits in {a}: {count_bits(a)}")
print(f"Total bits in {b}: {count_bits(b)}")
 
# This code is contributed by udit

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 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)

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