I am new to Kotlin (and Java). In order to pick up on the language I am trying to solve some problems from a website.
The problem is quite easy and straightfoward, the function has to count how many times the biggest value is included in an IntArray. My function also works for smaller arrays but seems to exceed the allowed time limit for larger ones (error: Your code did not execute within the time limits).
fun problem(inputArray: Array<Int>): Int {
// Write your code here
val n: Int = inputArray.count{it == inputArray.max()}
return n
}
So as I am trying to improve I am not looking for a faster solution, but for some hints on topics I could look at in order to find a faster solution myself.
Thanks a lot!
In an unordered array you to touch every element to calcuate inputArray.max(). So inputArray.count() goes over all elements and calls max() that goes over all elements.
So runtime goes up n^2 for n elements.
Store inputArray.max() in an extra variable, and you have a linear runtime.
val max = inputArray.max()
val n: Int = inputArray.count{ it == max }
Related
I start with a list of integers from 1 to 1000 in listOfRandoms.
I would like to left join on random from the createDatabase list.
I am currently using a find{} statement within a loop to do this but feel like this is too heavy. Is there not a better (quicker) way to achieve same result?
Psuedo Code
data class DatabaseRow(
val refKey: Int,
val random: Int
)
fun main() {
val createDatabase = (1..1000).map { i -> DatabaseRow(i, Random()) }
val listOfRandoms = (1..1000).map { j ->
val lookup = createDatabase.find { it.refKey == j }
lookup.random
}
}
As mentioned in comments, the question seems to be mixing up database and programming ideas, which isn't helping.
And it's not entirely clear which parts of the code are needed, and which can be replaced. I'm assuming that you already have the createDatabase list, but that listOfRandoms is open to improvement.
The ‘pseudo’ code compiles fine except that:
You don't give an import for Random(), but none of the likely ones return an Int. I'm going to assume that should be kotlin.random.Random.nextInt().
And because lookup is nullable, you can't simply call lookup.random; a quick fix is lookup!!.random, but it would be safer to handle the null case properly with e.g. lookup?.random ?: -1. (That's irrelevant, though, given the assumption above.)
I think the general solution is to create a Map. This can be done very easily from createDatabase, by calling associate():
val map = createDatabase.associate{ it.refKey to it.random }
That should take time roughly proportional to the size of the list. Looking up values in the map is then very efficient (approx. constant time):
map[someKey]
In this case, that takes rather more memory than needed, because both keys and values are integers and will be boxed (stored as separate objects on the heap). Also, most maps use a hash table, which takes some memory.
Since the key is (according to comments) “an ascending list starting from a random number, like 18123..19123”, in this particular case it can instead be stored in an IntArray without any boxing. As you say, array indexes start from 0, so using the key directly would need a huge array and use only the last few cells — but if you know the start key, you could simply subtract that from the array index each time.
Creating such an array would be a bit more complex, for example:
val minKey = createDatabase.minOf{ it.refKey }
val maxKey = createDatabase.maxOf{ it.refKey }
val array = IntArray(maxKey - minKey + 1)
for (row in createDatabase)
array[row.refKey - minKey] = row.random
You'd then access values with:
array[someKey - minKey]
…which is also constant-time.
Some caveats with this approach:
If createDatabase is empty, then minOf() will throw a NoSuchElementException.
If it has ‘holes’, omitting some keys inside that range, then the array will hold its default value of 0 — you can change that by using the alternative IntArray constructor which also takes a lambda giving the initial value.)
Trying to look up a value outside that range will give an ArrayIndexOutOfBoundsException.
Whether it's worth the extra complexity to save a bit of memory will depend on things like the size of the ‘database’, and how long it's in memory for; I wouldn't add that complexity unless you have good reason to think memory usage will be an issue.
I have a data class that describes a chef by their name and their skill level and two lists of chefs with various skill levels.
data class Chef(val name: String, val level: Int)
val listOfChefsOne = listOf(
Chef("Remy", 9),
Chef("Linguini", 7))
val listOfChefsTwo = listOf(
Chef("Mark", 6),
Chef("Maria", 8))
I'm to write a function that takes these two lists and creates a list of pairs
so that the two chefs in a pair skill level's add up to 15. The challenge is to do this using only built in list functions and not for/while loops.
println(pairChefs(listOfChefsOne, listOfChefsTwo))
######################################
[(Chef(name=Remy, level=9), Chef(name=Mark, level=6)),
(Chef(name=Linguini, level=7), Chef(name=Maria, level=8))]
As I mentioned previously I'm not to use any for or while loops in my implementation for the function. I've tried using the forEach function to create a list containing all possible pairs between two lists, but from there I've gotten lost as to how I can filter out only the correct pairs.
