Count Set Bits Brian Kernighan Algorithm in DSA Go - 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 number's binary form.

How does the time taken grow as the number gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func countSetBits(n int) int {
    count := 0
    for 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 using Brian Kernighan's method.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The loop runs and clears the rightmost set bit each time.
• How many times: The loop runs once for each set bit in n.

How Execution Grows With Input

Each loop iteration 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 (binary 1010)	2
100 (binary 1100100)	3
1000 (binary 1111101000)	6

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

Final Time Complexity

Time Complexity: O(k) where k is the number of set bits in n.

This means the algorithm runs faster when there are fewer 1s, and slower when there are more 1s in the number.

Common Mistake

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

[OK] Correct: The loop only runs once per set bit, not for every bit position, so it depends on the count of 1s, not total bits.

Interview Connect

Knowing this algorithm helps you show efficient bit manipulation skills, which are often appreciated in coding interviews.

Self-Check

What if we changed the input to always have only one set bit? 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 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 until it becomes zero and counts the shifts

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 clears the least significant set bit by doing n = n & (n-1) repeatedly.

2. Step 2: Effect of clearing bits

   Each operation removes one '1' bit, counting how many set bits exist until n 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 counts set bits [OK]

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

Common Mistakes:

• Counting total bits instead of set bits
• Using string conversion instead of bitwise operations
• Shifting without checking bits

2. Which of the following is the correct syntax for the Brian Kernighan loop in Go to count set bits of an integer n?

easy

A. while n > 0 { count++; n = n - 1 }

B. for n > 0 { count += n & 1; n = n >> 1 }

C. for n != 0 { count++; n = n & (n - 1) }

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

Solution

1. Step 1: Identify the Brian Kernighan loop condition

   The loop runs while n is not zero, clearing the least significant set bit each iteration.

2. Step 2: Confirm the bit clearing operation

   The operation n = n & (n - 1) clears the least significant set bit correctly.

3. Final Answer:

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

4. Quick Check:

   Loop with n & (n-1) clears bits [OK]

Hint: Use n = n & (n-1) inside for loop [OK]

Common Mistakes:

• Using shift instead of bit clear
• Using while instead of for in Go
• Counting bits by subtracting 1 only

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

func countSetBits(n int) int {
    count := 0
    for n != 0 {
        n = n & (n - 1)
        count++
    }
    return count
}

fmt.Println(countSetBits(13))

medium

A. 2

B. 3

C. 4

D. 5

Solution

1. Step 1: Convert 13 to binary

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

2. Step 2: Apply Brian Kernighan algorithm

   Each iteration clears one set bit, so count increments 3 times until n becomes 0.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Binary 1101 has 3 ones [OK]

Hint: Count ones in binary 13 = 3 [OK]

Common Mistakes:

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

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

func countSetBits(n int) int {
    count := 0
    for n > 0 {
        count = count + (n & 1)
        n = n & (n - 1)
    }
    return count
}

medium

A. Using n & 1 to count bits conflicts with n = n & (n - 1) clearing bits

B. Loop condition should be n != 0 instead of n > 0

C. Count variable should be initialized to 1

D. Missing return statement

Solution

1. Step 1: Analyze counting method

   The code adds n & 1 to count, which checks the least significant bit.

2. Step 2: Analyze bit clearing step

   Then it clears the least significant set bit with n = n & (n - 1), which may skip bits.

3. Step 3: Conflict explanation

   Using both methods together causes incorrect counting because n & 1 and clearing bits are mixed incorrectly.

4. Final Answer:

   Using n & 1 to count bits conflicts with n = n & (n - 1) clearing bits -> Option A

5. Quick Check:

   Mixing bit check and clear causes wrong count [OK]

Hint: Use either bit check or bit clear, not both [OK]

Common Mistakes:

• Mixing bitwise AND with bit clearing
• Incorrect loop condition
• Wrong initialization of count

5. Given a list of integers [7, 8, 15, 23], which Go code snippet correctly uses Brian Kernighan's algorithm to return a map of each integer to its count of set bits?

hard

A. func countSetBits(n int) int {
count := 0
for n != 0 {
n = n & (n - 1)
count++
}
return count
}

func mapSetBits(nums []int) map[int]int {
result := make(map[int]int)
for _, num := range nums {
result[countSetBits(num)] = num
}
return result
}

B. func countSetBits(n int) int {
count := 0
for n > 0 {
count += n & 1
n = n >> 1
}
return count
}

func mapSetBits(nums []int) map[int]int {
result := make(map[int]int)
for i := 0; i < len(nums); i++ {
result[i] = countSetBits(nums[i])
}
return result
}

C. func countSetBits(n int) int {
count := 0
for n != 0 {
n = n & (n - 1)
count++
}
return count
}

func mapSetBits(nums []int) map[int]int {
result := make(map[int]int)
for i, num := range nums {
result[i] = countSetBits(num)
}
return result
}

D. func countSetBits(n int) int {
count := 0
for n != 0 {
n = n & (n - 1)
count++
}
return count
}

func mapSetBits(nums []int) map[int]int {
result := make(map[int]int)
for _, num := range nums {
result[num] = countSetBits(num)
}
return result
}

Solution

1. Step 1: Check countSetBits function correctness

   func countSetBits(n int) int { count := 0 for n != 0 { n = n & (n - 1) count++ } return count } func mapSetBits(nums []int) map[int]int { result := make(map[int]int) for _, num := range nums { result[num] = countSetBits(num) } return result } uses Brian Kernighan's algorithm correctly to count set bits.

2. Step 2: Check mapSetBits function logic

   The mapSetBits function correctly maps each number (as key) to its set bit count (as value).

3. Step 3: Verify other options

   func countSetBits(n int) int { count := 0 for n > 0 { count += n & 1 n = n >> 1 } return count } func mapSetBits(nums []int) map[int]int { result := make(map[int]int) for i := 0; i < len(nums); i++ { result[i] = countSetBits(nums[i]) } return result } uses shift method and wrong map keys; func countSetBits(n int) int { count := 0 for n != 0 { n = n & (n - 1) count++ } return count } func mapSetBits(nums []int) map[int]int { result := make(map[int]int) for i, num := range nums { result[i] = countSetBits(num) } return result } uses index as key; func countSetBits(n int) int { count := 0 for n != 0 { n = n & (n - 1) count++ } return count } func mapSetBits(nums []int) map[int]int { result := make(map[int]int) for _, num := range nums { result[countSetBits(num)] = num } return result } reverses key and value.

4. Final Answer:

   The code that maps num to countSetBits(num) -> Option D

5. Quick Check:

   Map keys are numbers, values are counts [OK]

Hint: Map number to countSetBits result using correct keys [OK]

Common Mistakes:

• Using index as map key instead of number
• Swapping keys and values in map
• Using shift method instead of Brian Kernighan