Python Slicing | Extract ‘k’ bits from a given position

How to extract ‘k’ bits from a given position ‘p’ in a number? Examples:

Input : number = 171 k = 5 p = 2 Output : The extracted number is 21 171 is represented as 10101011 in binary, so, you should get only 10101 i.e. 21. Input : number = 72 k = 5 p = 1 Output : The extracted number is 8 72 is represented as 1001000 in binary, so, you should get only 01000 i.e 8.

We have existing solution for this problem please refer Extract ‘k’ bits from a given position in a number link. We can solve this problem quickly in python using slicing. Approach is simple,

1. Convert given number into it's binary using bin() function and remove first two characters '0b' from it, because bin function appends '0b' as prefix in output binary string.
2. We need to start extracting k bits from starting position p from right, that means end index of extracted sub-string will be end = (len(binary) - p) and start index will be start = end - k + 1 of original binary string.
3. Convert extracted sub-string into decimal again.

# Function to extract ‘k’ bits from a given # position in a number def extractKBits(num,k,p): # convert number into binary first binary = bin(num) # remove first two characters binary = binary[2:] end = len(binary) - p start = end - k + 1 # extract k bit sub-string kBitSubStr = binary[start : end+1] # convert extracted sub-string into decimal again print (int(kBitSubStr,2)) # Driver program if __name__ == "__main__": num = 171 k = 5 p = 2 extractKBits(num,k,p) 

Output:

21

Time Complexity: O(logn)
Space Complexity: O(logn)

Method#2: Using Bitwise Operators

Approach

This approach defines a function named extract_bits that takes three arguments: number, k, and p. It returns the decimal value of the k bits starting from position p (from right to left) in the binary representation of number. The function first right-shifts the number by p-1 bits to get the desired bits at the rightmost end of the number. Then, it creates a mask by left-shifting 1 by k bits and subtracting 1 to get a binary number consisting of k ones. It then bitwise-ANDs the shifted number with the mask to extract the k bits. Finally, it converts the extracted bits to a binary string and then to an integer to get the decimal value.

Algorithm

1. Right shift the given number by p-1 bits to get the desired bits at the rightmost end of the number.
2. Mask the rightmost k bits to get rid of any additional bits on the left.
3. Convert the resulting bits to decimal using the int() function.

def extract_bits(number, k, p): # Right shift the number by p-1 bits to get the desired bits at the rightmost end of the number shifted_number = number >> (p-1) # Mask the rightmost k bits to get rid of any additional bits on the left mask = (1 << k) - 1 extracted_bits = shifted_number & mask # Convert the extracted bits to decimal extracted_number = bin(extracted_bits)[2:] decimal_value = int(extracted_number, 2) return decimal_value number = 171 k = 5 p = 2 print(extract_bits(number, k, p)) 

21

Time Complexity: O(1)

Space Complexity: O(1)

Approach#3: Using lambda

This approach takes a decimal number num, the number of bits k, and the position of the bits p as input. It converts num to a binary string using the bin() function, removes the '0b' prefix from the string, and extracts a substring of length k starting from position p. Finally, the substring is converted back to an integer using the int() function with a base of 2.

Algorithm

1. Convert the decimal number to a binary string using bin().
2. Remove the '0b' prefix from the binary string.
3. Extract a substring of length k starting from position p.
4. Convert the substring back to an integer using int() with a base of 2.
5. Return the extracted integer.

extract_bits = lambda num, k, p: int(bin(num)[2:][p:p+k], 2) print(extract_bits(171, 5, 2)) 

21

Time complexity: O(1), The time complexity of bin() and int() is O(log n), where n is the input number, but the input num is limited to 32 bits (in most implementations of Python), so the time complexity can be considered constant.

Auxiliary Space: O(log n) The space complexity of bin() and int() is O(log n), where n is the input number, but the input num is limited to 32 bits (in most implementations of Python), so the space complexity can be considered constant.