NOT for inverting bits in Embedded C - Time & Space Complexity

Time Complexity: NOT for inverting bits

O(1)

Understanding Time Complexity

We want to understand how long it takes to invert bits using the NOT operator in embedded C.

How does the time needed change when we invert more bits?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

unsigned int invert_bits(unsigned int x) {
    return ~x;
}
    

This code uses the NOT operator (~) to flip all bits of the input number.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Single bitwise NOT operation on the input number.
• How many times: Exactly once, no loops or repeated steps.

How Execution Grows With Input

Explain the growth pattern intuitively.

Input Size (bits)	Approx. Operations
8	1 operation
16	1 operation
32	1 operation

Pattern observation: The operation count stays the same no matter how many bits the input has.

Final Time Complexity

Time Complexity: O(1)

This means the time to invert bits using NOT does not grow with input size; it takes the same time always.

Common Mistake

[X] Wrong: "Inverting bits with NOT takes longer for bigger numbers because there are more bits to flip."

[OK] Correct: The NOT operator works on the whole number at once in hardware, so it does not take more time for bigger inputs.

Interview Connect

Knowing that bitwise NOT runs in constant time helps you understand how low-level operations behave efficiently in embedded systems.

Self-Check

"What if we inverted bits one by one using a loop instead of the NOT operator? How would the time complexity change?"