Count Set Bits Brian Kernighan Algorithm in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the count variable to zero.

DSA Typescript

function countSetBits(n: number): number {
  let count = [1];
  while (n > 0) {
    n = n & (n - 1);
    count++;
  }
  return count;
}

Drag options to blanks, or click blank then click option'

A0

B1

Cn

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing count to 1 or n instead of 0.

The count variable must start at zero to count the number of set bits correctly.

2fill in blank

medium

Complete the code to update n by removing the rightmost set bit.

DSA Typescript

function countSetBits(n: number): number {
  let count = 0;
  while (n > 0) {
    n = n [1] (n - 1);
    count++;
  }
  return count;
}

Drag options to blanks, or click blank then click option'

A&

B|

C^

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using | or ^ instead of & causes wrong bit manipulation.

Using bitwise AND (&) between n and (n-1) removes the rightmost set bit.

3fill in blank

hard

Fix the error in the while loop condition to correctly count set bits.

DSA Typescript

function countSetBits(n: number): number {
  let count = 0;
  while (n [1] 0) {
    n = n & (n - 1);
    count++;
  }
  return count;
}

Drag options to blanks, or click blank then click option'

A==

B<

C>

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' or '==' causes infinite or no loops.

The loop should continue while n is greater than zero to count all set bits.

4fill in blank

hard

Fill both blanks to create a function that counts set bits using Brian Kernighan's algorithm.

DSA Typescript

function countSetBits(n: number): number {
  let count = [1];
  while (n [2] 0) {
    n = n & (n - 1);
    count++;
  }
  return count;
}

Drag options to blanks, or click blank then click option'

A0

B>

C!=

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting count at 1 or using wrong loop condition.

Initialize count to 0 and loop while n is greater than 0 to count all set bits.

5fill in blank

hard

Fill all three blanks to create a function that returns the count of set bits using Brian Kernighan's algorithm.

DSA Typescript

function countSetBits(n: number): number {
  let count = [1];
  while (n [2] 0) {
    n = n [3] (n - 1);
    count++;
  }
  return count;
}

Drag options to blanks, or click blank then click option'

A0

B>

C&

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Wrong initialization, loop condition, or bitwise operator.

Initialize count to 0, loop while n > 0, and update n with n & (n - 1) to remove rightmost set bit.

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 number.

C. It shifts the number to the right until it becomes zero without checking bits.

D. It adds all bits of the number without any bitwise operation.

Solution

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

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

2. Step 2: Effect of clearing bits

   Each clearing reduces the number of set bits by one until the number becomes zero, counting how many set bits were present.

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:

• Confusing bit count with digit count
• Using simple right shifts without bit checks
• Ignoring bitwise AND operation

2. Which of the following is the correct TypeScript syntax to clear the least significant set bit of a number n?

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: Identify the bitwise operation to clear least set bit

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

2. Step 2: Verify other options

   OR, XOR, and addition do not clear the least significant set bit correctly.

3. Final Answer:

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

4. Quick Check:

   AND with (n-1) clears least set bit [OK]

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

Common Mistakes:

• Using OR or XOR instead of AND
• Adding numbers instead of bitwise operations
• Missing parentheses around (n - 1)

3. What is the output of the following TypeScript code?

function countSetBits(n: number): number {
  let count = 0;
  while (n > 0) {
    n = n & (n - 1);
    count++;
  }
  return count;
}

console.log(countSetBits(13));

medium

A. 3

B. 4

C. 2

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 steps

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

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Number of 1s in 1101 = 3 [OK]

Hint: Count 1s in binary using n & (n-1) loop [OK]

Common Mistakes:

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

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

function countSetBits(n: number): number {
  let count = 0;
  while (n != 0) {
    n = n & n - 1;
    count++;
  }
  return count;
}

medium

A. Return statement is missing.

B. Using != instead of > in the while condition.

C. Incorrect initialization of count variable.

D. Missing parentheses around (n - 1) in the expression.

Solution

1. Step 1: Analyze the expression inside the loop

   The expression n = n & n - 1 lacks parentheses, so it is evaluated as (n & n) - 1, which is incorrect.

2. Step 2: Correct expression

   It should be n = n & (n - 1) to correctly clear the least significant set bit.

3. Final Answer:

   Missing parentheses around (n - 1) in the expression. -> Option D

4. Quick Check:

   Parentheses ensure correct order of operations [OK]

Hint: Use parentheses: n = n & (n - 1) [OK]

Common Mistakes:

• Using != 0 causing potential infinite loop for negatives
• Assuming both conditions always equivalent
• Operator precedence confusion causing wrong calculation

5. Given an array of numbers [5, 7, 10], which TypeScript code correctly uses Brian Kernighan's algorithm to return an array of their set bit counts?

hard

A. const counts = arr.map(n => {
let count = 0;
while (n > 0) {
n = n >> 1;
count++;
}
return count;
});

B. const counts = arr.map(n => n.toString(2).length);

C. const counts = arr.map(n => {
let count = 0;
while (n > 0) {
n = n & (n - 1);
count++;
}
return count;
});

D. const counts = arr.map(n => {
let count = 0;
for (let i = 0; i < 32; i++) {
if ((n & (1 << i)) !== 0) count++;
}
return count;
});

Solution

1. Step 1: Understand Brian Kernighan's algorithm usage

   const counts = arr.map(n => { let count = 0; while (n > 0) { n = n & (n - 1); count++; } return count; }); uses the correct loop with n = n & (n - 1) to count set bits for each number.

2. Step 2: Check other options for correctness

   const counts = arr.map(n => n.toString(2).length); counts total binary digits, not set bits. const counts = arr.map(n => { let count = 0; while (n > 0) { n = n >> 1; count++; } return count; }); counts total bits by shifting, not set bits. const counts = arr.map(n => { let count = 0; for (let i = 0; i < 32; i++) { if ((n & (1 << i)) !== 0) count++; } return count; }); uses a loop checking each bit, which works but is not Brian Kernighan's algorithm.

3. Final Answer:

   const counts = arr.map(n => { let count = 0; while (n > 0) { n = n & (n - 1); count++; } return count; }); -> Option C

4. Quick Check:

   Use n & (n-1) loop inside map for set bits [OK]

Hint: Use map with n & (n-1) loop for each number [OK]

Common Mistakes:

• Counting binary string length instead of set bits
• Using right shift without bit check
• Confusing bitwise loop with simple iteration