Labeled break and continue in Java - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of labeled break in nested loops

What is the output of the following Java code using a labeled break?

Java

public class Test {
    public static void main(String[] args) {
        outer:
        for (int i = 1; i <= 3; i++) {
            for (int j = 1; j <= 3; j++) {
                if (i * j > 3) {
                    break outer;
                }
                System.out.print(i * j + " ");
            }
        }
    }
}

A 1 2 3 2 4 6

B 1 2 3

C 1 2 3 2 4 6 3 6 9

D 1 2 3 2

💡 Hint

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

For i=1, j=1 prints 1, j=2 prints 2, j=3 prints 3. For i=2, j=1: 2*1=2 <=3 prints 2, j=2: 2*2=4 >3 break outer. Output: 1 2 3 2

❓ Predict Output

intermediate

Output of labeled continue in nested loops

What is the output of this Java code using a labeled continue?

Java

public class Test {
    public static void main(String[] args) {
        outer:
        for (int i = 1; i <= 2; i++) {
            for (int j = 1; j <= 3; j++) {
                if (j == 2) {
                    continue outer;
                }
                System.out.print(i * j + " ");
            }
        }
    }
}

A 1 3 2 3

B 1 2

C 1 3 3 6

D 1 3 2 4

💡 Hint

The labeled continue skips to the next iteration of the outer loop when j == 2, skipping the print for j=2 and remaining inner iterations.

i=1, j=1: prints 1, j=2: continue outer (skips j=3, print), goes to i=2. i=2, j=1: prints 2, j=2: continue outer (end). Output: 1 2

🔧 Debug

advanced

Identify error with labeled continue usage

What error does this Java code produce?

Java

public class Test {
    public static void main(String[] args) {
        outer:
        for (int i = 0; i < 3; i++) {
            if (i == 1) {
                continue outer;
            }
            System.out.print(i + " ");
        }
    }
}

A Output: 0 2

B Runtime exception: Illegal continue label

C Output: 0 1 2

D Compile-time error: continue label not allowed here

💡 Hint

Labeled continue can be used within the body of its own labeled loop and acts like an unlabeled continue.

No compile error. The labeled continue 'continue outer;' skips the print statement when i==1, equivalent to unlabeled continue. Prints 0 (i=0), skips i=1, prints 2 (i=2).

❓ Predict Output

advanced

Output with nested loops and labeled continue

What is the output of this Java program using labeled continue inside nested loops?

Java

public class Test {
    public static void main(String[] args) {
        outer:
        for (int i = 1; i <= 3; i++) {
            inner:
            for (int j = 1; j <= 3; j++) {
                if (i == j) {
                    continue outer;
                }
                System.out.print(i + "-" + j + " ");
            }
        }
    }
}

A 1-2 1-3 2-1 2-3 3-1 3-2

B 1-2 1-3 2-1 2-3 3-1

C 2-1 3-1 3-2

D 1-2 1-3 2-1 3-1

💡 Hint

The continue outer skips the rest of the inner loop and continues the next iteration of the outer loop when i == j.

i=1 j=1: i==j continue outer (no print, skips j=2,3) -> i=2. i=2 j=1: != print 2-1, j=2== continue outer (skips j=3) -> i=3. i=3 j=1 != print 3-1, j=2 != print 3-2, j=3== continue outer (end). Output: 2-1 3-1 3-2

🧠 Conceptual

expert

Effect of labeled break on loop variables

Consider the following Java code snippet. What is the final value of variable 'count' after execution?

Java

public class Test {
    public static void main(String[] args) {
        int count = 0;
        outer:
        for (int i = 0; i < 5; i++) {
            for (int j = 0; j < 5; j++) {
                if (i + j > 5) {
                    break outer;
                }
                count++;
            }
        }
        System.out.println(count);
    }
}

A 14

B 21

C 25

D 10

💡 Hint

The labeled break exits both loops immediately when i + j > 5.

Counts for i=0 all j=0-4 (5), i=1 all j=0-4 (5), i=2 j=0-3 (4, since j=4: 2+4=6>5 break outer). Total 14.

Practice

1. What does a labeled break statement do in Java nested loops?

easy

A. It skips the current iteration of the innermost loop.

B. It pauses the loop until a condition is met.

C. It restarts the innermost loop from the beginning.

D. It exits the loop identified by the label immediately.

Solution

1. Step 1: Understand labeled break usage

   A labeled break exits the loop that has the specified label, not just the innermost loop.

2. Step 2: Compare with normal break

   Normal break exits only the innermost loop, but labeled break can exit outer loops directly.

