Counter-based while loop in Java - Time & Space Complexity

Time Complexity: Counter-based while loop

O(n)

Understanding Time Complexity

We want to see how the time it takes to run a counter-based while loop changes as the number it counts to gets bigger.

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int count = 0;
while (count < n) {
    System.out.println(count);
    count++;
}
    

This code prints numbers from 0 up to n-1 using a while loop that increases the counter each time.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop runs and prints the current count.
• How many times: It repeats exactly n times, once for each number from 0 to n-1.

How Execution Grows With Input

As n gets bigger, the loop runs more times, so the total steps grow in a straight line with n.

Input Size (n)	Approx. Operations
10	About 10 prints and increments
100	About 100 prints and increments
1000	About 1000 prints and increments

Pattern observation: Doubling n doubles the work, so the growth is steady and linear.

Final Time Complexity

Time Complexity: O(n)

This means the time to finish grows directly in proportion to the size of n.

Common Mistake

[X] Wrong: "The loop runs a fixed number of times no matter what n is."

[OK] Correct: The loop depends on n and runs once for each number up to n, so bigger n means more steps.

Interview Connect

Understanding how loops grow with input size is a key skill. It helps you explain how your code will behave with bigger data, which is important in many real tasks.

Self-Check

"What if we changed the loop to count down from n to 0? How would the time complexity change?"

Practice

1. What is the main purpose of using a counter in a while loop in Java?

easy

A. To create an infinite loop

B. To store the result of the loop

C. To control how many times the loop runs

D. To skip loop iterations

Solution

1. Step 1: Understand the role of a counter

   A counter is a variable that keeps track of how many times the loop has run.

2. Step 2: Identify the purpose in a while loop

   The counter helps stop the loop after a fixed number of runs, preventing infinite loops.

3. Final Answer:

   To control how many times the loop runs -> Option C

4. Quick Check:

   Counter controls loop count = A [OK]

Hint: Counter limits loop repetitions to fixed times [OK]

Common Mistakes:

• Thinking counter stores results
• Assuming counter creates infinite loops
• Believing counter skips iterations

2. Which of the following is the correct way to initialize a counter variable for a while loop in Java?

easy

A. while (count = 0) {}

B. int count = 0;

C. int count == 0;

D. count := 0;

Solution

1. Step 1: Check Java variable initialization syntax

   Java uses int count = 0; to declare and initialize an integer variable.

2. Step 2: Identify incorrect options

   while (count = 0) {} uses assignment in while condition incorrectly, C uses double equals in declaration, D uses wrong assignment operator.

3. Final Answer:

   int count = 0; -> Option B

4. Quick Check:

   Correct Java int initialization = A [OK]

Hint: Use single equals (=) to assign variables in Java [OK]

Common Mistakes:

• Using == instead of = for assignment
• Trying to assign inside while condition
• Using := which is not Java syntax

3. What will be the output of this Java code?

int count = 1;
while (count <= 3) {
    System.out.print(count + " ");
    count++;
}

medium

A. 1 2 3

B. 0 1 2 3

C. 1 2 3 4

D. 3 2 1

Solution

1. Step 1: Analyze the loop start and condition

   The counter starts at 1 and runs while count is less than or equal to 3.

2. Step 2: Trace the output during each iteration

   It prints 1, then increments to 2, prints 2, increments to 3, prints 3, then increments to 4 and stops.

3. Final Answer:

   1 2 3 -> Option A

4. Quick Check:

   Prints 1 to 3 with spaces = B [OK]

Hint: Count from 1 to 3 prints 1 2 3 [OK]

Common Mistakes:

• Starting count at 0 instead of 1
• Printing after increment causing extra number
• Reversing the order of output

4. Identify the error in this Java code snippet:

int count = 1;
while (count < 5) {
    System.out.println(count);
}

medium

A. Counter variable is not incremented inside the loop

B. Loop condition should be count <= 5

C. Incorrect print statement syntax

D. Missing semicolon after while condition

Solution

1. Step 1: Check loop body for counter update

   The code does not increase count inside the loop, so count stays 1 forever.

2. Step 2: Understand effect of missing increment

   Without increment, the loop never ends, causing an infinite loop.

3. Final Answer:

   Counter variable is not incremented inside the loop -> Option A

4. Quick Check:

   Missing count++ causes infinite loop = D [OK]

Hint: Always update counter inside while loop [OK]

Common Mistakes:

• Forgetting to increment counter
• Changing loop condition incorrectly
• Assuming print syntax is wrong

5. You want to print all even numbers from 2 to 10 using a counter-based while loop in Java. Which code correctly does this?

hard

A. int count = 2;
while (count < 10) {
System.out.print(count + " " );
count += 2;
}

B. int count = 0;
while (count < 10) {
count += 2;
System.out.print(count + " " );
}

C. int count = 2;
while (count < 10) {
System.out.print(count + " " );
count++;
}

D. int count = 1;
while (count <= 10) {
if (count % 2 == 0) {
System.out.print(count + " " );
}
count++;
}

Solution

1. Step 1: Understand the goal to print even numbers 2 to 10

   We need to print 2,4,6,8,10 using a while loop and a counter.

2. Step 2: Analyze each option

   int count = 2; while (count < 10) { System.out.print(count + " " ); count += 2; } increments by 2 starting at 2, prints 2,4,6,8 misses 10. int count = 1; while (count <= 10) { if (count % 2 == 0) { System.out.print(count + " " ); } count++; } starts at 1, checks if even, prints if yes, increments by 1, correct. int count = 0; while (count < 10) { count += 2; System.out.print(count + " " ); } starts at 0, increments before print, prints 2,4,6,8 misses 10. int count = 2; while (count < 10) { System.out.print(count + " " ); count++; } increments by 1 but stops before 10, so misses 10.

3. Step 3: Choose the best option

   int count = 1; while (count <= 10) { if (count % 2 == 0) { System.out.print(count + " " ); } count++; } uses a counter-based loop with condition and checks evenness inside, which is a common pattern.

4. Final Answer:

   int count = 1; while (count <= 10) { if (count % 2 == 0) { System.out.print(count + " " ); } count++; } -> Option D

5. Quick Check:

   Check even inside loop and increment count = C [OK]

Hint: Check even inside loop, increment by 1 for all numbers [OK]

Common Mistakes:

• Incrementing by 1 but missing last number
• Starting count at 0 and printing before increment
• Not checking even condition inside loop