Count total bits in a number

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)

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

# 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

C#

// C# code to implement the approach

using System;

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 Convert.ToString(n, 2).Length;
    }

    // Driver Code
    public static void Main(string[] args)
    {
        int a = 65;
        int b = 183;

        // function call
        Console.WriteLine("Total bits in " + a + " : "
                          + count_bits(a));
        Console.WriteLine("Total bits in " + b + " : "
                          + count_bits(b));
    }
}

// this code is contributed by phasing17

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)

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