Sequence vs collection performance in Kotlin

Introduction

Sequences and collections help you handle groups of data. Knowing their performance helps you choose the best one for your task.

When you want to process large lists without using much memory.

When you need to perform many operations on data step-by-step.

When you want faster results by avoiding creating many temporary lists.

When you have small data and want simple, direct access.

When you want to chain multiple operations efficiently.

Syntax

Kotlin

val mySequence = sequenceOf(1, 2, 3)
val myList = listOf(1, 2, 3)

Sequences process elements one by one, which can save memory.

Collections like lists hold all elements in memory at once.

Examples

This creates a new list by doubling each number immediately.

Kotlin

val numbers = listOf(1, 2, 3, 4, 5)
val doubled = numbers.map { it * 2 }

This creates a sequence that will double numbers only when needed.

Kotlin

val numbersSeq = sequenceOf(1, 2, 3, 4, 5)
val doubledSeq = numbersSeq.map { it * 2 }

Filters even numbers from the sequence and converts to a list.

Kotlin

val result = numbersSeq.filter { it % 2 == 0 }.toList()

Sample Program

This program compares time taken to process a large list using collections and sequences. It doubles numbers and filters those divisible by 3.

Kotlin

import kotlin.system.measureTimeMillis

fun main() {
    val numbers = (1..1_000_000).toList()

    val timeList = measureTimeMillis {
        val doubled = numbers.map { it * 2 }
        val filtered = doubled.filter { it % 3 == 0 }
        println("List result size: ${filtered.size}")
    }

    val numbersSeq = numbers.asSequence()

    val timeSeq = measureTimeMillis {
        val doubledSeq = numbersSeq.map { it * 2 }
        val filteredSeq = doubledSeq.filter { it % 3 == 0 }.toList()
        println("Sequence result size: ${filteredSeq.size}")
    }

    println("List time: $timeList ms")
    println("Sequence time: $timeSeq ms")
}

Important Notes

Sequences are lazy: they do work only when needed, saving memory and sometimes time.

Collections do all work immediately, which can be faster for small data.

For big data or many chained operations, sequences usually perform better.

Summary

Sequences process data lazily, saving memory and improving performance on large data.

Collections hold all data in memory and process eagerly, good for small data.

Choose sequences for big or complex data tasks, collections for simple or small tasks.

Practice

1. What is the main advantage of using a Sequence over a Collection in Kotlin when working with large datasets?

easy

A. Sequences process data lazily, saving memory and improving performance.

B. Sequences store all data in memory for faster access.

C. Sequences automatically sort data for you.

D. Sequences are easier to write than collections.

Solution

1. Step 1: Understand how sequences process data

   Sequences process elements one by one only when needed, which is called lazy processing.

2. Step 2: Compare with collections

   Collections hold all elements in memory and process them eagerly, which can use more memory for large data.

3. Final Answer:

   Sequences process data lazily, saving memory and improving performance. -> Option A

4. Quick Check:

   Lazy processing = better for large data [OK]

Hint: Remember: sequences delay work, collections do it all at once [OK]

Common Mistakes:

• Thinking sequences store all data in memory
• Confusing lazy processing with sorting
• Assuming sequences are always simpler to write

2. Which of the following Kotlin code snippets correctly creates a sequence from a list?

easy

A. val seq = listOf(1, 2, 3).toSequence()

B. val seq = sequenceOf(listOf(1, 2, 3))

C. val seq = listOf(1, 2, 3).asSequence()

D. val seq = Sequence(listOf(1, 2, 3))

Solution

1. Step 1: Identify correct method to convert list to sequence

   The method asSequence() converts a collection to a sequence lazily.

2. Step 2: Check other options for correctness

   sequenceOf() creates a sequence from given elements, not a list; toSequence() and Sequence() constructor do not exist.

3. Final Answer:

   val seq = listOf(1, 2, 3).asSequence() -> Option C

4. Quick Check:

   Use asSequence() to convert collections [OK]

Hint: Use asSequence() to convert collections to sequences [OK]

Common Mistakes:

• Using non-existent toSequence() method
• Confusing sequenceOf() with asSequence()
• Trying to call Sequence() constructor directly

3. Consider the following Kotlin code:

val numbers = (1..5).asSequence()
    .map { it * 2 }
    .filter { it > 5 }
    .toList()
println(numbers)

What is the output?

medium

A. [12, 14, 16]

B. [2, 4, 6, 8, 10]

C. [1, 2, 3, 4, 5]

D. [6, 8, 10]

Solution

1. Step 1: Apply map transformation

   Numbers 1 to 5 are doubled: [2, 4, 6, 8, 10].

2. Step 2: Apply filter condition

   Keep only numbers greater than 5: [6, 8, 10].

3. Final Answer:

   [6, 8, 10] -> Option D

4. Quick Check:

   Map then filter = [6, 8, 10] [OK]

Hint: Map first, then filter to get final list [OK]

Common Mistakes:

• Filtering before mapping
• Forgetting to convert sequence to list
• Assuming original range is unchanged

4. What is wrong with this Kotlin code snippet?

val list = listOf(1, 2, 3, 4, 5)
val seq = list.asSequence()
val result = seq.map { it * 2 }
println(result)

medium

A. The map function does not exist for sequences.

B. The map operation is not executed because sequences are lazy.

C. The println statement causes a syntax error.

D. You cannot convert a list to a sequence using asSequence().

Solution

1. Step 1: Understand sequence laziness

   Sequences process operations lazily, so map returns a sequence but does not execute immediately.

2. Step 2: Check what is printed

   Printing result prints the sequence object, not the transformed data, because no terminal operation like toList() was called.

3. Final Answer:

   The map operation is not executed because sequences are lazy. -> Option B

4. Quick Check:

   Sequences need terminal operation to execute [OK]

Hint: Call toList() or similar to execute sequence operations [OK]

Common Mistakes:

• Expecting immediate execution of map on sequences
• Thinking asSequence() is invalid
• Assuming println shows transformed data

5. You have a large database query result as a Kotlin list of 1 million records. You want to filter records where age > 30, then map to names, and finally collect the first 10 names. Which approach is best for performance?

hard

A. Use a sequence: convert list to sequence, filter, map, then take(10) and toList().

B. Use a collection: filter, map on the list, then take first 10 elements.

C. Convert list to array, then filter and map using loops.

D. Sort the list first, then filter and map.

Solution

1. Step 1: Consider data size and operations

   With 1 million records, processing eagerly wastes memory and time.

2. Step 2: Use sequence for lazy processing

   Sequence processes elements one by one, stops after taking 10, saving resources.

3. Step 3: Compare other options

   Collections process all data eagerly; arrays and sorting add unnecessary overhead.

4. Final Answer:

   Use a sequence: convert list to sequence, filter, map, then take(10) and toList(). -> Option A

5. Quick Check:

   Lazy sequence + take(10) = best for large data [OK]

Hint: Use sequences with take() to limit processing early [OK]

Common Mistakes:

• Processing entire list eagerly wasting memory
• Sorting unnecessarily before filtering
• Using arrays without lazy evaluation