Set Clear Toggle a Specific Bit in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to set the bit at position 3 in the integer 'num'.

DSA Java

int result = num [1] (1 << 3);

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

A^

B&

C|

D<<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '&' which clears bits instead of setting.

Using '^' which toggles bits instead of setting.

Using '<<' which is a shift operator, not a bitwise operator.

Using the bitwise OR operator '|' sets the bit at position 3 without changing other bits.

2fill in blank

medium

Complete the code to clear the bit at position 2 in the integer 'num'.

DSA Java

int result = num [1] ~(1 << 2);

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

A|

B^

C~

D&

Attempts:

3 left

💡 Hint

Common Mistakes

Using '|' which sets bits instead of clearing.

Using '^' which toggles bits instead of clearing.

Using '~' alone which inverts all bits.

Using '&' with the negated mask clears the bit at position 2 while keeping others unchanged.

3fill in blank

hard

Fix the error in the code to toggle the bit at position 1 in 'num'.

DSA Java

int result = num [1] (1 << 1);

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

A^

B&

C~

D|

Attempts:

3 left

💡 Hint

Common Mistakes

Using '&' which clears bits instead of toggling.

Using '|' which sets bits instead of toggling.

Using '~' which inverts all bits.

The XOR '^' operator toggles the bit at position 1: if it was 0, it becomes 1; if 1, it becomes 0.

4fill in blank

hard

Fill both blanks to set the bit at position 'pos' in 'num'.

DSA Java

int result = num [1] (1 [2] pos);

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

A|

B&

C<<

D>>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '&' instead of '|' to set bits.

Using '>>' which shifts bits right, not left.

Mixing up the order of operators.

Shift 1 left by 'pos' bits using '<<' to create the mask, then OR '|' with 'num' to set that bit.

5fill in blank

hard

Fill all three blanks to clear the bit at position 'pos' in 'num'.

DSA Java

int result = num [1] ~ (1 [2] [3]);

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

A|

B&

C<<

Dpos

Attempts:

3 left

💡 Hint

Common Mistakes

Using '|' instead of '&' to clear bits.

Using '>>' instead of '<<' for shifting.

Forgetting to negate the mask.

Shift 1 left by 'pos' bits using '<<', then negate '~' to create mask, and AND '&' with 'num' to clear the bit.

Practice

1. What does the operation num | (1 << k) do to the integer num in terms of its bits?

easy

A. Toggles the k-th bit of num

B. Clears the k-th bit of num to 0

C. Sets the k-th bit of num to 1

D. Checks if the k-th bit of num is set

Solution

1. Step 1: Understand the bit shift operation

   1 << k creates a number with only the k-th bit set to 1 and all others 0.

2. Step 2: Apply bitwise OR with num

   Using num | (1 << k) sets the k-th bit of num to 1 without changing other bits.

3. Final Answer:

   Sets the k-th bit of num to 1 -> Option C

4. Quick Check:

   Set bit = OR with 1 shifted [OK]

Hint: Use OR with 1 shifted to set a bit [OK]

Common Mistakes:

• Confusing OR with AND operation
• Thinking it clears or toggles the bit
• Misunderstanding bit positions starting at 0

2. Which of the following Java statements correctly clears (sets to 0) the 3rd bit (0-based) of an integer variable num?

easy

A. num = num & (1 << 3);

B. num = num & ~(1 << 3);

C. num = num ^ (1 << 3);

D. num = num | (1 << 3);

Solution

1. Step 1: Understand clearing a bit

   To clear a bit, we AND with a mask that has 0 at that bit and 1s elsewhere.

2. Step 2: Create mask and apply

   ~(1 << 3) creates a mask with all 1s except 0 at bit 3. ANDing num with this clears bit 3.

3. Final Answer:

   num = num & ~(1 << 3); -> Option B

4. Quick Check:

   Clear bit = AND with NOT mask [OK]

Hint: Clear bit by AND with NOT of shifted 1 [OK]

Common Mistakes:

• Using OR instead of AND to clear bits
• Using XOR which toggles bits
• Using AND with shifted 1 which keeps only that bit

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

int num = 10; // binary 1010
int k = 1;
num = num ^ (1 << k);
System.out.println(num);

medium

A. 10

B. 14

C. 12

D. 8

Solution

1. Step 1: Understand initial value and bit positions

   num = 10 in binary is 1010. Bit positions (0-based) from right: bit0=0, bit1=1, bit2=0, bit3=1.

2. Step 2: Apply XOR to toggle bit 1

   1 << 1 is 0010. XOR toggles bit 1: 1010 ^ 0010 = 1000 (binary) which is 8 decimal.

3. Final Answer:

   8 -> Option D

4. Quick Check:

   XOR toggles bit 1: 10 -> 8 [OK]

Hint: XOR toggles the bit at position k [OK]

Common Mistakes:

• Confusing XOR with OR or AND
• Misreading bit positions starting at 1 instead of 0
• Incorrect binary to decimal conversion

4. The following Java code is intended to toggle the 2nd bit of num. What is the error?

int num = 5; // binary 0101
int k = 2;
num = num & (1 << k);
System.out.println(num);

medium

A. It clears all bits except the 2nd bit instead of toggling

B. It toggles the wrong bit

C. It causes a syntax error

D. It sets the 2nd bit to 1 always

Solution

1. Step 1: Analyze the operation used

   The code uses AND with 1 << k, which keeps only the k-th bit and clears others.

2. Step 2: Understand toggle operation

   Toggle requires XOR, not AND. AND here does not toggle but masks all bits except bit 2.

3. Final Answer:

   It clears all bits except the 2nd bit instead of toggling -> Option A

4. Quick Check:

   Toggle = XOR, not AND mask [OK]

Hint: Toggle bits using XOR, not AND [OK]

Common Mistakes:

• Using AND instead of XOR to toggle bits
• Confusing bit positions
• Expecting output without understanding bitwise operations

5. Given an integer num, write a Java expression to toggle the 0th bit only if it is currently set (1), otherwise leave num unchanged. Which expression achieves this?

hard

A. num = (num & (1 << 0)) != 0 ? num ^ (1 << 0) : num;

B. num = num & ~(1 << 0);

C. num = num ^ (1 << 0);

D. num = num | (1 << 0);

Solution

1. Step 1: Check if 0th bit is set

   Use (num & (1 << 0)) != 0 to test if bit 0 is 1.

2. Step 2: Toggle bit 0 only if set

   If bit 0 is set, toggle it with XOR; else keep num unchanged.

3. Final Answer:

   num = (num & (1 << 0)) != 0 ? num ^ (1 << 0) : num; -> Option A

4. Quick Check:

   Conditional toggle only if bit set [OK]

Hint: Use conditional check before toggling bit [OK]

Common Mistakes:

• Toggling bit unconditionally
• Using OR which always sets bit
• Clearing bit without condition