I just want to count the words in the array using .count but its giving the following error:
error: type mismatch: inferred type is String but (TypeVariable(T)) -> Boolean was expected
println(sentArr.count(x))
**code**:
fun main() {
var sentence = "Bob hit a ball, the hit BALL flew
far after it was hit."
var sentArr = sentence.split(" ")
var bannedWords = arrayOf("ball")
for(x in sentArr){
println(sentArr.count(x))
}
}
If I am correctly guessing what you're trying to do, this is what you're looking for:
for(x in sentArr){
println(sentArr.count { it == x })
}
This means to count how many of the Strings in the array are the same as x.
If you just want to count the words in the array, you can simply write sentArr.size or sentArr.count(). No need for a loop
Related
I'm wanting to slice a range which I can do in Javascfript but am struggling in kotlin.
my current code is:
internal class blah {
fun longestPalindrome(s: String): String {
var longestP = ""
for (i in 0..s.length) {
for (j in 1..s.length) {
var subS = s.slice(i, j)
if (subS === subS.split("").reversed().joinToString("") && subS.length > longestP.length) {
longestP = subS
}
}
}
return longestP
}
and the error I get is:
Type mismatch.
Required:
IntRange
Found:
Int
Is there a way around this keeping most of the code I have?
As the error message says, slice wants an IntRange, not two Ints. So, pass it a range:
var subS = s.slice(i..j)
By the way, there are some bugs in your code:
You need to iterate up to the length minus 1 since the range starts at 0. But the easier way is to grab the indices range directly: for (i in s.indices)
I assume j should be i or bigger, not 1 or bigger, or you'll be checking some inverted Strings redundantly. It should look like for (j in i until s.length).
You need to use == instead of ===. The second operator is for referential equality, which will always be false for two computed Strings, even if they are identical.
I know this is probably just practice, but even with the above fixes, this code will fail if the String contains any multi-code-unit code points or any grapheme clusters. The proper way to do this would be by turning the String into a list of grapheme clusters and then performing the algorithm, but this is fairly complicated and should probably rely on some String processing code library.
class Solution {
fun longestPalindrome(s: String): String {
var longestPal = ""
for (i in 0 until s.length) {
for (j in i + 1..s.length) {
val substring = s.substring(i, j)
if (substring == substring.reversed() && substring.length > longestPal.length) {
longestPal = substring
}
}
}
return longestPal
}
}
This code is now functioning but unfortunately is not optimized enough to get through all test cases.
I'm kind of stuck, I don't know how to make the second loop to start 1 position above the first loop in Kotlin.
I have an array (named myArray) with 10 elements, I need to Write a Kotlin program that prints the number that has the most consecutive repeated number in the array and also prints the number of times it appears in the sequence.
The program must parse the array from left to right so that if two numbers meet the condition, the one that appears first from left to right will be printed.
Longest: 3
Number: 8
fun main() {
val myArray: IntArray = intArrayOf(1,2,2,4,5,6,7,8,8,8)
for((index , value) in myArray.withIndex()){
var inx = index + 1
var count = 0
var longest = 0
var number = 0
for((inx,element) in myArray.withIndex()) {
if(value == element ){
count+=
}
}
if(longest < count){
longest = count
number = value
}
}
}
I'm against just dropping answers, but it is quite late for me, so I'll leave this answer here and edit it tomorrow with more info on how each part works. I hope that maybe in the meanwhile it will help you to gain some idea to where you might be going wrong.
val results = mutableMapOf<Int, Int>()
(0..myArray.size - 2).forEach { index ->
val current = myArray[index]
if (current == myArray[index + 1]) {
results[current] = (results[current] ?: 1) + 1
}
}
val (max, occurrences) = results.maxByOrNull { it.value } ?: run { println("No multiple occurrences"); return }
println("Most common consecutive number $max, with $occurrences occurrences")
Alternatively if the intArray would be a list, or if we allowed to change it to a list myArray.toList(), you could replace the whole forEach loop with a zipWithNext. But I'm pretty sure that this is a HW question, so I doubt this is the expected way of solving it.