I think the clue is in the question here!
I've tried using the forEach function to create a list containing all possible pairs between two lists, but from there I've gotten lost as to how I can filter out only the correct pairs.
There's a filter function that looks perfect for this…
To keep things clear, I'll split out a function for generating all possible pairs. (This is my own, but bears a reassuring resemblance to part of this answer! In any case, you said you'd already solved this bit.)
fun <A, B> Iterable<A>.product(other: Iterable<B>)
= flatMap{ a -> other.map{ b -> a to b }}
The result can then be:
val result = listOfChefsOne.product(listOfChefsTwo)
.filter{ (chef1, chef2) -> chef1.level + chef2.level == 15 }
Note that although this is probably the simplest and most readable way, it's not the most efficient for large lists. (It takes time and memory proportional to the product of the sizes of the two lists.) You could improve large-scale performance by using streams (which would take the same time but constant memory). But for this particular case, it might be even better to group one of the lists by level, then for each element of the other list, you could directly look up a Chef with 15 - its level. (That would time proportional to the sum of the sizes of the two lists, and space proportional to the size of the first list.)
Here is the pretty simple naive solution:
val result = listOfChefsOne.flatMap { chef1 ->
listOfChefsTwo.mapNotNull { chef2 ->
if (chef1.level + chef2.level == 15) {
chef1 to chef2
} else {
null
}
}
}
println(result) // prints [(Chef(name=Remy, level=9), Chef(name=Mark, level=6)), (Chef(name=Linguini, level=7), Chef(name=Maria, level=8))]
Currently, I am looking into Kotlin and have a question about Sequences vs. Collections.
I read a blog post about this topic and there you can find this code snippets:
List implementation:
val list = generateSequence(1) { it + 1 }
.take(50_000_000)
.toList()
measure {
list
.filter { it % 3 == 0 }
.average()
}
// 8644 ms
Sequence implementation:
val sequence = generateSequence(1) { it + 1 }
.take(50_000_000)
measure {
sequence
.filter { it % 3 == 0 }
.average()
}
// 822 ms
The point here is that the Sequence implementation is about 10x faster.
However, I do not really understand WHY that is. I know that with a Sequence, you do "lazy evaluation", but I cannot find any reason why that helps reducing the processing in this example.
However, here I know why a Sequence is generally faster:
val result = sequenceOf("a", "b", "c")
.map {
println("map: $it")
it.toUpperCase()
}
.any {
println("any: $it")
it.startsWith("B")
}
Because with a Sequence you process the data "vertically", when the first element starts with "B", you don't have to map for the rest of the elements. It makes sense here.
So, why is it also faster in the first example?
Let's look at what those two implementations are actually doing:
The List implementation first creates a List in memory with 50 million elements. This will take a bare minimum of 200MB, since an integer takes 4 bytes.
(In fact, it's probably far more than that. As Alexey Romanov pointed out, since it's a generic List implementation and not an IntList, it won't be storing the integers directly, but will be ‘boxing’ them — storing references to Int objects. On the JVM, each reference could be 8 or 16 bytes, and each Int could take 16, giving 1–2GB. Also, depending how the List gets created, it might start with a small array and keep creating larger and larger ones as the list grows, copying all the values across each time, using more memory still.)
Then it has to read all the values back from the list, filter them, and create another list in memory.
Finally, it has to read all those values back in again, to calculate the average.
The Sequence implementation, on the other hand, doesn't have to store anything! It simply generates the values in order, and as it does each one it checks whether it's divisible by 3 and if so includes it in the average.
(That's pretty much how you'd do it if you were implementing it ‘by hand’.)
You can see that in addition to the divisibility checking and average calculation, the List implementation is doing a massive amount of memory access, which will take a lot of time. That's the main reason it's far slower than the Sequence version, which doesn't!
