Labeled statements for nested loops in Swift - Time & Space Complexity

Time Complexity: Labeled statements for nested loops

O(n)

Understanding Time Complexity

When using labeled statements in nested loops, it helps control which loop to break or continue. We want to understand how this affects the total work the program does as input grows.

How does the number of steps change when loops are nested and labeled?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

outerLoop: for i in 1...n {
    for j in 1...n {
        if i * j > n {
            break outerLoop
        }
        print(i, j)
    }
}
    

This code uses a labeled break to exit the outer loop early when a condition is met inside the inner loop.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The nested loops where the inner loop runs for each outer loop iteration.
• How many times: The inner loop runs fewer times as the product i * j grows, but in the worst case, it can run up to n times for each i.

How Execution Grows With Input

Explain the growth pattern intuitively.

Input Size (n)	Approx. Operations
10	About 10 to 20 operations before break
100	About 100 to 200 operations before break
1000	About 1000 to 2000 operations before break

Pattern observation: The total steps grow roughly proportional to n, not n squared, because the break stops loops early.

Final Time Complexity

Time Complexity: O(n)

This means the total work grows linearly with the input size because the labeled break stops the loops early.

Common Mistake

[X] Wrong: "Nested loops always mean the program runs in n squared time."

[OK] Correct: Because the labeled break can stop loops early, the total steps can be much less than n squared.

Interview Connect

Understanding how labeled breaks affect loops shows you can reason about code flow and efficiency, a skill useful in many programming tasks.

Self-Check

"What if we replaced the labeled break with a simple break inside the inner loop? How would the time complexity change?"

Practice

1. What is the main purpose of using labeled statements in nested loops in Swift?

easy

A. To speed up the loop execution

B. To declare variables inside loops

C. To directly control which loop to break or continue

D. To automatically optimize memory usage

Solution

1. Step 1: Understand labeled statements

   Labeled statements give a name to loops so you can refer to them directly.

2. Step 2: Control flow in nested loops

   This naming allows breaking or continuing an outer loop from inside an inner loop without extra flags.

3. Final Answer:

   To directly control which loop to break or continue -> Option C

4. Quick Check:

   Labeled loops control flow directly = C [OK]

Hint: Labels let you break/continue outer loops easily [OK]

Common Mistakes:

• Thinking labels declare variables
• Assuming labels improve speed
• Confusing labels with memory optimization

2. Which of the following is the correct syntax to label a loop named 'outerLoop' in Swift?

easy

A. for outerLoop in 1...3 { }

B. outerLoop: for i in 1...3 { }

C. for i in 1...3 outerLoop: { }

D. label outerLoop for i in 1...3 { }

Solution

1. Step 1: Recall label syntax in Swift

   Labels are written as a name followed by a colon before the loop statement.

2. Step 2: Match syntax to options

   outerLoop: for i in 1...3 { } correctly places 'outerLoop:' before the for loop.

3. Final Answer:

   outerLoop: for i in 1...3 { } -> Option B

4. Quick Check:

   Label name + colon before loop = A [OK]

Hint: Label name ends with colon before loop starts [OK]

Common Mistakes:

• Placing label after the loop
• Using 'label' keyword (not valid)
• Putting label inside loop braces

3. What will be the output of the following Swift code?

outerLoop: for i in 1...2 {
    for j in 1...3 {
        if j == 2 {
            break outerLoop
        }
        print("\(i),\(j)")
    }
}

medium

A. 1,1
1,2
1,3
2,1
2,2
2,3

B. 1,1
1,2

C. 1,1
2,1

D. 1,1

Solution

1. Step 1: Understand loop and break behavior

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

2. Step 2: Trace iterations

   First iteration: i=1, j=1 prints "1,1". When j=2, break outerLoop triggers, stopping all loops immediately.

3. Final Answer:

   1,1 -> Option D

4. Quick Check:

   Break outer loop at j=2 stops all = D [OK]

Hint: Break with label stops outer loop immediately [OK]

Common Mistakes:

• Thinking break only stops inner loop
• Expecting all prints before break
• Ignoring label effect on break

4. Identify the error in this Swift code using labeled statements:

outer: for i in 1...2 {
    for j in 1...2 {
        if j == 2 {
            break outerLoop
        }
        print(i, j)
    }
}

medium

A. Label 'outerLoop' is not defined

B. Syntax error: missing colon after label

C. Cannot use break inside nested loops

D. No error, code runs fine

Solution

1. Step 1: Check label definition

   The loop is labeled 'outer', but break tries to use 'outerLoop' which is undefined.

2. Step 2: Understand label usage

   Labels must match exactly. Using a different name causes a compile error.

3. Final Answer:

   Label 'outerLoop' is not defined -> Option A

4. Quick Check:

   Break label must match defined label = B [OK]

Hint: Break label must exactly match loop label name [OK]

Common Mistakes:

• Using different label names
• Forgetting colon after label
• Thinking break can't be used in nested loops

5. Given the task to print all pairs (i, j) where i and j range from 1 to 3, but stop all loops immediately when i + j equals 4, which Swift code correctly uses labeled statements to achieve this?

hard

A. outerLoop: for i in 1...3 {
for j in 1...3 {
if i + j == 4 {
break outerLoop
}
print("\(i),\(j)")
}
}

B. for i in 1...3 {
for j in 1...3 {
if i + j == 4 {
break
}
print(i, j)
}
}

C. outerLoop: for i in 1...3 {
for j in 1...3 {
if i + j == 4 {
continue outerLoop
}
print(i, j)
}
}

D. for i in 1...3 {
for j in 1...3 {
if i + j == 4 {
continue
}
print(i, j)
}
}

Solution

1. Step 1: Understand the stopping condition

   We want to stop all loops immediately when i + j == 4, so we must break the outer loop.

2. Step 2: Use labeled break correctly

   outerLoop: for i in 1...3 { for j in 1...3 { if i + j == 4 { break outerLoop } print("\(i),\(j)") } } uses 'break outerLoop' to stop the outer loop from inside the inner loop, which is correct.

3. Step 3: Check other options

   for i in 1...3 { for j in 1...3 { if i + j == 4 { break } print(i, j) } } breaks only inner loop; C uses continue which skips iterations but doesn't stop loops; D continues inner loop only.

4. Final Answer:

   Code that breaks outerLoop when i + j == 4 -> Option A

5. Quick Check:

   Labeled break stops all loops = A [OK]

Hint: Use labeled break to stop outer loop from inner loop [OK]

Common Mistakes:

• Using break without label stops only inner loop
• Using continue instead of break
• Not labeling the outer loop