Counter-based while loop in Java - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of a simple counter-based while loop

What is the output of the following Java code?

Java

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

A 1 2 3

B 0 1 2 3

C 1 2 3 4

D 1 2

💡 Hint

Think about how the counter changes and when the loop stops.

The loop starts at 1 and runs while count is less than or equal to 3. It prints 1, then 2, then 3, then stops.

❓ Predict Output

intermediate

Counting down with a while loop

What will this Java code print?

Java

int count = 5;
while (count > 2) {
    System.out.print(count + " ");
    count--;
}

A 5 4 3 2

B 5 4 3

C 5 4 3 2 1

D 4 3 2

💡 Hint

Look at the condition and how the counter decreases.

The loop runs while count is greater than 2, so it prints 5, 4, and 3, then stops before printing 2.

❓ Predict Output

advanced

Sum of numbers using a while loop

What is the value of sum after this code runs?

Java

int count = 1;
int sum = 0;
while (count <= 4) {
    sum += count;
    count++;
}
System.out.println(sum);

A 10

B 9

C 11

D 0

💡 Hint

Add numbers 1 through 4 together.

The loop adds 1 + 2 + 3 + 4 = 10 to sum.

❓ Predict Output

advanced

Loop with incorrect counter update

What happens when this code runs?

Java

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

A Prints nothing

B Prints 1 2 3 and stops

C Syntax error due to missing increment

D Infinite loop printing 1 repeatedly

💡 Hint

What happens if the counter never changes inside the loop?

Without increasing count, the condition is always true and the loop never ends, printing 1 forever.

🧠 Conceptual

expert

Number of iterations in a counter-based while loop

How many times will the loop body execute in this code?

Java

int count = 2;
while (count < 10) {
    count += 3;
}

A 4

B 6

C 3

D 5

💡 Hint

Trace the values: start with count=2, add 3 each time until the condition fails.

count=2 (<10), body executes (count=5); 5<10, body (count=8); 8<10, body (count=11); 11<10 false, stop. Executes 3 times.

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