Labeled break and continue in Kotlin - Time & Space Complexity

Time Complexity: Labeled break and continue

O(n * m)

Understanding Time Complexity

We want to understand how using labeled break and continue affects the time a program takes to run.

Specifically, we ask: How does the number of steps change as the input grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fun searchMatrix(matrix: Array, target: Int) {
    outer@ for (row in matrix) {
        for (value in row) {
            if (value == target) break@outer
            if (value > target) continue@outer
        }
    }
}
    

This code searches for a target number in a 2D matrix, using labeled break and continue to control loops.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops over rows and values in the matrix.
• How many times: Outer loop runs for each row, inner loop runs for each value in that row until break or continue.

How Execution Grows With Input

As the matrix size grows, the loops run more times, but labeled break and continue can stop inner loops early.

Input Size (n x m)	Approx. Operations
10 x 10	Up to 100 checks
100 x 100	Up to 10,000 checks
1000 x 1000	Up to 1,000,000 checks

Pattern observation: The number of operations grows roughly with the total number of elements, but early breaks can reduce work.

Final Time Complexity

Time Complexity: O(n * m)

This means the time grows roughly with the total number of elements in the matrix.

Common Mistake

[X] Wrong: "Using labeled break or continue makes the loops run faster always."

[OK] Correct: Labeled break and continue only skip some steps but do not change the worst-case number of checks if the target is not found early.

Interview Connect

Understanding how labeled break and continue affect loop execution helps you explain control flow clearly and reason about performance in real code.

Self-Check

"What if we removed the labeled continue and used only break? How would the time complexity change?"

Practice

1. What does the break@label statement do in Kotlin nested loops?

easy

A. It stops the loop identified by the label immediately.

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

C. It restarts the entire program from the beginning.

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

Solution

1. Step 1: Understand labeled break usage

   The break@label statement is used to exit a specific loop marked by the label, not just the innermost loop.

2. Step 2: Compare with other options

   Options B and D describe continue or pause behavior, which is incorrect. It restarts the entire program from the beginning. is unrelated to loop control.

3. Final Answer:

   It stops the loop identified by the label immediately. -> Option A

4. Quick Check:

   break@label stops labeled loop [OK]

Hint: break@label exits the labeled loop immediately [OK]

Common Mistakes:

• Confusing break@label with continue@label
• Thinking break@label only stops innermost loop
• Assuming break@label pauses instead of stops

2. Which of the following is the correct syntax to continue the outer loop labeled outerLoop in Kotlin?

easy

A. continue#outerLoop

B. continue outerLoop

C. continue(outerLoop)

D. continue@outerLoop

Solution

1. Step 1: Recall Kotlin labeled continue syntax

   Kotlin uses the @ symbol to refer to labels with break or continue, so the correct syntax is continue@label.

2. Step 2: Eliminate incorrect options

   Options B, C, and D use invalid syntax for labels in Kotlin.

3. Final Answer:

   continue@outerLoop -> Option D

4. Quick Check:

   Use continue@label to continue labeled loop [OK]

Hint: Use @ symbol with label for continue or break [OK]

Common Mistakes:

• Omitting @ symbol before label
• Using parentheses or other symbols with label
• Confusing syntax with other languages

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

outer@ for (i in 1..3) {
    inner@ for (j in 1..3) {
        if (i == 2 && j == 2) break@outer
        print("$i$j ")
    }
}

medium

A. 11 12 13 21 22

B. 11 12 13 21

C. 11 12 13 21 22 23

D. 11 12 13

Solution

1. Step 1: Trace nested loops and break condition

   The outer loop runs i=1 to 3, inner loop j=1 to 3. When i=2 and j=2, break@outer stops the outer loop entirely.

2. Step 2: List printed values before break

   For i=1, j=1..3 prints 11 12 13. For i=2, j=1 prints 21, then at j=2 break@outer stops all loops. So output is "11 12 13 21".

3. Final Answer:

   11 12 13 21 -> Option B

4. Quick Check:

   break@outer stops all loops at i=2,j=2 [OK]

Hint: break@label stops all loops at labeled point [OK]

Common Mistakes:

• Assuming break@outer only stops inner loop
• Including values after break condition
• Confusing break@outer with continue@outer

4. Identify the error in this Kotlin code snippet:

outer@ for (i in 1..2) {
    for (j in 1..2) {
        if (i == 1) continue outer
        println("i=$i, j=$j")
    }
}

medium

A. continue@outer should be used instead of continue outer.

B. continue cannot be used with a label on an inner loop.

C. Label 'outer' is not defined.

D. No error; code runs correctly.

Solution

1. Step 1: Check labeled continue syntax

   Kotlin requires the @ symbol when using labels with break or continue, so it must be continue@outer.

2. Step 2: Verify label definition and usage

   The label 'outer' is correctly defined, so Label 'outer' is not defined. is wrong. Using continue with label without @ causes syntax error.

3. Final Answer:

   continue@outer should be used instead of continue outer. -> Option A

4. Quick Check:

   Use continue@label syntax to avoid errors [OK]

Hint: Always use @ with label in continue or break [OK]

Common Mistakes:

• Omitting @ symbol in labeled continue
• Thinking label must be on inner loop
• Assuming no error when syntax is wrong

5. Given nested loops labeled outer and inner, how can you skip only the current iteration of the outer loop when a condition in the inner loop is met?

hard

A. Use break@outer inside inner loop.

B. Use continue without label inside inner loop.

C. Use continue@outer inside inner loop.

D. Use break without label inside inner loop.

Solution

1. Step 1: Understand labeled continue effect

   Using continue@outer skips the current iteration of the outer loop, moving to the next iteration of outer.

2. Step 2: Compare with break and unlabeled continue

   break@outer would exit the outer loop entirely, not just skip iteration. Unlabeled continue affects only inner loop.

3. Final Answer:

   Use continue@outer inside inner loop. -> Option C

4. Quick Check:

   continue@outer skips outer loop iteration [OK]

Hint: Use continue@label to skip labeled loop iteration [OK]

Common Mistakes:

• Using break@outer instead of continue@outer
• Using unlabeled continue to skip outer loop
• Confusing break and continue effects