Count Set Bits Brian Kernighan Algorithm in DSA Rust - Build from Scratch

Count Set Bits using Brian Kernighan Algorithm

📖 Scenario: Imagine you are working with a system that needs to count how many lights are turned on in a row of switches. Each switch can be either off (0) or on (1). We represent the switches as bits in a number. Your task is to count how many switches are on using a smart method called Brian Kernighan's algorithm.

🎯 Goal: You will write a Rust program that counts the number of set bits (1s) in a given unsigned integer using Brian Kernighan's algorithm. This method efficiently counts bits by turning off the rightmost set bit one by one until the number becomes zero.

📋 What You'll Learn

Create a variable called num with the exact value 29u32

Create a variable called count and initialize it to 0

Use a while loop that runs while num is greater than zero

Inside the loop, update num to num & (num - 1) to turn off the rightmost set bit

Increment count by 1 inside the loop

Print the final value of count

💡 Why This Matters

🌍 Real World

Counting set bits is useful in network masks, permissions, and hardware control where bits represent states or flags.

💼 Career

Understanding bit manipulation and efficient algorithms is important for systems programmers, embedded developers, and performance engineers.

1

Create the number variable

Create a variable called num and set it to the unsigned 32-bit integer 29u32.

DSA Rust

# Create a variable called num and set it to 29u32
# Your code here

Hint

Use let mut num = 29u32; to create a mutable variable named num.

2

Initialize the count variable

Create a variable called count and initialize it to 0 inside the main function after num.

DSA Rust

fn main() {
    let mut num = 29u32;
    # Create a variable called count and set it to 0
    # Your code here
}

Hint

Use let mut count = 0; to create a mutable counter variable.

3

Implement Brian Kernighan's loop

Use a while loop that runs while num > 0. Inside the loop, update num to num & (num - 1) and increment count by 1.

DSA Rust

fn main() {
    let mut num = 29u32;
    let mut count = 0;
    # Use a while loop to count set bits using Brian Kernighan's algorithm
    # Your code here
}

Hint

The loop should run while num is greater than zero. Each time, turn off the rightmost set bit and increase count by one.

4

Print the count of set bits

Print the value of count using println! inside the main function after the loop.

DSA Rust

fn main() {
    let mut num = 29u32;
    let mut count = 0;
    while num > 0 {
        num = num & (num - 1);
        count += 1;
    }
    # Print the count variable
    # Your code here
}

Hint

Use println!("{}", count); to display the number of set bits.

Practice

1. What does the Brian Kernighan algorithm do when counting set bits in a number?

easy

A. It repeatedly clears the least significant set bit until the number becomes zero

B. It counts the total number of bits in the number

C. It shifts the number right until it becomes zero

D. It converts the number to a string and counts '1' characters

Solution

1. Step 1: Understand the algorithm's main operation

   The Brian Kernighan algorithm works by clearing the least significant set bit in each iteration.

2. Step 2: Effect of clearing bits

   Each time the least significant set bit is cleared, the count of set bits increases by one until the number becomes zero.

3. Final Answer:

   It repeatedly clears the least significant set bit until the number becomes zero -> Option A

4. Quick Check:

   Clearing bits one by one counts set bits [OK]

Hint: Clear least set bit repeatedly to count set bits [OK]

Common Mistakes:

• Counting total bits instead of set bits
• Using string conversion instead of bit operations
• Shifting without checking bits

2. Which of the following Rust code snippets correctly implements the Brian Kernighan algorithm to count set bits?

easy

A. fn count_set_bits(mut n: u32) -> u32 { let mut count = 0; while n > 0 { count += n & 1; n >>= 1; } count }

B. fn count_set_bits(n: u32) -> u32 { n.count_ones() }

C. fn count_set_bits(n: u32) -> u32 { n.trailing_zeros() }

D. fn count_set_bits(mut n: u32) -> u32 { let mut count = 0; while n != 0 { n &= n - 1; count += 1; } count }

Solution

1. Step 1: Identify the Brian Kernighan algorithm code

   fn count_set_bits(mut n: u32) -> u32 { let mut count = 0; while n != 0 { n &= n - 1; count += 1; } count } uses the loop with n &= n - 1 to clear the least significant set bit repeatedly.

2. Step 2: Verify other options

   fn count_set_bits(n: u32) -> u32 { n.count_ones() } uses built-in count_ones(), fn count_set_bits(mut n: u32) -> u32 { let mut count = 0; while n > 0 { count += n & 1; n >>= 1; } count } uses bit shifting, and fn count_set_bits(n: u32) -> u32 { n.trailing_zeros() } uses trailing_zeros(), which are not Brian Kernighan's method.

