Break and continue behavior in Kotlin - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of break in nested loops

What is the output of this Kotlin code snippet?

Kotlin

fun main() {
    for (i in 1..3) {
        for (j in 1..3) {
            if (j == 2) break
            print("$i$j ")
        }
    }
}

A 11 21 22 31 32

B 11 21 31

C 11 12 13 21 22 23 31 32 33

D 11 12 21 22 31 32

💡 Hint

Remember that break exits the nearest enclosing loop.

The inner loop breaks when j == 2, so it only prints j=1 for each i. So output is 11 21 31.

❓ Predict Output

intermediate

Effect of continue in a single loop

What will this Kotlin code print?

Kotlin

fun main() {
    for (i in 1..5) {
        if (i % 2 == 0) continue
        print(i)
    }
}

A 135

B 12345

C 246

D 1245

💡 Hint

continue skips the current iteration when the condition is true.

The loop skips even numbers, so only odd numbers 1,3,5 print.

❓ Predict Output

advanced

Output with labeled break in nested loops

What is the output of this Kotlin code?

Kotlin

fun main() {
    outer@ for (i in 1..3) {
        for (j in 1..2) {
            if (i * j > 2) break@outer
            print("$i$j ")
        }
    }
}

A 11 12 21 22

B 11 12 13 21 22 23 31 32 33

C 11 12 21 22 31

D 11 12 21

💡 Hint

The labeled break exits the outer loop when condition is met.

When i*j > 2, break@outer stops all loops. It stops after printing 11 12 21.

❓ Predict Output

advanced

Continue with label in nested loops

What will this Kotlin code print?

Kotlin

fun main() {
    outer@ for (i in 1..3) {
        for (j in 1..3) {
            if (j == 2) continue@outer
            print("$i$j ")
        }
    }
}

A 11 21 31

B 11 12 21 22 31 32

C 11 13 21 23 31 33

D 11 13 22 23 32 33

💡 Hint

continue@outer skips to next iteration of outer loop when j == 2.

When j == 2, continue@outer skips to the next iteration of the outer loop, skipping the remaining inner loop iterations (j=3). Thus, only j=1 prints for each i: 11 21 31.

🧠 Conceptual

expert

Behavior of break and continue in Kotlin loops

Which statement about break and continue in Kotlin is correct?

A Using break or continue without labels always exits all nested loops.

B A break statement can only exit the innermost loop and cannot use labels.

C A continue statement skips the rest of the current iteration and can use labels to skip outer loops.

D Labels in Kotlin can only be used with break, not with continue.

💡 Hint

Think about how labels affect loop control in Kotlin.

In Kotlin, continue can use labels to skip to the next iteration of an outer loop. Break can also use labels to exit outer loops.

Practice

1. What does the break statement do inside a Kotlin loop?

easy

A. Stops the entire loop immediately

B. Skips the current iteration and continues the loop

C. Restarts the loop from the beginning

D. Pauses the loop until a condition is met

Solution

1. Step 1: Understand the role of break

   The break statement immediately stops the whole loop when executed.

2. Step 2: Compare with other loop controls

   Unlike continue, which skips only the current step, break ends the loop fully.

3. Final Answer:

   Stops the entire loop immediately -> Option A

4. Quick Check:

   break stops loop = B [OK]

Hint: Remember: break = stop loop, continue = skip step [OK]

Common Mistakes:

• Confusing break with continue
• Thinking break skips only one iteration
• Assuming break restarts the loop

2. Which of the following is the correct syntax to skip the current iteration in a Kotlin for loop?

easy

A. continue

B. break

C. skip

D. stop

Solution

1. Step 1: Identify the keyword to skip iteration

   In Kotlin, continue is used to skip the current loop iteration and move to the next.

2. Step 2: Check other options for correctness

   break stops the loop, skip and stop are not valid Kotlin keywords.

3. Final Answer:

   continue -> Option A

4. Quick Check:

   Skip iteration = continue [OK]

Hint: Use continue to skip steps inside loops [OK]

Common Mistakes:

• Using break instead of continue
• Using non-existent keywords like skip or stop
• Confusing syntax with other languages

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

for (i in 1..5) {
    if (i == 3) break
    print(i)
}

medium

A. Error

B. 12345

C. 345

D. 12

Solution

1. Step 1: Analyze the loop and break condition

   The loop runs from 1 to 5. When i equals 3, break stops the loop immediately.

2. Step 2: Determine printed values before break

   Values 1 and 2 are printed before hitting i == 3, then loop stops.

3. Final Answer:

   12 -> Option D

4. Quick Check:

   Loop stops at 3, prints 1 and 2 [OK]

Hint: Break stops loop before printing 3 [OK]

Common Mistakes:

• Assuming 3 is printed before break
• Thinking loop continues after break
• Confusing break with continue

4. Find the error in this Kotlin code snippet:

for (i in 1..5) {
    if (i == 3) continue
    if (i == 6) break
    print(i)
}

medium

A. The loop will run infinitely

B. continue cannot be used inside for loops

C. The break condition is unreachable

D. Syntax error in if statements

Solution

1. Step 1: Check loop range and conditions

   The loop runs from 1 to 5, so i == 6 never happens.

2. Step 2: Analyze break condition reachability

   Since i never reaches 6, the break statement is never executed, making it unreachable code.

3. Final Answer:

   The break condition is unreachable -> Option C

4. Quick Check:

   Break at i=6 unreachable in 1..5 loop [OK]

Hint: Check loop range matches break condition [OK]

Common Mistakes:

• Assuming break will run at i=6
• Thinking continue is invalid in loops
• Believing loop runs infinitely

5. Consider this Kotlin code:

var sum = 0
for (i in 1..10) {
    if (i % 2 == 0) continue
    if (sum > 10) break
    sum += i
}
print(sum)

What is the output?

hard

A. 25

B. 11

C. 9

D. 16

Solution

1. Step 1: Understand loop behavior with continue and break

   The loop adds only odd numbers (because even numbers are skipped by continue). It stops when sum > 10.

2. Step 2: Calculate sum step-by-step

   Odd numbers: 1,3,5,7,9
   i=1: sum=0 <=10, sum=1
   i=3: sum=1 <=10, sum=4
   i=5: sum=4 <=10, sum=9
   i=7: sum=9 <=10, sum=16
   i=9: sum=16 >10, break (no add 9).

3. Final Answer:

   11 -> Option B

4. Quick Check:

   Sum odd numbers until sum > 10 = 11 [OK]

Hint: Add odds, stop when sum exceeds 10 [OK]

Common Mistakes:

• Adding even numbers mistakenly
• Breaking before sum exceeds 10
• Miscounting sum steps