For loop with ranges in Kotlin - Time & Space Complexity

Time Complexity: For loop with ranges

O(n)

Understanding Time Complexity

We want to understand how the time it takes to run a for loop with ranges changes as the range size grows.

How does the number of steps grow when the range gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

for (i in 1..n) {
    println(i)
}
    

This code prints numbers from 1 up to n using a for loop with a range.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The loop runs once for each number from 1 to n.
• How many times: Exactly n times, where n is the size of the range.

How Execution Grows With Input

As n grows, the number of times the loop runs grows the same way.

Input Size (n)	Approx. Operations
10	10
100	100
1000	1000

Pattern observation: The number of operations grows directly with n. If n doubles, the work doubles.

Final Time Complexity

Time Complexity: O(n)

This means the time to run the loop grows in a straight line with the size of the range.

Common Mistake

[X] Wrong: "The loop runs in constant time because it just prints numbers."

[OK] Correct: Each print takes time, and since the loop runs n times, the total time grows with n.

Interview Connect

Understanding how loops grow with input size helps you explain code efficiency clearly and confidently in interviews.

Self-Check

"What if we changed the loop to run from 1 to n squared (1..n*n)? How would the time complexity change?"

Practice

1. What does the following Kotlin code do?
for (i in 1..5) { println(i) }

easy

A. Prints numbers from 1 to 4

B. Prints numbers from 1 to 5 inclusive

C. Prints numbers from 0 to 5

D. Prints numbers from 5 down to 1

Solution

1. Step 1: Understand the range operator

   The operator 1..5 creates a range including numbers 1, 2, 3, 4, and 5.

2. Step 2: Loop through the range

   The for loop runs from 1 to 5, printing each number.

3. Final Answer:

   Prints numbers from 1 to 5 inclusive -> Option B

4. Quick Check:

   Range 1..5 includes 5 [OK]

Hint: .. means inclusive range from start to end [OK]

Common Mistakes:

• Thinking .. excludes the last number
• Confusing range with step or downTo
• Assuming it counts down automatically

2. Which of the following is the correct syntax for a for loop counting down from 5 to 1 in Kotlin?

easy

A. for (i in 5 downTo 1) { println(i) }

B. for (i in 1..5 step -1) { println(i) }

C. for (i in 5..1) { println(i) }

D. for (i in 1 downTo 5) { println(i) }

Solution

1. Step 1: Identify correct descending range syntax

   Kotlin uses downTo to count down from a higher number to a lower number.

2. Step 2: Check each option

   for (i in 5 downTo 1) { println(i) } uses 5 downTo 1, which is correct. for (i in 5..1) { println(i) } uses an invalid range (5..1). for (i in 1..5 step -1) { println(i) } tries to use negative step incorrectly. for (i in 1 downTo 5) { println(i) } reverses the range incorrectly.

3. Final Answer:

   for (i in 5 downTo 1) { println(i) } -> Option A

4. Quick Check:

   Use downTo for descending loops [OK]

Hint: Use downTo to count down in Kotlin loops [OK]

Common Mistakes:

• Using .. for descending ranges
• Trying negative step with .. (not allowed)
• Swapping start and end in downTo

3. What is the output of this Kotlin code?

var sum = 0
for (i in 1..5 step 2) {
  sum += i
}
println(sum)

medium

A. 6

B. 15

C. 9

D. 10

Solution

1. Step 1: Understand the loop range and step

   The loop runs from 1 to 5, stepping by 2: values are 1, 3, 5.

2. Step 2: Calculate the sum of these values

   sum = 0 + 1 + 3 + 5 = 9

3. Final Answer:

   9 -> Option C

4. Quick Check:

   1 + 3 + 5 = 9 [OK]

Hint: Add numbers in steps to find sum quickly [OK]

Common Mistakes:

• Adding all numbers 1 to 5 instead of stepping
• Using step as a multiplier
• Confusing step with downTo

4. Find the error in this Kotlin code snippet:

for (i in 10..1 step 2) {
  println(i)
}

medium

A. println cannot be used inside for loop

B. step cannot be used with ranges

C. Variable i is not declared

D. Range 10..1 is invalid for counting down

Solution

1. Step 1: Analyze the range 10..1

   The range 10..1 tries to count up from 10 to 1, which is invalid because 10 is greater than 1.

2. Step 2: Understand correct descending syntax

   To count down, use downTo instead of .., e.g., 10 downTo 1 step 2.

3. Final Answer:

   Range 10..1 is invalid for counting down -> Option D

4. Quick Check:

   Use downTo for descending ranges [OK]

Hint: .. always counts up; use downTo to count down [OK]

Common Mistakes:

• Using .. for descending ranges
• Thinking step fixes descending range
• Ignoring Kotlin's range direction rules

5. You want to print all even numbers from 2 to 20 inclusive using a for loop in Kotlin. Which code snippet correctly does this?

hard

A. for (i in 2..20 step 2) { println(i) }

B. for (i in 2..19) { if (i % 2 == 0) println(i) }

C. for (i in 2 downTo 20 step 2) { println(i) }

D. for (i in 1..10 step 2) { println(i * 2) }

Solution

1. Step 1: Identify the range and step for even numbers

   Even numbers from 2 to 20 can be generated by starting at 2 and stepping by 2 up to 20.

2. Step 2: Check each option

   for (i in 2..20 step 2) { println(i) } uses 2..20 step 2, which prints 2,4,6,...,20 correctly. for (i in 2..19) { if (i % 2 == 0) println(i) } ends at 19 so prints evens up to 18, missing 20. for (i in 2 downTo 20 step 2) { println(i) } counts down incorrectly. for (i in 1..10 step 2) { println(i * 2) } prints 2,6,10,..., not all even numbers.

3. Final Answer:

   for (i in 2..20 step 2) { println(i) } -> Option A

4. Quick Check:

   Use step 2 from 2 to 20 for evens [OK]

Hint: Use step 2 starting at 2 to print even numbers [OK]

Common Mistakes:

• Using downTo instead of .. for ascending
• Forgetting to step by 2
• Multiplying inside loop incorrectly