Count Set Bits Brian Kernighan Algorithm in DSA Rust - Time & Space Complexity

Time Complexity: Count Set Bits Brian Kernighan Algorithm

O(k)

Understanding Time Complexity

We want to understand how fast the Brian Kernighan algorithm counts the number of 1s in a binary number.

How does the time it takes change as the number gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn count_set_bits(mut n: u32) -> u32 {
    let mut count = 0;
    while n != 0 {
        n &= n - 1; // Remove the rightmost set bit
        count += 1;
    }
    count
}
    

This code counts how many bits are set to 1 in the number by repeatedly removing the rightmost 1 bit.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop runs once for each set bit in the number.
• How many times: Equal to the number of 1 bits in the input number.

How Execution Grows With Input

The number of loop runs depends on how many 1s are in the number, not its size in bits.

Input Size (n)	Approx. Operations (loop runs)
10 (binary 1010)	2
100 (binary 1100100)	3
1000 (binary 1111101000)	6

Pattern observation: The loop count grows with the number of 1 bits, which can be much less than total bits.

Final Time Complexity

Time Complexity: O(k)

This means the time grows with the number of set bits k in the number, not the total bits.

Common Mistake

[X] Wrong: "The loop runs once for every bit in the number (always 32 or 64 times)."

[OK] Correct: The loop only runs as many times as there are 1 bits, which can be much fewer than total bits.

Interview Connect

Understanding this algorithm shows you can optimize bit counting by focusing on set bits, a useful skill in many coding problems.

Self-Check

"What if we changed the input to always have half of its bits set? How would the time complexity change?"

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