Python Bin | 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

We have existing solution for this problem please refer Count total bits in a number link.

Approach#1:  We can solve this problem quickly in Python using bin() function. Convert number into it's binary using bin() function and remove starting two characters '0b' of output binary string because bin function appends '0b' as prefix in output string. Now print length of binary string that will be the count of bits in binary representation of input number.
 

# Function to count total bits in a number

def countTotalBits(num):
    
     # convert number into it's binary and 
     # remove first two characters .
     binary = bin(num)[2:]
     print(len(binary)) 

# Driver program
if __name__ == "__main__":
    num = 13
    countTotalBits(num)

4

Approach#2: We can solve this problem by simply using bit_length function of integer. This function returns the number of bits required to represent the number in binary form. 

# Function to count total bits in a number

def countTotalBits(num):
    
     # bit_length function return 
     # total bits in number 
     B_len = num.bit_length()
     print("Total bits in binary are : ", B_len) 

# Driver program
if __name__ == "__main__":
    num = 13
    countTotalBits(num)

Total bits in binary are :  4

Approach#3: Using math

approach to count the total number of bits in a number is to use the logarithmic function log2() from the math module. Since the logarithmic function base 2 tells us the number of bits required to represent a number in binary, we can use it to calculate the total number of bits in a number.

Algorithm

1. Import the math module.
2. Define a function count_bits that takes an integer num as input.
3. Calculate the logarithm base 2 of the input number using the log2() function from the math module.
4. Add 1 to the result of the logarithm calculation to account for the sign bit (if the number is negative).
5. Return the integer value of the result.

import math

def count_bits(num):
    return int(math.log2(num)) + 1

num=13
print(count_bits(num))

4

Time complexity:
The time complexity of this function is O(1) because the log2() function and the addition operation both have constant time complexity.

Auxiliary Space:
The space complexity of this function is O(1) because the only extra space used is for storing the integer value of the result, which has a constant size regardless of the input size.

METHOD 4:Using defaultdict

APPRAOCH:

 This program counts the total number of bits in a given number using a defaultdict data structure.

ALGORITHM:

1.Convert the given number to its binary representation using the built-in bin() function.
2.Create a defaultdict to count the number of '0's and '1's in the binary representation.
3.Iterate over the binary representation and increment the corresponding count in the defaultdict.
4.Add the counts of '0's and '1's to get the total number of bits in the binary representation.

from collections import defaultdict

# Function to count the number of bits
def count_bits(num):
    binary = bin(num)[2:]
    bit_count = defaultdict(int)
    for bit in binary:
        bit_count[bit] += 1
    return bit_count['1'] + bit_count['0']


# Driver Code
num = 13
bits = count_bits(num)

print("Total bits in", num, ":", bits)

Total bits in 13 : 4

Time Complexity: O(log n), where n is the given number.

Space Complexity: O(log n), where n is the given number, due to the storage of the binary representation in memory and the creation of the defaultdict.