myList.zipWithNext { a, b ->
if (a == b) results[a] = (results[a] ?: 1) + 1
}
I'm trying to average the grades from the input below. How do I slice out the name and just pass the integers to my calcAvg function? The grades are of variable length so my guess is I need to somehow slice my collection from element 1 to the size of the list?
input: Jack:90,80,70,90,50 Jill:80,100,30 Mary:20,100,90,80
fun main(){
print("Enter your grades: ")
val studentGrades = mutableListOf<String>()
while (true){
val input = readLine()!!
if (input.isNullOrBlank()) break
else
studentGrades += input
}
for (i in studentGrades){
val splitRecords = i.split(" ")
for (j in splitRecords){
val splitName = j.split(":", ",")//split the name from the grades
}
}
}
fun calcAvg(list: List<Double>): Double{
return list.average()
}
You're almost there, I think.
After splitting out individual records*, I'd split on just ":" to separate the name from the grades list. That should give you a list of two items: the name, and the grades list, so you can treat those separately. A naive way would be:
val nameAndGradesList = j.split(":")
val name = nameAndGrades[0]
val gradesList = nameAndGrades[1]
But there's a simpler way of doing that, using a destructuring declaration:
val (name, gradesList) = j.split(":")
Then you can separate the grades list into individual grades, and convert them to numbers:
val grades = gradesList.split(",").map{ it.toDouble() }
I guess you can take it from there…
Note that all this code assumes every record is in the right format; if not, it's liable to do strange things (e.g. crash by throwing an exception, or — even worse — give wrong results). In production code, you'd usually want to check for that, and handle invalid records in an appropriate and predictable way.
(* You wouldn't need the first split if each record were on a separate line, which is the usual way.)
I am currently learning kotlin and therefore following the kotlin track on exercism. The following exercise required me to calculate the Hamming difference between two Strings (so basically just counting the number of differences).
I got to the solution with the following code:
object Hamming {
fun compute(dnaOne: String, dnaTwo: String): Int {
if (dnaOne.length != dnaTwo.length) throw IllegalArgumentException("left and right strands must be of equal length.")
var counter = 0
for ((index, letter) in dnaOne.toCharArray().withIndex()) {
if (letter != dnaTwo.toCharArray()[index]) {
counter++
}
}
return counter
}
}
however, in the beginning I tried to do dnaOne.split("").withIndex() instead of dnaOne.toCharArray().withIndex() which did not work, it would literally stop after the first iteration and the following example
Hamming.compute("GGACGGATTCTG", "AGGACGGATTCT") would return 1 instead of the correct integer 9 (which only gets returned when using toCharArray)
I would appreciate any explanation
I was able to simplify this by using the built-in CharSequence.zip function because StringimplementsCharSequence` in Kotlin.
According to the documentation for zip:
Returns a list of pairs built from the characters of this and the [other] char sequences with the same index
The returned list has length of the shortest char sequence.
Which means we will get a List<Pair<Char,Char>> back (a list of pairs of letters in the same positions). Now that we have this, we can use Iterable.count to determine how many of them are different.
I implemented this as an extension function on String rather than in an object:
fun String.hamming(other: String): Int =
if(this.length != other.length) {
throw IllegalArgumentException("String lengths must match")
} else {
this.zip(other).count { it.first != it.second }
}
This also becomes a single expression now.
And to call this:
val ham = "GGACGGATTCTG".hamming("AGGACGGATTCT")
println("Hamming distance: $ham")
I am pretty confused with both functions fold() and reduce() in Kotlin, can anyone give me a concrete example that distinguishes both of them?
fold takes an initial value, and the first invocation of the lambda you pass to it will receive that initial value and the first element of the collection as parameters.
For example, take the following code that calculates the sum of a list of integers:
listOf(1, 2, 3).fold(0) { sum, element -> sum + element }
The first call to the lambda will be with parameters 0 and 1.