3. Final Answer:

   It exits the loop identified by the label immediately. -> Option D

4. Quick Check:

   Labeled break = exit labeled loop [OK]

Hint: Labeled break jumps out of the named loop instantly [OK]

Common Mistakes:

• Confusing labeled break with continue
• Thinking it only affects the innermost loop
• Assuming it pauses instead of exits

2. Which of the following is the correct syntax for a labeled continue statement in Java?

easy

A. continue: labelName;

B. continue labelName;

C. continue (labelName);

D. continue->labelName;

Solution

1. Step 1: Recall labeled continue syntax

   The correct syntax is the keyword continue followed by the label name and a semicolon.

2. Step 2: Eliminate incorrect forms

   Java does not use colons, parentheses, or arrows with continue statements.

3. Final Answer:

   continue labelName; -> Option B

4. Quick Check:

   Correct labeled continue syntax = continue labelName; [OK]

Hint: Use 'continue' followed by label and semicolon [OK]

Common Mistakes:

• Adding colon after continue
• Using parentheses around label
• Using arrow symbols instead of space

3. What is the output of the following Java code?

outer:
for (int i = 1; i <= 3; i++) {
    inner:
    for (int j = 1; j <= 3; j++) {
        if (j == 2) continue outer;
        System.out.print(i + "-" + j + " ");
    }
}

medium

A. 1-1 2-1 3-1 3-2

B. 1-1 1-2 1-3 2-1 2-2 2-3 3-1 3-2 3-3

C. 1-1 2-1 3-1

D. 1-1 2-1 2-2 3-1 3-2

Solution

1. Step 1: Analyze loops and continue

   When j == 2, the statement continue outer; skips the rest of the current outer loop iteration and moves to the next i.

2. Step 2: Trace output

   For i=1, j=1 prints "1-1 ". At j=2, continue outer skips to i=2. Similarly for i=2 and i=3, only j=1 prints.

3. Final Answer:

   1-1 2-1 3-1 -> Option C

4. Quick Check:

   Labeled continue skips to next outer loop iteration [OK]

Hint: Labeled continue skips to next iteration of labeled loop [OK]

Common Mistakes:

• Thinking continue skips only inner loop
• Expecting all inner loop values to print
• Ignoring label effect on continue

4. Identify the error in this Java code snippet using labeled break:

outer:
for (int i = 0; i < 3; i++) {
    for (int j = 0; j < 3; j++) {
        if (i == j) break;
    }
}

medium

A. The label 'outer' is not used with break or continue.

B. The break statement should be 'break outer;' to use the label.

C. Labels cannot be used with for loops.

D. The code will cause a syntax error due to missing semicolon.

Solution

1. Step 1: Check label usage

   The label 'outer' is declared but never referenced by break or continue statements.

2. Step 2: Understand label necessity

   Labels must be used with break or continue to have effect; otherwise, they are redundant but not syntax errors.

3. Final Answer:

   The label 'outer' is not used with break or continue. -> Option A

4. Quick Check:

   Unused label = no effect but no syntax error [OK]

Hint: Use label only if break or continue references it [OK]

Common Mistakes:

• Assuming unused label causes syntax error
• Thinking break without label breaks outer loop
• Confusing label usage rules

5. Given nested loops, how would you use labeled break and continue to print only pairs (i, j) where i < j, skipping all others efficiently? Choose the correct code snippet.

hard

A. outer:
for (int i = 1; i <= 3; i++) {
inner:
for (int j = 1; j <= 3; j++) {
if (i >= j) continue inner;
System.out.print("(" + i + "," + j + ") ");
}
}

B. outer:
for (int i = 1; i <= 3; i++) {
inner:
for (int j = 1; j <= 3; j++) {
if (i >= j) break outer;
System.out.print("(" + i + "," + j + ") ");
}
}

C. outer:
for (int i = 1; i <= 3; i++) {
inner:
for (int j = 1; j <= 3; j++) {
if (i >= j) continue outer;
System.out.print("(" + i + "," + j + ") ");
}
}

D. outer:
for (int i = 1; i <= 3; i++) {
inner:
for (int j = 1; j <= 3; j++) {
if (i >= j) break inner;
System.out.print("(" + i + "," + j + ") ");
}
}

Solution

1. Step 1: Understand the condition and loops

   We want to print pairs where i < j only. If i >= j, skip to next j (inner loop iteration).

2. Step 2: Analyze each option's control flow

   outer: for (int i = 1; i <= 3; i++) { inner: for (int j = 1; j <= 3; j++) { if (i >= j) continue inner; System.out.print("(" + i + "," + j + ") "); } } uses continue inner; to skip the current iteration and move to next j, efficiently skipping invalid pairs.

3. Final Answer:

   Uses continue inner; to skip when i >= j. -> Option A

4. Quick Check:

   Labeled continue to inner loop skips invalid pairs [OK]

Hint: Use 'continue inner;' to skip to next inner loop iteration [OK]

Common Mistakes:

• Using break outer which stops all loops
• Using continue outer which skips valid pairs as well
• Using break inner which stops inner loop prematurely