**π‘ Problem Formulation:** This article tackles the challenge of calculating the total number of set bits (bits with the value 1) in the binary representation of all numbers from 1 to an input number `n`

. For instance, if `n = 3`

, the binary representations are `1`

(1 set bit), `10`

(1 set bit), and `11`

(2 set bits), leading to a total of 4 set bits.

## Method 1: Brute Force Iteration

The brute force method involves iterating through each number from 1 to `n`

, converting it into binary, and counting the number of set bits. This is the most straightforward approach and easiest to implement, but it is not efficient for large values of `n`

.

Here’s an example:

def count_set_bits(n): count = 0 for i in range(1, n+1): count += bin(i).count('1') return count print(count_set_bits(3))

Output: 4

In the given code, we define a function `count_set_bits(n)`

that initializes a counter `count`

. It loops through numbers 1 to `n`

, converting each to binary using `bin()`

and counting ‘1’s with `count('1')`

. Finally, it returns the total count. This is simple but becomes slow as `n`

increases.

## Method 2: Brian Kernighan’s Algorithm

Brian Kernighan’s algorithm is a classic technique that efficiently counts the number of set bits by repeatedly flipping the least significant set bit of a number to 0 and counting the operations performed. This method is more efficient as it avoids checking every single bit.

Here’s an example:

def count_set_bits(n): total_count = 0 for i in range(1, n+1): count = 0 while i: i &= i-1 count += 1 total_count += count return total_count print(count_set_bits(3))

Output: 4

The function `count_set_bits(n)`

uses Brian Kernighan’s approach to iterate through each integer, using bitwise AND with the number and the number decremented by one. This operation turns off the rightmost set bit, and the loop count tracks the number of set bits. After looping, it adds this count to the total count for all numbers.

## Method 3: Dynamic Programming with Memorization

The dynamic programming approach utilizes the fact that the number of set bits in a number is 1 plus the number of set bits in the number with its least significant bit unset. This allows reuse of calculations by storing the results for previous numbers.

Here’s an example:

def count_set_bits(n): set_bits = [0] * (n+1) for i in range(1, n+1): set_bits[i] = set_bits[i & (i-1)] + 1 return sum(set_bits) print(count_set_bits(3))

Output: 4

Here, `count_set_bits(n)`

initializes an array `set_bits`

to store the count of set bits for all numbers up to `n`

. It updates the count for each number by adding 1 to the count of the number with its least significant bit turned off. The sum of the array is the total count.

## Method 4: Using Lookup Table

A lookup table can be used to improve performance. This method involves pre-computing the number of set bits for an 8-bit number and using this table to calculate set bits for larger numbers.

Here’s an example:

lookup_table = [bin(i).count('1') for i in range(256)] def count_set_bits(n): return sum(lookup_table[i & 0xff] + lookup_table[(i >> 8) & 0xff] for i in range(1, n+1)) print(count_set_bits(3))

Output: 4

The function `count_set_bits(n)`

utilizes a pre-built `lookup_table`

for 8-bit numbers. It iterates through the range 1 to `n`

, applying bitwise AND with 0xff and right shifting to find set bits using the table, summing them for the total count. Efficient for 32-bit integers.

## Bonus One-Liner Method 5: Using Python’s Built-in Functions

This succinct method uses Python’s built-in `sum()`

and `bin()`

functions in a one-liner to count the number of set bits from 1 to `n`

. It’s a compact and readable approach, though not the most performant.

Here’s an example:

print(sum(bin(i).count('1') for i in range(1, n+1)))

Output: 4 for `n = 3`

The one-liner sums the count of ‘1’s in the binary representation of each number from 1 to `n`

. While its clarity is an advantage, it does not offer any performance improvements over the brute force method.

## Summary/Discussion

**Method 1: Brute Force Iteration.**It is simple to implement. However, it has poor performance for larger numbers as it checks every bit of every number.**Method 2: Brian Kernighan’s Algorithm.**More efficient than brute force for large`n`

. But it still processes each number individually.**Method 3: Dynamic Programming with Memorization.**Utilizes previously calculated counts for efficiency. May use more memory for the set bits array.**Method 4: Using Lookup Table.**Provides significant speedup for sets of numbers. Requires initial space and time to generate the lookup table.**Method 5: One-Liner with Built-in Functions.**Very readable code. Performance is equivalent to the brute force method and not suited for a very large`n`

.