Why Count Set Bits Brian Kernighan Algorithm in DSA Java?

The Big Idea

What if you could count all the 1s in a number's binary form without checking every single bit?

The Scenario

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

The Problem

Manually checking each bit is slow and boring. It wastes time and can easily lead to mistakes, especially with big numbers or many numbers. This makes your program slow and frustrating.

The Solution

The Brian Kernighan Algorithm quickly jumps from one 1-bit to the next by removing the rightmost 1 each time. This clever trick skips all the 0s and counts only the 1s, making the process much faster and simpler.

Before vs After

✗ Before

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

✓ After

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

What It Enables

This algorithm lets you count set bits in numbers lightning fast, enabling efficient solutions in areas like cryptography, error detection, and data compression.

Real Life Example

When your phone processes signals or encrypts messages, it often needs to count set bits quickly to work smoothly and securely without delays.

Key Takeaways

Manual bit counting is slow and error-prone.

Brian Kernighan's method removes the rightmost set bit each step.

This makes counting set bits much faster and efficient.

Practice

1. What does the Brian Kernighan algorithm do when counting set bits in an integer?

easy

A. It counts the total number of digits in the integer

B. It converts the number to binary string and counts '0's

C. It shifts the bits to the right without changing the number

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

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 clearing 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 D

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 digits instead of set bits
• Shifting bits without counting
• Counting zeros instead of ones

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

easy

A. int count = 0; for(int i=0; i<32; i++) { if((n & (1 << i)) != 0) count++; }

B. int count = 0; while(n > 0) { n = n >> 1; count++; }

C. int count = 0; while(n > 0) { n = n & (n - 1); count++; }

D. int count = 0; while(n > 0) { n = n | (n - 1); count++; }

Solution

1. Step 1: Identify the correct bit clearing operation

   The Brian Kernighan algorithm uses n = n & (n - 1) to clear the least significant set bit.

2. Step 2: Verify the loop and count increment

   Loop runs while n > 0 and increments count each time a bit is cleared.

3. Final Answer:

   int count = 0; while(n > 0) { n = n & (n - 1); count++; } -> Option C

4. Quick Check:

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

Hint: Use n = n & (n - 1) inside loop to count bits [OK]

Common Mistakes:

• Using right shift instead of bit clearing
• Using bitwise OR instead of AND
• Counting bits by checking each bit manually

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

int n = 13; // binary 1101
int count = 0;
while (n > 0) {
  n = n & (n - 1);
  count++;
}
System.out.println(count);

medium

A. 2

B. 3

C. 4

D. 5

Solution

1. Step 1: Convert 13 to binary and count set bits

   13 in binary is 1101, which has three set bits (1,1,0,1).

2. Step 2: Apply Brian Kernighan algorithm iterations

   Each iteration clears one set bit: - Iteration 1: 1101 & 1100 = 1100 - Iteration 2: 1100 & 1011 = 1000 - Iteration 3: 1000 & 0111 = 0000 Loop runs 3 times, so count = 3.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Number of set bits in 13 = 3 [OK]

Hint: Count set bits in binary 1101 = 3 [OK]

Common Mistakes:

• Counting total bits instead of set bits
• Miscounting set bits in binary
• Confusing bitwise AND with OR

4. Identify the error in the following Java code that attempts 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. Using n + 1 instead of n - 1 in bit clearing step

B. Incorrect loop condition should be n > 0

C. Missing initialization of count variable

D. Using bitwise OR instead of AND

Solution

1. Step 1: Check the bit clearing operation

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

2. Step 2: Effect of incorrect operation

   Using n + 1 does not clear the least significant set bit and leads to incorrect counting or infinite loop.

3. Final Answer:

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

4. Quick Check:

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

Hint: Always use n & (n - 1) to clear bits [OK]

Common Mistakes:

• Using n + 1 instead of n - 1
• Wrong loop condition causing infinite loop
• Confusing AND with OR operators

5. Given an array of integers, which Java method correctly uses Brian Kernighan's algorithm to return the total count of set bits in all elements?

int totalSetBits(int[] arr) {
  int total = 0;
  for (int num : arr) {
    while (num > 0) {
      num = num & (num - 1);
      total++;
    }
  }
  return total;
}

What is the output of totalSetBits(new int[]{3, 5, 7})?

hard

A. 7

B. 6

C. 8

D. 5

Solution

1. Step 1: Count set bits for each element

   3 (binary 11) has 2 set bits, 5 (binary 101) has 2 set bits, 7 (binary 111) has 3 set bits.

2. Step 2: Sum all set bits

   Total = 2 + 2 + 3 = 7.

3. Step 3: Verify code logic

   The code correctly counts set bits for each number and sums them in total.

4. Final Answer:

   7 -> Option A

5. Quick Check:

   Sum of set bits in 3,5,7 = 7 [OK]

Hint: Sum set bits of each number using Brian Kernighan [OK]

Common Mistakes:

• Miscounting set bits in numbers
• Adding counts incorrectly
• Using wrong bit clearing logic