Count Set Bits Brian Kernighan Algorithm in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the Brian Kernighan algorithm do in the context of counting set bits?

It efficiently counts the number of 1s (set bits) in the binary representation of a number by repeatedly turning off the rightmost set bit until the number becomes zero.

intermediate

Explain why the operation n = n & (n - 1) removes the rightmost set bit.

Subtracting 1 from n flips all bits after the rightmost set bit including that bit. ANDing n with (n - 1) clears the rightmost set bit because it turns that bit from 1 to 0 while leaving other bits unchanged.

intermediate

What is the time complexity of the Brian Kernighan algorithm for counting set bits?

O(k), where k is the number of set bits in the number. This is faster than checking all bits if the number has few set bits.

beginner

Show the Rust code snippet for counting set bits using 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
}

beginner

Why is Brian Kernighan's algorithm preferred over checking each bit one by one?

Because it only loops as many times as there are set bits, making it faster especially for numbers with few set bits compared to checking all bits individually.

What does the expression n & (n - 1) do in Brian Kernighan's algorithm?

A Removes the rightmost set bit from n

B Adds 1 to n

C Flips all bits of n

D Checks if n is zero

The operation n & (n - 1) clears the rightmost set bit in n.

What is the main advantage of Brian Kernighan's algorithm for counting set bits?

A Uses recursion

B Runs in time proportional to the number of set bits

C Runs in constant time

D Checks all bits one by one

It loops only as many times as there are set bits, making it efficient.

In Rust, which keyword is used to declare a mutable variable?

A let

B var

C mut

D const

The keyword 'mut' makes a variable mutable in Rust.

What will be the output of count_set_bits(13) using Brian Kernighan's algorithm? (13 in binary is 1101)

A 1

B 2

C 4

D 3

13 has three set bits: 1, 1, 0, 1.

Which of these is NOT true about Brian Kernighan's algorithm?

A It counts unset bits

B It uses bitwise AND operation

C It modifies the input number

D It is efficient for sparse set bits

The algorithm counts set bits, not unset bits.

Describe how Brian Kernighan's algorithm counts the number of set bits in a number.

Think about how the number changes each time you remove a set bit.

Write the Rust function to count set bits using Brian Kernighan's algorithm and explain each step.

Focus on how the loop reduces 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