While loop in C - Time & Space Complexity

Time Complexity: While loop

O(n)

Understanding Time Complexity

We want to see how the time a while loop takes changes as the input size grows.

How does the number of steps increase when the loop runs more times?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int i = 0;
while (i < n) {
    // some constant time work
    i++;
}
    

This code runs a loop that repeats until i reaches n, doing a small task each time.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop running the code inside.
• How many times: It runs once for each number from 0 up to n - 1, so n times.

How Execution Grows With Input

Each time n gets bigger, the loop runs more times, adding more steps.

Input Size (n)	Approx. Operations
10	About 10 steps
100	About 100 steps
1000	About 1000 steps

Pattern observation: The number of steps grows directly with n; if n doubles, steps double.

Final Time Complexity

Time Complexity: O(n)

This means the time to finish grows in a straight line with the size of n.

Common Mistake

[X] Wrong: "The while loop always takes the same time no matter what n is."

[OK] Correct: The loop runs n times, so bigger n means more steps and more time.

Interview Connect

Understanding how loops grow with input helps you explain your code clearly and shows you know how programs scale.

Self-Check

"What if we changed the loop to increase i by 2 each time? How would the time complexity change?"

Practice

1. What does a while loop do in C programming?

easy

A. Repeats a block of code while a condition is true

B. Executes code only once regardless of condition

C. Runs code a fixed number of times without checking condition

D. Stops the program immediately

Solution

1. Step 1: Understand the purpose of a while loop

   A while loop runs repeatedly as long as its condition remains true.

2. Step 2: Compare options with the definition

   Only Repeats a block of code while a condition is true correctly describes this behavior.

3. Final Answer:

   Repeats a block of code while a condition is true -> Option A

4. Quick Check:

   While loop = repeat while condition true [OK]

Hint: While loops run only if condition is true before each repeat [OK]

Common Mistakes:

• Thinking while loop runs at least once even if condition is false
• Confusing while loop with for loop fixed iterations
• Believing while loop stops program immediately

2. Which of the following is the correct syntax for a while loop in C?

easy

A. while (condition) { /* code */ }

B. while condition { /* code */ }

C. while [condition] { /* code */ }

D. while: (condition) { /* code */ }

Solution

1. Step 1: Recall C while loop syntax

   The condition must be inside parentheses and the code block inside braces.

2. Step 2: Check each option

   Only while (condition) { /* code */ } uses correct parentheses and braces syntax.

3. Final Answer:

   while (condition) { /* code */ } -> Option A

4. Quick Check:

   Correct syntax uses parentheses and braces [OK]

Hint: Remember: while(condition) with parentheses and braces [OK]

Common Mistakes:

• Omitting parentheses around condition
• Using square brackets instead of parentheses
• Adding colon after while keyword

3. What is the output of this C code?

int i = 1;
while (i <= 3) {
    printf("%d ", i);
    i++;
}

medium

A. 0 1 2 3

B. 1 2 3 4

C. 3 2 1

D. 1 2 3

Solution

1. Step 1: Trace the loop variable values

   i starts at 1, prints 1, then increments to 2, prints 2, then 3, and stops after printing 3 because i becomes 4 and condition fails.

2. Step 2: Confirm printed output

   The printed numbers are "1 2 3 " with spaces after each number.

3. Final Answer:

   1 2 3 -> Option D

4. Quick Check:

   Prints 1 to 3 with spaces [OK]

Hint: Check loop start, end condition, and increments carefully [OK]

Common Mistakes:

• Including 4 in output by mistake
• Starting from 0 instead of 1
• Printing numbers in reverse order

4. Identify the error in this code snippet:

int i = 0;
while (i < 5)
    printf("%d ", i);

medium

A. Syntax error due to missing braces

B. Missing semicolon after while condition

C. Infinite loop because i is never updated

D. Variable i is not declared

Solution

1. Step 1: Analyze loop behavior

   Variable i starts at 0 and condition is i < 5, but i is never incremented inside the loop.

2. Step 2: Understand consequence

   Since i never changes, the condition stays true forever, causing an infinite loop.

3. Final Answer:

   Infinite loop because i is never updated -> Option C

4. Quick Check:

   Loop variable must change to avoid infinite loop [OK]

Hint: Always update loop variable inside while loop [OK]

Common Mistakes:

• Assuming loop ends without variable update
• Thinking missing braces cause syntax error here
• Ignoring infinite loop possibility

5. How can you modify this code to print all even numbers from 2 to 10 using a while loop?

int num = 1;
while (num <= 10) {
    printf("%d ", num);
    num++;
}

hard

A. Use a for loop instead of while loop

B. Initialize num to 2 and increment by 2 inside the loop

C. Initialize num to 0 and increment by 1 inside the loop

D. Keep num at 1 and print only if num % 2 == 0 without incrementing

Solution

1. Step 1: Set starting point and increment for even numbers

   Start num at 2 (first even number) and increase by 2 each time to get only even numbers.

2. Step 2: Modify loop accordingly

   Change initialization to num = 2 and inside loop do num += 2 instead of num++.

3. Final Answer:

   Initialize num to 2 and increment by 2 inside the loop -> Option B

4. Quick Check:

   Start at 2, add 2 each time for evens [OK]

Hint: Start at first even, add 2 each loop to get evens [OK]

Common Mistakes:

• Not incrementing num causing infinite loop
• Printing odd numbers by starting at 1
• Using for loop when question asks for while loop