Why Bit Manipulation and When It Beats Arithmetic in DSA Javascript - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to double a number using bit manipulation.

DSA Javascript

const double = (num) => num [1] 1;

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

A-

B>>

C<<

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using right shift (>>) instead of left shift (<<).

Using addition (+) which works but is not bit manipulation.

Using the left shift operator (<<) by 1 moves all bits one place to the left, effectively multiplying the number by 2.

2fill in blank

medium

Complete the code to check if a number is even using bit manipulation.

DSA Javascript

const isEven = (num) => (num [1] 1) === 0;

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

A^

B&

C|

D>>

Attempts:

3 left

💡 Hint

Common Mistakes

Using bitwise OR (|) which does not check parity.

Using right shift (>>) which changes the number.

The bitwise AND (&) with 1 checks the least significant bit. If it is 0, the number is even.

3fill in blank

hard

Fix the error in the code to halve a number using bit manipulation.

DSA Javascript

const half = (num) => num [1] 1;

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

A<<

B+

C-

D>>

Attempts:

3 left

💡 Hint

Common Mistakes

Using left shift (<<) which doubles the number instead of halving.

Using addition (+) which does not halve the number.

Right shifting (>>) by 1 moves bits one place to the right, effectively dividing the number by 2 and discarding the remainder.

4fill in blank

hard

Fill both blanks to toggle the third bit of a number (bit index 2).

DSA Javascript

const toggleThirdBit = (num) => num [1] (1 [2] 2);

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

A^

B<<

C&

D|

Attempts:

3 left

💡 Hint

Common Mistakes

Using AND (&) which clears bits instead of toggling.

Using OR (|) which sets bits instead of toggling.

We use left shift (<<) to create a mask with 1 at bit 2, then XOR (^) toggles that bit in the number.

5fill in blank

hard

Fill all three blanks to create a mask that sets bits 0 and 3, then apply it to a number.

DSA Javascript

const setBits = (num) => num [1] ([2] [3] 3);

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

A|

B1

C<<

D&

Attempts:

3 left

💡 Hint

Common Mistakes

Using AND (&) which clears bits instead of setting.

Using XOR (^) which toggles bits instead of setting.

We create a mask by shifting 1 left by 3 and OR it with 1 (bit 0). Then OR (|) applies the mask to set those bits in num.

Practice

1. Why is bit manipulation often faster than arithmetic operations like multiplication or division?

easy

A. Because bit manipulation works directly on binary digits using CPU instructions

B. Because arithmetic operations use less memory

C. Because bit manipulation requires more CPU cycles

D. Because arithmetic operations are always done in parallel

Solution

1. Step 1: Understand how bit manipulation works

   Bit manipulation uses CPU instructions that directly change bits (0s and 1s) in numbers.

2. Step 2: Compare with arithmetic operations

   Arithmetic like multiplication or division involves more complex calculations and takes more CPU cycles.

3. Final Answer:

   Because bit manipulation works directly on binary digits using CPU instructions -> Option A

4. Quick Check:

   Bit manipulation = faster CPU-level operations [OK]

Hint: Bit ops use CPU bits directly, so they run faster [OK]

Common Mistakes:

• Thinking arithmetic uses less memory
• Believing bit manipulation is slower
• Assuming arithmetic is always parallel

2. Which of the following is the correct JavaScript syntax to check if the 3rd bit (from right, zero-based) of a number n is set using bit manipulation?

easy

A. if ((n & (1 << 3)) !== 0) { /* bit is set */ }

B. if ((n | (1 << 3)) === 0) { /* bit is set */ }

C. if ((n & (3 << 1)) !== 0) { /* bit is set */ }

D. if ((n ^ (1 << 3)) !== 0) { /* bit is set */ }

Solution

1. Step 1: Understand bit checking syntax

   To check if a bit is set, use bitwise AND (&) with 1 shifted left by the bit position.

2. Step 2: Apply to 3rd bit (zero-based)

   Shift 1 by 3 places: (1 << 3). Then check if (n & (1 << 3)) is not zero.

3. Final Answer:

   if ((n & (1 << 3)) !== 0) { /* bit is set */ } -> Option A

4. Quick Check:

   Bit check = (n & (1 << 3)) !== 0 [OK]

Hint: Use & with 1 shifted to bit position to check bit [OK]

Common Mistakes:

• Using bitwise OR (|) instead of AND (&)
• Shifting wrong number of bits
• Using XOR (^) which toggles bits

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

let x = 5; // binary 0101
let y = x << 2;
console.log(y);

medium

A. 10

B. 40

C. 20

D. 8

Solution

1. Step 1: Understand left shift operation

   Left shifting by 2 moves bits 2 places left, multiplying the number by 2^2 = 4.

2. Step 2: Calculate shifted value

   5 * 4 = 20, so y = 20.

3. Final Answer:

   20 -> Option C

4. Quick Check:

   5 << 2 = 20 [OK]

Hint: Left shift by n multiplies by 2^n [OK]

Common Mistakes:

• Confusing left shift with right shift
• Forgetting shift multiplies by powers of two
• Assuming shift adds bits instead of moving

4. Find the error in this JavaScript code that tries to clear the 2nd bit of a number num:

num = num & (1 << 2);

medium

A. It should use bitwise OR (|) instead of AND (&)

B. It should use bitwise XOR (^) to clear the bit

C. There is no error; code correctly clears the bit

D. It should use bitwise AND with the negation of the bit mask

Solution

1. Step 1: Understand clearing a bit

   To clear a bit, AND with a mask that has all bits 1 except the target bit 0.

2. Step 2: Fix the mask

   Mask should be ~(1 << 2), so code should be: num = num & ~(1 << 2).

3. Final Answer:

   It should use bitwise AND with the negation of the bit mask -> Option D

4. Quick Check:

   Clear bit = num & ~(1 << bit) [OK]

Hint: Clear bit by AND with negated mask ~(1 << bit) [OK]

Common Mistakes:

• Using OR (|) which sets bits instead of clearing
• Using XOR (^) which toggles bits
• Not negating the mask before AND

5. You want to quickly check if a number n is even or odd without using modulus (%) or division. Which bit manipulation expression correctly does this and why is it faster than arithmetic?

hard

A. (n ^ 1) === 0 because XOR toggles the last bit

B. (n & 1) === 0 because it checks the least significant bit directly

C. (n | 1) === n because OR sets the last bit

D. (n >> 1) === 0 because shifting right divides by 2

Solution

1. Step 1: Understand even/odd check with bits

   The least significant bit (rightmost) is 0 for even numbers and 1 for odd numbers.

2. Step 2: Use bitwise AND to check LSB

   Expression (n & 1) extracts the last bit. If it equals 0, number is even.

3. Step 3: Why faster than arithmetic

   Bitwise AND is a simple CPU instruction, faster than modulus or division.

4. Final Answer:

   (n & 1) === 0 because it checks the least significant bit directly -> Option B

5. Quick Check:

   Even check = (n & 1) === 0 [OK]

Hint: Check even with (n & 1) === 0 for fast bit test [OK]

Common Mistakes:

• Using right shift (>>) which changes number value
• Using OR (|) which sets bits instead of checking
• Using XOR (^) which toggles bits