Why Count Set Bits Brian Kernighan Algorithm in DSA C?

The Big Idea

What if you could count only the important bits and skip the rest, saving time and effort?

The Scenario

Imagine you have a long list of numbers, and you want to know how many bits are set to 1 in each number. Doing this by checking every bit one by one feels like counting every grain of sand on a beach by hand.

The Problem

Checking each bit manually means going through all bits even if most are zero. This is slow and wastes time, especially for big numbers. It's easy to make mistakes counting so many bits one by one.

The Solution

The Brian Kernighan Algorithm quickly jumps from one set bit to the next by turning off the rightmost set bit each time. This way, it only counts the bits that are 1, skipping all zeros, making the process much faster and less error-prone.

Before vs After

✗ Before

int countSetBits(int number) {
    int count = 0;
    while (number > 0) {
        if (number & 1) count++;
        number >>= 1;
    }
    return count;
}

✓ After

int countSetBits(int number) {
    int count = 0;
    while (number) {
        number &= (number - 1);
        count++;
    }
    return count;
}

What It Enables

This algorithm enables fast and efficient counting of set bits, which is essential in tasks like error detection, cryptography, and network data processing.

Real Life Example

In networking, counting set bits helps quickly find how many devices are active in a network mask, improving speed and accuracy in managing connections.

Key Takeaways

Manual bit counting is slow and error-prone.

Brian Kernighan's method skips zeros and counts only set bits.

This makes counting bits faster and more efficient.

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 digits in the binary representation.

C. It shifts the number right and counts zeros.

D. It converts the number to decimal and counts digits.

Solution

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

   The algorithm uses n = n & (n-1) to clear the least significant set bit in each iteration.

2. Step 2: Effect of clearing bits

   Each operation reduces the number of set bits 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 = It repeatedly clears the least significant set bit until the number becomes zero. [OK]

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

Common Mistakes:

• Counting zeros instead of ones
• Shifting without clearing bits
• Confusing decimal digits with bits

2. Which of the following is the correct syntax to clear the least significant set bit in C using Brian Kernighan's method?

easy

A. n = n & (n - 1);

B. n = n | (n - 1);

C. n = n & (n + 1);

D. n = n ^ (n - 1);

Solution

1. Step 1: Recall the bit clearing operation

   The Brian Kernighan algorithm clears the least significant set bit by n = n & (n - 1).

2. Step 2: Verify other options

   Options A, B, and D do not clear the least significant set bit correctly.

3. Final Answer:

   n = n & (n - 1); -> Option A

4. Quick Check:

   Use n & (n-1) to clear bits [OK]

Hint: Use n & (n-1) to clear least set bit [OK]

Common Mistakes:

• Using +1 instead of -1
• Using OR or XOR instead of AND
• Confusing bitwise operators

3. What is the output of the following C code snippet?

unsigned int n = 13; // binary 1101
int count = 0;
while (n) {
  n = n & (n - 1);
  count++;
}
printf("%d", count);

medium

A. 5

B. 2

C. 4

D. 3

Solution

1. Step 1: Analyze initial value and binary form

   13 in binary is 1101, which has three set bits (1s).

2. Step 2: Count iterations clearing bits

   Each loop clears one set bit: 1101 -> 1100 -> 1000 -> 0000, so loop runs 3 times.

3. Final Answer:

   3 -> Option D

4. Quick Check:

   Number of set bits in 13 = 3 [OK]

Hint: Count set bits by clearing one bit per loop [OK]

Common Mistakes:

• Counting total bits instead of set bits
• Off-by-one in loop count
• Misreading binary representation

4. Identify the error in this C code snippet that tries to count set bits using Brian Kernighan's algorithm:

int countSetBits(int n) {
  int count = 0;
  while (n != 0) {
    n = n & (n + 1);
    count++;
  }
  return count;
}

medium

A. Missing initialization of count variable.

B. Using n + 1 instead of n - 1 in bit clearing step.

C. Incorrect return type of the function.

D. Loop condition should be while (n > 0).

Solution

1. Step 1: Check the bit clearing operation

   The code uses n = n & (n + 1), which does not clear the least significant set bit correctly.

2. Step 2: Correct operation for clearing bits

   The correct operation is n = n & (n - 1) to clear the least significant set bit.

3. Final Answer:

   Using n + 1 instead of n - 1 in bit clearing step. -> Option B

4. Quick Check:

   Use n & (n-1) to clear bits, not n & (n+1) [OK]

Hint: Use n & (n-1) to clear bits, not n & (n+1) [OK]

Common Mistakes:

• Using +1 instead of -1
• Ignoring loop condition correctness
• Forgetting to initialize count

5. Given an array of unsigned integers, which approach using Brian Kernighan's algorithm is most efficient to find the total count of set bits in all elements?

hard

A. Shift each number bit by bit and count set bits using bitwise AND with 1.

B. Convert each number to binary string and count '1's using string functions.

C. For each element, repeatedly clear least significant set bit and count until zero, then sum counts.

D. Use a lookup table for each byte and sum counts for all bytes in each number.

Solution

1. Step 1: Identify Brian Kernighan's approach

   For each element, repeatedly clear least significant set bit and count until zero, then sum counts.: For each element, repeatedly clear the least significant set bit until zero and count (using n = n & (n-1)), then sum the counts across elements. This is the direct application of Brian Kernighan's algorithm.

2. Step 2: Why it is most efficient using Brian Kernighan

   B (strings) is slow; C (shifts) is O(word size) always; D (lookup) is fast but a separate method, not Brian Kernighan's. Brian Kernighan excels for sparse bits (O(set bits) time).

3. Final Answer:

   For each element, repeatedly clear least significant set bit and count until zero, then sum counts. -> Option C

4. Quick Check:

   Brian Kernighan on array: apply n & (n-1) loop per element [OK]

Hint: Brian Kernighan per element: clear LSB until zero, sum counts [OK]

Common Mistakes:

• Using string conversion which is slow
• Bit shifting each bit without optimization
• Confusing lookup table with Brian Kernighan method