Having the ability to pass in an initial value is useful if you have to provide some sort of default value or parameter for your operation. For example, if you were looking for the maximum value inside a list, but for some reason want to return at least 10, you could do the following:
listOf(1, 6, 4).fold(10) { max, element ->
if (element > max) element else max
}
reduce doesn't take an initial value, but instead starts with the first element of the collection as the accumulator (called sum in the following example).
For example, let's do a sum of integers again:
listOf(1, 2, 3).reduce { sum, element -> sum + element }
The first call to the lambda here will be with parameters 1 and 2.
You can use reduce when your operation does not depend on any values other than those in the collection you're applying it to.
The major functional difference I would call out (which is mentioned in the comments on the other answer, but may be hard to understand) is that reduce will throw an exception if performed on an empty collection.
listOf<Int>().reduce { x, y -> x + y }
// java.lang.UnsupportedOperationException: Empty collection can't be reduced.
This is because .reduce doesn't know what value to return in the event of "no data".
Contrast this with .fold, which requires you to provide a "starting value", which will be the default value in the event of an empty collection:
val result = listOf<Int>().fold(0) { x, y -> x + y }
assertEquals(0, result)
So, even if you don't want to aggregate your collection down to a single element of a different (non-related) type (which only .fold will let you do), if your starting collection may be empty then you must either check your collection size first and then .reduce, or just use .fold
val collection: List<Int> = // collection of unknown size
val result1 = if (collection.isEmpty()) 0
else collection.reduce { x, y -> x + y }
val result2 = collection.fold(0) { x, y -> x + y }
assertEquals(result1, result2)
Another difference that none of the other answers mentioned is the following:
The result of a reduce operation will always be of the same type (or a super type) as the data that is being reduced.
We can see that from the definition of the reduce method:
public inline fun <S, T : S> Iterable<T>.reduce(operation: (acc: S, T) -> S): S {
val iterator = this.iterator()
if (!iterator.hasNext()) throw UnsupportedOperationException("Empty collection can't be reduced.")
var accumulator: S = iterator.next()
while (iterator.hasNext()) {
accumulator = operation(accumulator, iterator.next())
}
return accumulator
}
On the other hand, the result of a fold operation can be anything, because there are no restrictions when it comes to setting up the initial value.
So, for example, let us say that we have a string that contains letters and digits. We want to calculate the sum of all the digits.
We can easily do that with fold:
val string = "1a2b3"
val result: Int = string.fold(0, { currentSum: Int, char: Char ->
if (char.isDigit())
currentSum + Character.getNumericValue(char)
else currentSum
})
//result is equal to 6
reduce - The reduce() method transforms a given collection into a single result.
val numbers: List<Int> = listOf(1, 2, 3)
val sum: Int = numbers.reduce { acc, next -> acc + next }
//sum is 6 now.
fold - What would happen in the previous case of an empty list? Actually, there’s no right value to return, so reduce() throws a RuntimeException
In this case, fold is a handy tool. You can put an initial value by it -
val sum: Int = numbers.fold(0, { acc, next -> acc + next })
Here, we’ve provided initial value. In contrast, to reduce(), if the collection is empty, the initial value will be returned which will prevent you from the RuntimeException.
Simple Answer
Result of both reduce and fold is "a list of items will be transformed into a single item".
In case of fold,we provide 1 extra parameter apart from list but in case of reduce,only items in list will be considered.
Fold
listOf("AC","Fridge").fold("stabilizer") { freeGift, itemBought -> freeGift + itemBought }
//output: stabilizerACFridge
In above case,think as AC,fridge bought from store & they give stabilizer as gift(this will be the parameter passed in the fold).so,you get all 3 items together.Please note that freeGift will be available only once i.e for the first iteration.
Reduce
In case of reduce,we get items in list as parameters and can perform required transformations on it.
listOf("AC","Fridge").reduce { itemBought1, itemBought2 -> itemBought1 + itemBought2 }
//output: ACFridge
The difference between the two functions is that fold() takes an initial value and uses it as the accumulated value on the first step, whereas the first step of reduce() uses the first and the second elements as operation arguments on the first step.