Count Set Bits Brian Kernighan Algorithm in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Brian Kernighan's Algorithm for Counting Set Bits

What is the output of this Rust code that counts set bits using Brian Kernighan's algorithm?

DSA Rust

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

fn main() {
    let number = 29; // binary: 11101
    println!("{}", count_set_bits(number));
}

A 3

B 4

C 5

D 6

💡 Hint

Count how many 1s are in the binary form of 29 (11101).

Brian Kernighan's algorithm repeatedly removes the rightmost set bit. For 29 (11101), the set bits are at positions 0, 2, 3, and 4, totaling 4.

❓ Predict Output

intermediate

Result of Counting Set Bits for Zero

What does the following Rust code print when counting set bits of zero using Brian Kernighan's algorithm?

DSA Rust

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

fn main() {
    let number = 0;
    println!("{}", count_set_bits(number));
}

A 0

B 1

C None

D Runtime error

💡 Hint

Zero has no set bits in binary.

Since zero has no bits set to 1, the count remains zero and the function returns 0.

🔧 Debug

advanced

Identify the Error in Brian Kernighan's Algorithm Implementation

What error will this Rust code produce when trying to count set bits using Brian Kernighan's algorithm?

DSA Rust

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

fn main() {
    let number = 15;
    println!("{}", count_set_bits(number));
}

A Logical error: infinite loop

B RuntimeError: integer overflow

C SyntaxError: missing semicolon

D Output: 4

💡 Hint

Check the line where n is updated inside the loop.

The line 'n = n & (n - 1)' is missing a semicolon at the end, causing a syntax error.

❓ Predict Output

advanced

Output of Counting Set Bits for Maximum 8-bit Value

What is the output of this Rust code counting set bits of 255 (all bits set in 8-bit) using Brian Kernighan's algorithm?

DSA Rust

fn count_set_bits(mut n: u8) -> u8 {
    let mut count = 0;
    while n > 0 {
        n &= n - 1;
        count += 1;
    }
    count
}

fn main() {
    let number = 255u8; // binary: 11111111
    println!("{}", count_set_bits(number));
}

A 8

B 7

C 0

D Runtime error

💡 Hint

255 in binary has all 8 bits set to 1.

255 in 8-bit binary is 11111111, which has 8 set bits, so the output is 8.

🧠 Conceptual

expert

Why Brian Kernighan's Algorithm is Efficient for Counting Set Bits

Which statement best explains why Brian Kernighan's algorithm is more efficient than checking each bit individually?

A It uses a lookup table for all possible bit patterns.

B It uses recursion to count bits faster than loops.

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

D It reduces the number of iterations to the number of set bits instead of total bits.

💡 Hint

Think about how many times the loop runs compared to the total bits.

Brian Kernighan's algorithm loops only as many times as there are set bits, skipping zero bits, making it faster especially for sparse bit patterns.

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