3. Final Answer:

   fn count_set_bits(mut n: u32) -> u32 { let mut count = 0; while n != 0 { n &= n - 1; count += 1; } count } -> Option D

4. Quick Check:

   n &= n - 1 loop is Brian Kernighan's method [OK]

Hint: Use n &= n - 1 in loop to clear bits [OK]

Common Mistakes:

• Using built-in functions instead of algorithm
• Shifting bits instead of clearing
• Using trailing_zeros which counts zeros, not ones

3. What is the output of the following Rust code?

fn main() {
    let mut n = 29u32; // binary 11101
    let mut count = 0;
    while n != 0 {
        n &= n - 1;
        count += 1;
    }
    println!("{}", count);
}

medium

A. 3

B. 4

C. 5

D. 6

Solution

1. Step 1: Convert 29 to binary and count set bits

   29 in binary is 11101, which has four set bits (1s).

2. Step 2: Apply Brian Kernighan algorithm steps

   Each iteration clears one set bit until n becomes zero, so count increments 4 times.

3. Final Answer:

   4 -> Option B

4. Quick Check:

   Number of 1s in 11101 = 4 [OK]

Hint: Count 1s in binary representation [OK]

Common Mistakes:

• Counting total bits instead of set bits
• Miscounting binary digits
• Confusing zero bits with set bits

4. Identify the error in this Rust implementation of Brian Kernighan's algorithm:

fn count_set_bits(mut n: u32) -> u32 {
    let mut count = 0;
    while n > 0 {
        n = n - 1;
        count += 1;
    }
    count
}

medium

A. It uses n > 0 instead of n != 0 in the loop condition

B. It uses a while loop instead of a for loop

C. It decrements n by 1 instead of clearing the least significant set bit

D. It does not initialize count to zero

Solution

1. Step 1: Analyze the loop operation on n

   The code subtracts 1 from n each time, which does not clear the least significant set bit correctly.

2. Step 2: Understand Brian Kernighan's method

   The correct operation is n &= n - 1 to clear the least significant set bit, not n = n - 1.

3. Final Answer:

   It decrements n by 1 instead of clearing the least significant set bit -> Option C

4. Quick Check:

   Use n &= n - 1 to clear bits, not n = n - 1 [OK]

Hint: Use bitwise AND to clear bits, not subtraction alone [OK]

Common Mistakes:

• Using subtraction instead of bitwise AND
• Wrong loop condition causing infinite loop
• Not incrementing count properly

5. Given a list of unsigned integers [7, 8, 15, 23], which Rust code using Brian Kernighan's algorithm correctly returns a vector of their set bit counts?

hard

A. fn count_bits_list(nums: &[u32]) -> Vec<u32> { nums.iter().map(|&n| { let mut x = n; let mut c = 0; while x != 0 { x &= x - 1; c += 1; } c }).collect() }

B. fn count_bits_list(nums: &[u32]) -> Vec<u32> { nums.iter().map(|&n| n.count_ones()).collect() }

C. fn count_bits_list(nums: &[u32]) -> Vec<u32> { nums.iter().map(|&n| n.trailing_zeros()).collect() }

D. fn count_bits_list(nums: &[u32]) -> Vec<u32> { nums.iter().map(|&n| n.leading_zeros()).collect() }

Solution

1. Step 1: Identify Brian Kernighan's algorithm in code

   fn count_bits_list(nums: &[u32]) -> Vec<u32> { nums.iter().map(|&n| { let mut x = n; let mut c = 0; while x != 0 { x &= x - 1; c += 1; } c }).collect() } uses the loop with x &= x - 1 to count set bits for each number.

2. Step 2: Check other options

   Options B, C, and D use built-in functions not related to Brian Kernighan's algorithm.

3. Step 3: Confirm output correctness

   fn count_bits_list(nums: &[u32]) -> Vec<u32> { nums.iter().map(|&n| { let mut x = n; let mut c = 0; while x != 0 { x &= x - 1; c += 1; } c }).collect() } returns a vector with counts of set bits for each number in the list.

4. Final Answer:

   fn count_bits_list(nums: &[u32]) -> Vec<u32> { nums.iter().map(|&n| { let mut x = n; let mut c = 0; while x != 0 { x &= x - 1; c += 1; } c }).collect() } -> Option A

5. Quick Check:

   Loop with x &= x - 1 counts bits per number [OK]

Hint: Map each number with Brian Kernighan loop to count bits [OK]

Common Mistakes:

• Using built-in count_ones instead of algorithm
• Using trailing_zeros or leading_zeros which count zeros
• Not collecting results into a vector