5 Best Ways to Count Set Bits in a Range with Python

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πŸ’‘ Problem Formulation: We often face the task of counting set bits (1s) in binary representations of numbers within a certain range. For example, given the range [2, 7], we would convert each number to binary (10, 11, 100, 101, 110, 111) and count all the 1s, resulting in 12. This article demonstrates 5 different methods to efficiently compute the number of set bits in a given range using Python.

Method 1: Brute Force Iteration

This method involves iterating through each number in the range, converting it to its binary form, and then counting the 1s. It’s straightforward but potentially inefficient for large ranges as it does a bit-count for each number individually.

Here’s an example:

def count_set_bits(start, end):
    total_count = 0
    for num in range(start, end + 1):
        total_count += bin(num).count('1')
    return total_count

print(count_set_bits(2, 7))

Output:

12

This code defines a function count_set_bits() that takes in the start and end of the range. It iterates through the range, converts each number to binary using bin(), and counts the ‘1’ characters with the count() string method, summing up to the total count.

Method 2: Brian Kernighan’s Algorithm

Brian Kernighan’s algorithm is an efficient way to count set bits of a number. The principle is to repeatedly flip the least significant set bit to zero and count the operations required. This method significantly reduces the number of operations compared to the brute force approach.

Here’s an example:

def count_set_bits_in_number(n):
    count = 0
    while n:
        n &= n - 1  # Flip the least significant set bit
        count += 1
    return count

def count_set_bits(start, end):
    return sum(count_set_bits_in_number(num) for num in range(start, end + 1))

print(count_set_bits(2, 7))

Output:

12

In this snippet, the count_set_bits_in_number() function implements Brian Kernighan’s algorithm, counting the set bits in a single number. The count_set_bits() function then uses this to count the set bits in a range, using a list comprehension and the sum() function to combine the counts.

Method 3: Dynamic Programming

Using dynamic programming, you can store the results of subproblems (count of set bits in smaller ranges) and use these to efficiently solve the larger problem. This avoids unnecessary recomputations, offering a speed boost when counting many ranges.

Here’s an example:

bit_counts = {0: 0}  # Base case: count of set bits for 0 is 0

def count_set_bits_to_n(n):
    if n in bit_counts:
        return bit_counts[n]
    highest_power_of_2 = n.bit_length() - 1
    # Number of bits set in [0, 2^highest_power_of_2)
    prev_count = count_set_bits_to_n((1 << highest_power_of_2) - 1)
    # Count of bits set in the remainder
    remainder_count = n - (1 << highest_power_of_2) + 1
    bit_counts[n] = prev_count + remainder_count + count_set_bits_to_n(n - (1 < 0 else 0)

print(count_set_bits(2, 7))

Output:

12

This code snippet makes use of dynamic programming by caching previously calculated counts in a dictionary bit_counts. The count_set_bits_to_n() function calculates the count up to n, factoring in the highest power of 2 within n and recursively solves for the next count. The main function, count_set_bits(), uses this helper to calculate the count for a range.

Method 4: Lookup Table

A lookup table can greatly speed up the counting process by precomputing the set bits for numbers within a smaller subset and using these precomputed values to find the set bits for larger numbers. This method is particularly effective for fixed-size bit representations.

Here’s an example:

PRECOMPUTED_BITS = [bin(i).count('1') for i in range(256)]

def count_set_bits_in_number(n):
    count = 0
    while n:
        count += PRECOMPUTED_BITS[n & 0xff]  # Lookup for the last 8 bits
        n >>= 8  # Move to the next 8 bits
    return count

def count_set_bits(start, end):
    return sum(count_set_bits_in_number(num) for num in range(start, end + 1))

print(count_set_bits(2, 7))

Output:

12

This example uses a precomputed list PRECOMPUTED_BITS that contains the count of set bits for numbers 0 through 255. The function count_set_bits_in_number() operates on 8 bits at a time using bitwise and and right shift operators, looking up the set bit count from PRECOMPUTED_BITS.

Bonus One-Liner Method 5: Built-in Python Function

Python’s standard library has a built-in function called popcount() (or bit_count() in older versions) in the int class, which returns the number of set bits of a number directly. This one-liner approach uses a built-in optimized function for counting set bits.

Here’s an example:

def count_set_bits(start, end):
    return sum(num.bit_count() for num in range(start, end + 1))

print(count_set_bits(2, 7))

Output:

12

In this brief code snippet, the count_set_bits() function makes use of Python’s bit_count() method on integers, which does the job of counting set bits in an optimized and concise manner. The sum is done over the range of interest, and the result is printed out.

Summary/Discussion

  • Method 1: Brute Force Iteration. Simple to understand and implement. Inefficient for large ranges due to its linear complexity.
  • Method 2: Brian Kernighan’s Algorithm. More efficient than brute force. Still has to process each number individually, which can be slow for large ranges.
  • Method 3: Dynamic Programming. Reduces the number of computations by caching the results. Complexity can become an issue for very large numbers.
  • Method 4: Lookup Table. Fast for fixed-size integers, with quick lookup times. Precomputation can take extra space and initialization time.
  • Bonus Method 5: Built-in Python Function. Most straightforward and efficient for Python users. Relies on Python’s built-in features, so less educational value for understanding bit manipulation.