Bit Manipulation Basics AND OR XOR NOT Left Right Shift in DSA C - Time & Space Complexity

Time Complexity: Bit Manipulation Basics AND OR XOR NOT Left Right Shift

O(b)

Understanding Time Complexity

Bit manipulation operations are very fast and work directly on the bits of numbers.

We want to understand how the time needed changes as the size of the input number grows.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

unsigned int bitOps(unsigned int x, unsigned int y) {
    unsigned int a = x & y;      // AND
    unsigned int b = x | y;      // OR
    unsigned int c = x ^ y;      // XOR
    unsigned int d = ~x;         // NOT
    unsigned int e = x << 1;     // Left shift
    unsigned int f = y >> 1;     // Right shift
    return a + b + c + d + e + f;
}
    

This code performs basic bitwise operations on two unsigned integers.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Each bitwise operation processes all bits of the input numbers.
• How many times: The operations run once each, but internally they handle each bit of the numbers.

How Execution Grows With Input

Each bitwise operation touches every bit of the input numbers.

Input Size (bits)	Approx. Operations
8	About 8 steps per operation
32	About 32 steps per operation
64	About 64 steps per operation

Pattern observation: The time grows linearly with the number of bits in the input.

Final Time Complexity

Time Complexity: O(b)

This means the time grows linearly with the number of bits in the input numbers.

Common Mistake

[X] Wrong: "Bitwise operations take constant time no matter the input size."

[OK] Correct: While bitwise operations are very fast, they actually process each bit, so the time depends on the number of bits in the input.

Interview Connect

Understanding bit manipulation time helps you write efficient low-level code and reason about performance in systems programming and algorithms.

Self-Check

"What if we used a loop to apply bitwise operations on an array of n numbers? How would the time complexity change?"

Practice

1. What is the result of the bitwise AND operation between 5 and 3? (5 = 0101, 3 = 0011 in binary)

easy

A. 0

B. 1

C. 6

D. 7

Solution

1. Step 1: Convert numbers to binary

   5 in binary is 0101, and 3 in binary is 0011.

2. Step 2: Perform bitwise AND operation

   0101 AND 0011 = 0001 (only bits set in both numbers remain set).

3. Final Answer:

   1 -> Option B

4. Quick Check:

   5 & 3 = 1 [OK]

Hint: AND keeps bits set in both numbers only [OK]

Common Mistakes:

• Confusing AND with OR operation
• Not converting decimal to binary correctly
• Mixing bit positions during calculation

2. Which of the following is the correct syntax for left shifting the integer x by 2 bits in C?

easy

A. x >> 2;

B. x <<< 2;

C. x << 2;

D. x >> 2 <<;

Solution

1. Step 1: Understand left shift operator in C

   The left shift operator is written as << and shifts bits to the left.

2. Step 2: Check syntax correctness

   x << 2; uses x << 2; which correctly shifts bits of x left by 2 positions.

3. Final Answer:

   x << 2; -> Option C

4. Quick Check:

   Left shift syntax = x << 2; [OK]

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

Common Mistakes:

• Using >>> which is invalid in C
• Confusing left shift with right shift
• Incorrect operator placement or syntax

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

int x = 6; // binary 0110
int y = x ^ 3; // 3 = 0011
printf("%d", y);

medium

A. 9

B. 7

C. 1

D. 5

Solution

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

   6 in binary is 0110, 3 is 0011. XOR operation: 0110 ^ 0011 = 0101.

2. Step 2: Calculate decimal value of XOR result

   0101 in binary equals 5 decimal.

3. Final Answer:

   5 -> Option D

4. Quick Check:

   6 ^ 3 = 5 [OK]

Hint: XOR flips bits where bits differ [OK]

Common Mistakes:

• Mixing XOR with OR or AND
• Incorrect binary conversion
• Misreading bit positions

4. Identify the error in the following C code snippet:

int x = 8;
int y = ~x << 1;
printf("%d", y);

medium

A. Using ~ operator without parentheses causes wrong shift

B. Missing semicolon after declaration

C. Shift operator << cannot be used with ~ operator

D. No error, code runs correctly

Solution

1. Step 1: Understand operator precedence

   The bitwise NOT (~) has higher precedence than shift (<<), so ~x is evaluated first.

2. Step 2: Check if parentheses needed

   Without parentheses, ~x is computed, then shifted left. This is correct but may cause confusion if intended differently.

3. Step 3: Identify common mistake

   If intention was to shift x first then NOT, parentheses are needed. Otherwise, no syntax error but logic may be wrong.

4. Final Answer:

   Using ~ operator without parentheses causes wrong shift -> Option A

5. Quick Check:

   Operator precedence matters; use parentheses [OK]

Hint: Use parentheses to control bitwise operation order [OK]

Common Mistakes:

• Ignoring operator precedence
• Assuming shift applies before NOT
• Thinking code has syntax error when it doesn't

5. Given an integer n, which expression correctly checks if the 3rd bit (from right, zero-based) is set (1)?

hard

A. (n & (1 << 3)) != 0

B. (n | (1 << 3)) == 0

C. (n ^ (1 << 3)) == 0

D. (n >> 3) & 1 == 0

Solution

1. Step 1: Understand bit position and mask

   The 3rd bit (zero-based) means bit at position 3. Mask is 1 shifted left by 3: (1 << 3) = 8.

2. Step 2: Check if bit is set using AND

   Using (n & (1 << 3)) checks if that bit is 1. If result is not zero, bit is set.

3. Final Answer:

   (n & (1 << 3)) != 0 -> Option A

4. Quick Check:

   Bit check uses AND with mask [OK]

Hint: Use AND with (1 << bit_position) to check bit [OK]

Common Mistakes:

• Using OR or XOR instead of AND for checking
• Wrong bit position shift
• Comparing shifted bit to zero incorrectly