Labeled break and continue in Java - Time & Space Complexity

Time Complexity: Labeled break and continue

O(n^2)

Understanding Time Complexity

We want to see how using labeled break and continue affects how long a program runs.

Specifically, we ask: how does the program's steps grow as input gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public class Example {
    public static void labeledLoop(int n) {
        outer: for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (j == i) continue outer;
            }
        }
    }
}
    

This code uses a labeled continue to jump out of the inner loop and continue the outer loop early.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops with an early jump using labeled continue.
• How many times: Outer loop runs n times; for each i, inner loop body executes (i+1) times; total ~ n²/2.

How Execution Grows With Input

Because the inner loop stops early each time, total steps grow roughly like n²/2.

Input Size (n)	Approx. Operations
10	About 55 steps
100	About 5,050 steps
1000	About 500,500 steps

Pattern observation: The total work grows quadratically with n.

Final Time Complexity

Time Complexity: O(n^2)

This means the program's steps increase quadratically as input size grows.

Common Mistake

[X] Wrong: "Early exit from inner loop makes overall time O(n)."

[OK] Correct: The inner loop stops at j==i each time, but total operations is sum_{i=0}^{n-1} (i+1) ≈ n²/2, still O(n^2).

Interview Connect

Understanding how labeled break and continue affect loops shows you can read code carefully and explain its speed clearly.

Self-Check

"What if we replaced the labeled continue with a labeled break? How would the time complexity change?"

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