Seeing this, you might ask why we don't use Sequences everywhere… But this is a fairly extreme example. Setting up and then iterating the Sequence has some overhead of its own, and for smallish lists that can outweigh the memory overhead. So Sequences only have a clear advantage in cases when the lists are very large, are processed strictly in order, there are several intermediate steps, and/or many items are filtered out along the way (especially if the Sequence is infinite!).
In my experience, those conditions don't occur very often. But this question shows how important it is to recognise them when they do!
Leveraging lazy-evaluation allows avoiding the creation of intermediate objects that are irrelevant from the point of the end goal.
Also, the benchmarking method used in the mentioned article is not super accurate. Try to repeat the experiment with JMH.
Initial code produces a list containing 50_000_000 objects:
val list = generateSequence(1) { it + 1 }
.take(50_000_000)
.toList()
then iterates through it and creates another list containing a subset of its elements:
.filter { it % 3 == 0 }
... and then proceeds with calculating the average:
.average()
Using sequences allows you to avoid doing all those intermediate steps. The below code doesn't produce 50_000_000 elements, it's just a representation of that 1...50_000_000 sequence:
val sequence = generateSequence(1) { it + 1 }
.take(50_000_000)
adding a filtering to it doesn't trigger the calculation itself as well but derives a new sequence from the existing one (3, 6, 9...):
.filter { it % 3 == 0 }
and eventually, a terminal operation is called that triggers the evaluation of the sequence and the actual calculation:
.average()
Some relevant reading:
Kotlin: Beware of Java Stream API Habits
Kotlin Collections API Performance Antipatterns
I am just trying the new kotlin language. I came across sequences which generate infinite list. I generated a sequence and tried printing the first 10 elements. But the below code didnt print anything:
fun main(args: Array<String>) {
val generatePrimeFrom2 = generateSequence(3){ it + 2 }
print(generatePrimeFrom2.toList().take(10))
}
But when I changed take(10).toList() in the print statement it work fine. Why is it so ?
This code worked fine for me:
fun main(args: Array<String>) {
val generatePrimeFrom2 = generateSequence(3){ it + 2 }
print(generatePrimeFrom2.take(10).toList())
}
The generateSequence function generates a sequence that is either infinite or finishes when the lambda passed to it returns null. In your case, it is { it + 2 }, which never returns null, so the sequence is infinite.
When you call .toList() on a sequence, it will try to collect all sequence elements and thus will never stop if the sequence is infinite (unless the index overflows or an out-of-memory error happens), so it does not print anything because it does not finish.
In the second case, on contrary, you limit the number of elements in the sequence with .take(10) before trying to collect its items. Then the .toList() call simply collects the 10 items and finishes.
It may become more clear if you check this Q&A about differences between Sequence<T> and Iterable<T>: (link)
Here is the hint -> generate infinite list. In the first solution you first want to create a list (wait infinity) then take first 10 elements.
On the second snippet, from infinite list you take only first 10 elements and change it to list
generatePrimeFrom2.toList() tries to compute/create an infinite-length list.
generatePrimeFrom2.toList().take(10) then takes the first 10 elements from the infinite-length list.
It does not print because it is calculating that infinite-length list.
Whereas, generatePrimeFrom2.take(10) only tries to compute the first 10 elements.
generatePrimeFrom2.take(10).toList() converts the first 10 elements to the list.
You know, generateSequence(3){ it + 2 } does not have the end. So it has the infinite length.
Sequences do not have actual values, calculated when they are needed, but Lists have to have actual values.
I came across sequences which generate infinite list.
This is not actually correct. The main point is that a sequence is not a list. It is a lazily evaluated construct and only the items you request will actually become "materialized", i.e., their memory allocated on the heap.
That's why it's not interchangeable to write
infiniteSeq.toList().take(10)
and
infiniteSeq.take(10).toList()
The former will try to instantiate infinitely many items—and, predictably, fail at it.
