Stride for custom step in Swift - Time & Space Complexity

Time Complexity: Stride for custom step

O(n / step)

Understanding Time Complexity

When using stride with a custom step in Swift, it is important to understand how the number of steps affects the time it takes to run the code.

We want to know how the execution time grows as the range and step size change.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

let start = 0
let end = 100
let step = 5

for i in stride(from: start, to: end, by: step) {
    print(i)
}
    

This code prints numbers from 0 up to but not including 100, stepping by 5 each time.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop using stride repeats printing.
• How many times: It repeats once for each step from start to end, increasing by the step size.

How Execution Grows With Input

As the range between start and end grows, the number of loop steps increases proportionally to how many steps fit in that range.

Input Size (n)	Approx. Operations
10	2 (steps of 5: 0, 5)
100	20 (steps of 5: 0, 5, 10, ..., 95)
1000	200 (steps of 5: 0, 5, 10, ..., 995)

Pattern observation: The number of operations grows roughly in direct proportion to the input size divided by the step size.

Final Time Complexity

Time Complexity: O(n / step)

This means the time grows linearly with the number of steps taken, which depends on the range size divided by the step size.

Common Mistake

[X] Wrong: "The loop always runs n times regardless of the step size."

[OK] Correct: The step size controls how many times the loop runs; bigger steps mean fewer iterations.

Interview Connect

Understanding how loops with custom steps scale helps you explain performance clearly and shows you can reason about code efficiency in real situations.

Self-Check

"What if we changed the step size to 1? How would the time complexity change?"

Practice

1. What does the Swift function stride(from: 1, to: 10, by: 2) do?

easy

A. Generates numbers starting at 1, decreasing by 2, up to but not including 10.

B. Generates numbers starting at 1, increasing by 2, including 10.

C. Generates numbers starting at 1, increasing by 2, up to but not including 10.

D. Generates numbers starting at 1, decreasing by 2, including 10.

Solution

1. Step 1: Understand the parameters of stride

   The function uses from: 1 as the start, to: 10 as the end (exclusive), and by: 2 as the step size.

2. Step 2: Determine the sequence generated

   It generates numbers starting at 1, adding 2 each time, stopping before reaching 10, so the sequence is 1, 3, 5, 7, 9.

3. Final Answer:

   Generates numbers starting at 1, increasing by 2, up to but not including 10. -> Option C

4. Quick Check:

   stride(from:1, to:10, by:2) excludes 10 [OK]

Hint: Remember: 'to' excludes the end value in stride [OK]

Common Mistakes:

• Confusing 'to' with 'through' and including the end value
• Mixing up positive and negative step directions
• Assuming the step is always 1

2. Which of the following is the correct syntax to create a stride in Swift that counts down from 10 to 2 by steps of 2, including 2?

easy

A. stride(from: 10, through: 2, by: -2)

B. stride(from: 10, to: 2, by: -2)

C. stride(from: 2, through: 10, by: 2)

D. stride(from: 2, to: 10, by: 2)

Solution

1. Step 1: Identify counting down with inclusion

   Counting down from 10 to 2 means the step is negative (-2). Including 2 means using through.

2. Step 2: Match syntax to requirements

   Only stride(from: 10, through: 2, by: -2) matches counting down with inclusion of 2.

3. Final Answer:

   stride(from: 10, through: 2, by: -2) -> Option A

4. Quick Check:

   Use 'through' to include end, negative step for countdown [OK]

Hint: Use 'through' to include end; negative step for countdown [OK]

Common Mistakes:

• Using 'to' instead of 'through' when including the end
• Using positive step when counting down
• Swapping start and end values

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

for i in stride(from: 0, to: 5, by: 2) {
    print(i)
}

medium

A. 0 2 4

B. 0 2 4 6

C. 0 2 4 5

D. 2 4

Solution

1. Step 1: Analyze the stride parameters

   The loop starts at 0, goes up to but excludes 5, stepping by 2.

2. Step 2: List the values printed

   Values are 0, 2, 4. It stops before reaching 5.

3. Final Answer:

   0 2 4 -> Option A

4. Quick Check:

   Stride 'to' excludes end, so 5 is not printed [OK]

Hint: Remember 'to' excludes end value in stride loops [OK]

Common Mistakes:

• Including the end value 5 mistakenly
• Adding an extra step beyond the end
• Confusing step size with end value

4. Identify the error in this Swift code snippet:

for i in stride(from: 1, through: 10, by: 0) {
    print(i)
}

medium

A. Using 'through' instead of 'to' is invalid syntax.

B. Step value cannot be zero; it causes a runtime error.

C. Start value must be less than end value for positive step.

D. The loop will run infinitely without errors.

Solution

1. Step 1: Check the step parameter

   The step is zero, which is invalid because it causes no progress in the loop.

2. Step 2: Understand the effect of zero step

   A zero step causes a runtime error or infinite loop because the loop never advances.

3. Final Answer:

   Step value cannot be zero; it causes a runtime error. -> Option B

4. Quick Check:

   Step must not be zero in stride [OK]

Hint: Step cannot be zero; always non-zero positive or negative [OK]

Common Mistakes:

• Thinking 'through' is invalid syntax
• Assuming zero step means step of 1
• Ignoring runtime errors caused by zero step

5. You want to create a Swift loop that prints all odd numbers from 1 to 15 inclusive, but only those greater than 5. Which stride and condition combination achieves this?

hard

A. for i in stride(from: 1, to: 15, by: 2) { if i >= 5 { print(i) } }

B. for i in stride(from: 5, to: 15, by: 2) { print(i) }

C. for i in stride(from: 1, through: 15, by: 2) { if i >= 5 { print(i) } }

D. for i in stride(from: 1, through: 15, by: 2) { if i > 5 { print(i) } }

Solution

1. Step 1: Identify stride to generate odd numbers 1 to 15 inclusive

   Using stride(from: 1, through: 15, by: 2) generates 1,3,5,...,15 including 15.

2. Step 2: Filter numbers greater than 5

   The condition if i > 5 prints only numbers greater than 5, i.e., 7,9,11,13,15.

3. Final Answer:

   for i in stride(from: 1, through: 15, by: 2) { if i > 5 { print(i) } } -> Option D

4. Quick Check:

   Stride with 'through' includes 15; condition filters >5 [OK]

Hint: Use 'through' to include end; filter inside loop for conditions [OK]

Common Mistakes:

• Using 'to' excludes 15, missing last number
• Incorrect condition filtering numbers equal to 5
• Starting stride at 5 misses odd numbers less than 5