Type checking with is operator in Kotlin - Time & Space Complexity

Time Complexity: Type checking with is operator

O(n)

Understanding Time Complexity

We want to understand how long it takes to check an object's type using the is operator in Kotlin.

How does the time needed change as we check more objects?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fun countStrings(items: List<Any>): Int {
    var count = 0
    for (item in items) {
        if (item is String) {
            count++
        }
    }
    return count
}

This code counts how many items in a list are strings by checking each item's type.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through each item in the list and checking its type with is.
• How many times: Once for every item in the list.

How Execution Grows With Input

As the list gets bigger, the number of type checks grows directly with the number of items.

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

Pattern observation: The work doubles when the list size doubles, growing in a straight line.

Final Time Complexity

Time Complexity: O(n)

This means the time to check types grows directly with the number of items in the list.

Common Mistake

[X] Wrong: "Type checking with is is instant and does not depend on list size."

[OK] Correct: Each item must be checked one by one, so more items mean more checks and more time.

Interview Connect

Understanding how type checks scale helps you write efficient code and explain your reasoning clearly in interviews.

Self-Check

"What if we checked types only for the first half of the list? How would the time complexity change?"

Practice

1. What does the is operator do in Kotlin?

easy

A. Checks if a variable is of a specific type

B. Changes the type of a variable

C. Creates a new variable

D. Deletes a variable

Solution

1. Step 1: Understand the purpose of is operator

   The is operator is used to check if a variable holds a value of a certain type.

2. Step 2: Differentiate from other options

   It does not change, create, or delete variables; it only checks type.

3. Final Answer:

   Checks if a variable is of a specific type -> Option A

4. Quick Check:

   Type check = A [OK]

Hint: Remember: is means 'is this type?' check [OK]

Common Mistakes:

• Thinking is changes the variable type
• Confusing is with variable creation
• Assuming is deletes variables

2. Which of the following is the correct syntax to check if obj is a String in Kotlin?

easy

A. if (obj instanceof String)

B. if (obj is String)

C. if (obj type String)

D. if (obj == String)

Solution

1. Step 1: Recall Kotlin syntax for type checking

   Kotlin uses is keyword to check types, not instanceof or other keywords.

2. Step 2: Match the correct syntax

   The correct syntax is if (obj is String).

3. Final Answer:

   if (obj is String) -> Option B

4. Quick Check:

   Correct syntax = C [OK]

Hint: Use is keyword, not Java's instanceof [OK]

Common Mistakes:

• Using Java syntax instanceof in Kotlin
• Writing obj type String which is invalid
• Comparing with == instead of type check

3. What will be the output of this Kotlin code?

fun checkType(x: Any) {
    if (x is Int) {
        println(x + 5)
    } else {
        println("Not an Int")
    }
}

checkType(10)
checkType("Hello")

medium

A. Error at runtime

B. 10
Not an Int

C. 15
Hello5

D. 15
Not an Int

Solution

1. Step 1: Analyze the first call checkType(10)

   10 is an Int, so inside the if block, x + 5 equals 15, printed as "15".

2. Step 2: Analyze the second call checkType("Hello")

   "Hello" is not an Int, so else block runs printing "Not an Int".

3. Final Answer:

   15 Not an Int -> Option D

4. Quick Check:

   Int check prints sum, else prints message = A [OK]

Hint: If is true, use variable as that type inside block [OK]

Common Mistakes:

• Assuming string "Hello" can be added to 5
• Expecting runtime error instead of else output
• Confusing output order

4. Identify the error in this Kotlin code snippet:

fun printLength(obj: Any) {
    if (obj is String) {
        println(obj.length())
    }
}

medium

A. No error, code is correct

B. Missing else block after if

C. Using parentheses with length property

D. Incorrect use of is operator

Solution

1. Step 1: Check usage of length on String

   In Kotlin, length is a property, not a function, so it should not have parentheses.

2. Step 2: Confirm other parts are correct

   The is operator is used correctly and else block is optional.

3. Final Answer:

   Using parentheses with length property -> Option C

4. Quick Check:

   Property access no parentheses = D [OK]

Hint: Remember: Kotlin properties have no parentheses [OK]

Common Mistakes:

• Adding parentheses to properties like length
• Thinking else block is mandatory
• Misusing is operator syntax

5. Given a list of Any elements, how can you create a new list containing only the strings using the is operator in Kotlin?

hard

A. val strings = list.filterIsInstance<String>()

B. val strings = list.filter { it is String }

C. val strings = list.map { if (it is String) it else null }

D. val strings = list.filter { it == String }

Solution

1. Step 1: Understand filtering by type

   To get only strings, Kotlin provides filterIsInstance<String>() which filters elements by type safely.

2. Step 2: Evaluate other options

   val strings = list.filter { it is String } filters by is String but keeps type as Any, not cast to String. val strings = list.map { if (it is String) it else null } maps to nullable list, not filtering. val strings = list.filter { it == String } compares with type itself, which is invalid.

3. Final Answer:

   val strings = list.filterIsInstance<String>() -> Option A

4. Quick Check:

   Use filterIsInstance for type filtering [OK]

Hint: Use filterIsInstance to get typed list quickly [OK]

Common Mistakes:

• Using filter without casting
• Mapping to nullable instead of filtering
• Comparing element to type object