I wrote the following code:
val src = (0 until 1000000).toList()
val dest = ArrayList<Double>(src.size / 2 + 1)
for (i in src)
{
if (i % 2 == 0) dest.add(Math.sqrt(i.toDouble()))
}
IntellJ (in my case AndroidStudio) is asking me if I want to replace the for loop with operations from stdlib. This results in the following code:
val src = (0 until 1000000).toList()
val dest = ArrayList<Double>(src.size / 2 + 1)
src.filter { it % 2 == 0 }
.mapTo(dest) { Math.sqrt(it.toDouble()) }
Now I must say, I like the changed code. I find it easier to write than for loops when I come up with similar situations. However upon reading what filter function does, I realized that this is a lot slower code compared to the for loop. filter function creates a new list containing only the elements from src that match the predicate. So there is one more list created and one more loop in the stdlib version of the code. Ofc for small lists it might not be important, but in general this does not sound like a good alternative. Especially if one should chain more methods like this, you can get a lot of additional loops that could be avoided by writing a for loop.
My question is what is considered good practice in Kotlin. Should I stick to for loops or am I missing something and it does not work as I think it works.
If you are concerned about performance, what you need is Sequence. For example, your above code will be
val src = (0 until 1000000).toList()
val dest = ArrayList<Double>(src.size / 2 + 1)
src.asSequence()
.filter { it % 2 == 0 }
.mapTo(dest) { Math.sqrt(it.toDouble()) }
In the above code, filter returns another Sequence, which represents an intermediate step. Nothing is really created yet, no object or array creation (except a new Sequence wrapper). Only when mapTo, a terminal operator, is called does the resulting collection is created.
If you have learned java 8 stream, you may found the above explaination somewhat familiar. Actually, Sequence is roughly the kotlin equivalent of java 8 Stream. They share similiar purpose and performance characteristic. The only difference is Sequence isn't designed to work with ForkJoinPool, thus a lot easier to implement.
When there is multiple steps involved or the collection may be large, it's suggested to use Sequence instead of plain .filter {...}.mapTo{...}. I also suggest you to use the Sequence form instead of your imperative form because it's easier to understand. Imperative form may become complex, thus hard to understand, when there are 5 or more steps involved in the data processing. If there is just one step, you don't need a Sequence, because it just creates garbage and gives you nothing useful.
You're missing something. :-)
In this particular case, you can use an IntProgression:
val progression = 0 until 1_000_000 step 2
You can then create your desired list of squares in various ways:
// may make the list larger than necessary
// its internal array is copied each time the list grows beyond its capacity
// code is very straight forward
progression.map { Math.sqrt(it.toDouble()) }
// will make the list the exact size needed
// no copies are made
// code is more complicated
progression.mapTo(ArrayList(progression.last / 2 + 1)) { Math.sqrt(it.toDouble()) }
// will make the list the exact size needed
// a single intermediate list is made
// code is minimal and makes sense
progression.toList().map { Math.sqrt(it.toDouble()) }
My advice would be to choose whichever coding style you prefer. Kotlin is both object-oriented and functional language, meaning both of your propositions are correct.
Usually, functional constructs favor readability over performance; however, in some cases, procedural code will also be more readable. You should try to stick with one style as much as possible, but don't be afraid to switch some code if you feel like it's better suited to your constraints, either readability, performance, or both.
The converted code does not need the manual creation of the destination list, and can be simplified to:
val src = (0 until 1000000).toList()
val dest = src.filter { it % 2 == 0 }
.map { Math.sqrt(it.toDouble()) }
And as mentioned in the excellent answer by #glee8e you can use a sequence to do a lazy evaluation. The simplified code for using a sequence:
val src = (0 until 1000000).toList()
val dest = src.asSequence() // change to lazy
.filter { it % 2 == 0 }
.map { Math.sqrt(it.toDouble()) }
.toList() // create the final list
Note the addition of the toList() at the end is to change from a sequence back to a final list which is the one copy made during the processing. You can omit that step to remain as a sequence.
It is important to highlight the comments by #hotkey saying that you should not always assume that another iteration or a copy of a list causes worse performance than lazy evaluation. #hotkey says:
Sometimes several loops. even if they copy the whole collection, show good performance because of good locality of reference. See: Kotlin's Iterable and Sequence look exactly same. Why are two types required?
And excerpted from that link:
... in most cases it has good locality of reference thus taking advantage of CPU cache, prediction, prefetching etc. so that even multiple copying of a collection still works good enough and performs better in simple cases with small collections.
#glee8e says that there are similarities between Kotlin sequences and Java 8 streams, for detailed comparisons see: What Java 8 Stream.collect equivalents are available in the standard Kotlin library?