Bit Manipulation Basics AND OR XOR NOT Left Right Shift in DSA C# - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this bitwise AND operation?

Consider the following C# code snippet. What will be printed?

DSA C#

int a = 12;  // binary: 1100
int b = 10;  // binary: 1010
int c = a & b;
Console.WriteLine(c);

A 10

B 4

C 2

D 8

💡 Hint

Remember bitwise AND compares each bit of two numbers and returns 1 only if both bits are 1.

12 in binary is 1100 and 10 is 1010. Bitwise AND compares bits: 1&1=1, 1&0=0, 0&1=0, 0&0=0. So result is 1000 in binary which is 8 decimal.

Check carefully each bit position.

❓ Predict Output

intermediate

What is the result of this left shift operation?

What will be printed by this C# code?

DSA C#

int x = 5;  // binary: 0101
int y = x << 2;
Console.WriteLine(y);

A 10

B 20

C 40

D 80

💡 Hint

Left shift by 2 means multiply by 2 twice (multiply by 4).

5 in binary is 0101. Left shifting by 2 moves bits two places left: 0101 becomes 010100 which is 20 decimal.

❓ Predict Output

advanced

What is the output of this XOR operation?

Given the code below, what will be printed?

DSA C#

int p = 14;  // binary: 1110
int q = 9;   // binary: 1001
int r = p ^ q;
Console.WriteLine(r);

A 3

B 15

C 7

D 5

💡 Hint

XOR returns 1 only when bits differ.

14 (1110) XOR 9 (1001) compares bits: 1^1=0, 1^0=1, 1^0=1, 0^1=1. Result is 0111 binary which is 7 decimal.

❓ Predict Output

advanced

What is the output after applying NOT operator?

What will this C# code print? Assume 32-bit integers.

DSA C#

int n = 0;
int m = ~n;
Console.WriteLine(m);

A -1

B 0

C 1

D 2147483647

💡 Hint

NOT flips all bits. For 0, all bits are 0, flipping gives all 1s which is -1 in two's complement.

0 in binary is all zeros. Applying NOT flips all bits to 1. In 32-bit signed integers, this is -1.

❓ Predict Output

expert

What is the output of this combined bitwise operation?

What will be printed by this C# code snippet?

DSA C#

int a = 6;   // binary: 0110
int b = 3;   // binary: 0011
int c = (a | b) & (~a << 1);
Console.WriteLine(c);

A 2

B 4

C 6

D 0

💡 Hint

Break down each operation step by step: OR, NOT, left shift, then AND.

a | b = 0110 | 0011 = 0111 (7 decimal)

~a = ~0110 = 1001 (in 4 bits, but in 32-bit signed int it's more complex, but for bits 4 least significant bits: 1001)

~a << 1 shifts 1001 left by 1: 10010 (binary 18 decimal)

Now (a | b) & (~a << 1) = 0111 & 10010 = 00010 (binary 2 decimal)

Practice

1. What does the bitwise AND operator (&) do when applied to two bits?

easy

A. Returns 0 only if both bits are 0

B. Returns 1 if either bit is 1

C. Flips the bits from 0 to 1 and 1 to 0

D. Returns 1 only if both bits are 1

Solution

1. Step 1: Understand bitwise AND operation

   The AND operator compares each bit of two numbers and returns 1 only if both bits are 1.

2. Step 2: Apply to example bits

   For bits 1 & 1, result is 1; for 1 & 0 or 0 & 1 or 0 & 0, result is 0.

3. Final Answer:

   Returns 1 only if both bits are 1 -> Option D

4. Quick Check:

   AND = 1 only if both bits 1 [OK]

Hint: AND returns 1 only when both bits are 1 [OK]

Common Mistakes:

• Confusing AND with OR operator
• Thinking AND flips bits
• Assuming AND returns 1 if either bit is 1

2. Which of the following is the correct syntax to perform a left shift by 2 on an integer variable x in C#?

easy

A. x >> 2;

B. x << 2;

C. x <<< 2;

D. x >> 2 <<;

Solution

1. Step 1: Recall left shift operator syntax in C#

   The left shift operator is << and used as x << n to shift bits left by n places.

2. Step 2: Identify correct syntax among options

   x << 2; uses x << 2; which is correct. x >> 2; is right shift, C uses invalid operator, D is invalid syntax.

3. Final Answer:

   x << 2; -> Option B

4. Quick Check:

   Left shift syntax is x << n [OK]

Hint: Use << for left shift, >> for right shift [OK]

Common Mistakes:

• Using >>> which is invalid in C#
• Confusing left shift with right shift
• Adding extra symbols after shift operator

3. What is the output of the following C# code?

int a = 5; // binary 0101
int b = 3; // binary 0011
int c = a ^ b;
Console.WriteLine(c);

medium

A. 2

B. 7

C. 6

D. 8

Solution

1. Step 1: Convert numbers to binary and apply XOR

   5 in binary is 0101, 3 is 0011. XOR (^) returns 1 if bits differ, else 0.

2. Step 2: Calculate XOR bit by bit

   0101 ^ 0011 = 0110 (binary), which is 6 in decimal.

3. Final Answer:

   6 -> Option C

4. Quick Check:

   5 ^ 3 = 6 [OK]

Hint: XOR sets bits where inputs differ [OK]

Common Mistakes:

• Confusing XOR with OR or AND
• Calculating decimal sum instead of XOR
• Misreading binary values

4. Identify the error in the following C# code snippet that attempts to invert bits of an integer num:

int num = 12;
int inverted = ~num();
Console.WriteLine(inverted);

medium

A. Using parentheses after variable name num()

B. Using bitwise NOT operator incorrectly

C. Missing semicolon after int num = 12

D. Console.WriteLine cannot print integers

Solution

1. Step 1: Check usage of variable num

   The code uses num() which looks like a function call but num is an integer variable, so parentheses are invalid.

2. Step 2: Confirm bitwise NOT operator usage

   The bitwise NOT operator ~ is correctly used before the variable, so no error there.

3. Final Answer:

   Using parentheses after variable name num() -> Option A

4. Quick Check:

   Variables don't use () [OK]

Hint: Variables are not functions; no () after variable names [OK]

Common Mistakes:

• Thinking ~num() is valid syntax
• Assuming missing semicolon error
• Believing Console.WriteLine can't print integers

5. Given an integer n = 8, which expression correctly checks if n is a power of two using bit manipulation in C#?

hard

A. (n & (n - 1)) == 0

B. (n | (n - 1)) == 0

C. (n ^ (n - 1)) == 0

D. (~n & (n - 1)) == 0

Solution

1. Step 1: Understand the power of two property

   A number is a power of two if it has exactly one bit set to 1 in binary.

2. Step 2: Apply bitwise trick

   For such numbers, n & (n - 1) clears the lowest set bit. If result is 0, n was a power of two.

3. Final Answer:

   (n & (n - 1)) == 0 -> Option A

4. Quick Check:

   Power of two check uses n & (n-1) == 0 [OK]

Hint: Power of two if n & (n-1) equals zero [OK]

Common Mistakes:

• Using OR or XOR instead of AND
• Forgetting to compare result to zero
• Misunderstanding bitwise NOT usage