Count Set Bits Brian Kernighan Algorithm in DSA Java - 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 taken change as the number gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code counts how many bits are set to 1 in the binary form of the number n by repeatedly turning off the rightmost set bit.

Identify Repeating Operations

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

How Execution Grows With Input

Each loop removes one set bit, so the number of operations depends on how many 1s are in n.

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

Pattern observation: The time grows with the number of 1s, not the total bits or value size.

Final Time Complexity

Time Complexity: O(k)

This means the time depends on the count of set bits (k) in the number, not the total number of bits.

Common Mistake

[X] Wrong: "The loop runs for all bits of the number, so it is O(log n)."

[OK] Correct: The loop only runs once per set bit, so if the number has few 1s, it runs faster than checking every bit.

Interview Connect

Knowing this algorithm shows you can optimize bit counting by focusing on set bits, a useful skill for efficient low-level coding.

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 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