Writing HIGH and LOW to output pins in Embedded C - Time & Space Complexity

Time Complexity: Writing HIGH and LOW to output pins

O(n)

Understanding Time Complexity

When writing HIGH or LOW to output pins, it's important to know how the time taken changes as we do more writes.

We want to see how the number of pin writes affects the total time the program takes.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

for (int i = 0; i < n; i++) {
    digitalWrite(pin, HIGH);
    digitalWrite(pin, LOW);
}
    

This code writes HIGH then LOW to the same output pin repeatedly n times.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Writing HIGH and LOW to the output pin.
• How many times: Each write happens inside a loop that runs n times, so 2 writes per loop, total 2*n writes.

How Execution Grows With Input

Each time we increase n, the number of writes grows directly with n.

Input Size (n)	Approx. Operations
10	20 writes
100	200 writes
1000	2000 writes

Pattern observation: The total operations grow in a straight line as n increases.

Final Time Complexity

Time Complexity: O(n)

This means the time taken grows directly in proportion to the number of writes we do.

Common Mistake

[X] Wrong: "Writing HIGH and LOW inside the loop takes constant time no matter how many times we run it."

[OK] Correct: Each write takes some time, so doing it many times adds up and the total time grows with n.

Interview Connect

Understanding how repeated pin writes affect time helps you write efficient embedded code and explain your reasoning clearly in interviews.

Self-Check

"What if we only wrote HIGH once before the loop and LOW once after? How would the time complexity change?"

Practice

1. Which function is used to set a pin as an output before writing HIGH or LOW to it?

easy

A. setPinMode(pin, HIGH)

B. digitalWrite(pin, OUTPUT)

C. pinMode(pin, OUTPUT)

D. writePin(pin, OUTPUT)

Solution

1. Step 1: Understand pin configuration

   Before writing HIGH or LOW, the pin must be set as an output to control voltage.

2. Step 2: Identify the correct function

   The function to set pin mode is pinMode(pin, OUTPUT), which configures the pin for output.

3. Final Answer:

   pinMode(pin, OUTPUT) -> Option C

4. Quick Check:

   Set pin as output = pinMode(pin, OUTPUT) [OK]

Hint: Set pin mode with pinMode before writing HIGH or LOW [OK]

Common Mistakes:

• Using digitalWrite to set pin mode
• Confusing HIGH/LOW with OUTPUT/INPUT
• Forgetting to set pin mode before writing

2. Which of the following is the correct syntax to turn pin 5 ON (HIGH) in embedded C?

easy

A. digitalWrite(5, 1);

B. digitalWrite(5, ON);

C. digitalWrite(5, OUTPUT);

D. digitalWrite(5, HIGH);

Solution

1. Step 1: Identify the function to write pin state

   The function digitalWrite(pin, value) sets the pin voltage HIGH or LOW.

2. Step 2: Use correct constants

   Use HIGH to turn the pin ON. The correct syntax is digitalWrite(5, HIGH);.

3. Final Answer:

   digitalWrite(5, HIGH); -> Option D

4. Quick Check:

   Turn pin ON = digitalWrite(pin, HIGH) [OK]

Hint: Use HIGH constant, not ON or numbers, for clarity [OK]

Common Mistakes:

• Using ON instead of HIGH
• Using OUTPUT instead of HIGH
• Using numeric 1 instead of HIGH constant

3. What will be the output state of pin 3 after running this code?

pinMode(3, OUTPUT);
digitalWrite(3, LOW);
digitalWrite(3, HIGH);

medium

A. Pin 3 will be LOW

B. Pin 3 will be HIGH

C. Pin 3 will be in input mode

D. Pin 3 will be undefined

Solution

1. Step 1: Set pin 3 as output

   The code sets pin 3 as output with pinMode(3, OUTPUT);.

2. Step 2: Write LOW then HIGH to pin 3

   First, pin 3 is set LOW, then immediately set HIGH, so final state is HIGH.

3. Final Answer:

   Pin 3 will be HIGH -> Option B

4. Quick Check:

   Last digitalWrite sets pin state = HIGH [OK]

Hint: Last digitalWrite sets the pin state [OK]

Common Mistakes:

• Assuming pin stays LOW after writing HIGH
• Forgetting pinMode sets output mode
• Thinking pin state is undefined without delay

4. Identify the error in this code snippet:

pinMode(7, OUTPUT);
digitalWrite(7, HIGH)
digitalWrite(7, LOW);

medium

A. Missing semicolon after digitalWrite(7, HIGH)

B. pinMode should be called after digitalWrite

C. HIGH and LOW are not valid constants

D. Pin number 7 cannot be used

Solution

1. Step 1: Check syntax line by line

   The second line digitalWrite(7, HIGH) is missing a semicolon at the end.

2. Step 2: Confirm other lines are correct

   pinMode(7, OUTPUT); and digitalWrite(7, LOW); are correct syntax.

3. Final Answer:

   Missing semicolon after digitalWrite(7, HIGH) -> Option A

4. Quick Check:

   Every statement must end with semicolon [OK]

Hint: Check each line ends with a semicolon [OK]

Common Mistakes:

• Forgetting semicolon after function calls
• Calling pinMode after digitalWrite
• Misusing HIGH/LOW constants

5. You want to toggle pin 10 ON and OFF every second in a loop. Which code snippet correctly sets pin 10 HIGH then LOW with a 1-second delay between states?

hard

A. pinMode(10, OUTPUT);
digitalWrite(10, HIGH);
delay(1000);
digitalWrite(10, LOW);
delay(1000);

B. digitalWrite(10, HIGH);
delay(1000);
digitalWrite(10, LOW);
delay(1000);

C. pinMode(10, INPUT);
digitalWrite(10, HIGH);
delay(1000);

D. pinMode(10, OUTPUT);
digitalWrite(10, LOW);
digitalWrite(10, HIGH);

Solution

1. Step 1: Set pin 10 as output

   Use pinMode(10, OUTPUT); to configure pin 10 for output.

2. Step 2: Write HIGH, wait, then LOW, wait

   Use digitalWrite(10, HIGH); then delay(1000); to wait 1 second, then digitalWrite(10, LOW); and delay again.

3. Final Answer:

   pinMode(10, OUTPUT); digitalWrite(10, HIGH); delay(1000); digitalWrite(10, LOW); delay(1000); -> Option A

4. Quick Check:

   Set output, write states with delay between [OK]

Hint: Always set pinMode before toggling with delays [OK]

Common Mistakes:

• Forgetting to set pinMode as OUTPUT
• Missing delay between HIGH and LOW
• Setting pinMode to INPUT instead of OUTPUT