I am trying to solve the following question on LeetCode; Write a function that takes an unsigned integer and returns the number of '1' bits it has. Constraints: The input must be a binary string of length 32.
I have written the following code for that which works fine for inputs 00000000000000000000000000001011 and 00000000000000000000000010000000 (provided internally by the website) but give output 0 for input 11111111111111111111111111111101 and in my local compiler for the last input it says "out of range"
class Solution {
// you need treat n as an unsigned value
fun hammingWeight(n:Int):Int {
var num = n
var setCountBit = 0
while (num > 0) {
setCountBit++
num= num and num-1
}
return setCountBit
}
}
To correctly convert binary string to Int and avoid "out of range error", you need to do the following (I believe LeetCode does the same under the hood):
fun binaryStringToInt(s: String): Int = s.toUInt(radix = 2).toInt()
"11111111111111111111111111111101" is equivalent to 4294967293. This is greater than Int.MAX_VALUE, so it will be represented as negative number after .toInt() convertion (-3 in this case).
Actually, this problem could be solved with one-liner in Kotlin 1.4:
fun hammingWeight(n: Int): Int = n.countOneBits()
But LeetCode uses Kotlin 1.3.10, so you need to adjust your solution to handle negative Ints as well.
Please change the type of your input variable from Int to a type like Double .At the moment The given value is bigger than the maximum value that a type Int number can store.
Related
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'm trying to find the simplest way to convert a digit (0..9) into the respective character '0'..'9' in Kotlin.
My initial attempt was to write the following code:
fun convertToCharacter() {
val number = 0
val character = number.toChar()
println(character)
}
Of course, after running, I quickly saw that this produces \u0000, and not '0' like I expected. Then, remembering from how to do this in Java, I modified the code to add '0', but then this would not compile.
fun convertToCharacter() {
val number = 0
val character = number.toChar() + '0'
println(character)
}
What is the appropriate way to convert a number into its respective character counterpart in Kotlin? Ideally, I'm trying to avoid pulling up the ASCII table to accomplish this (I know I can add 48 to the number since 48 -> '0' in ASCII).
val character = '0' + number
is the shortest way, given that the number is in range 0..9
Kotlin stdlib provides this function since 1.5.0.
fun Int.digitToChar(): Char
Returns the Char that represents this decimal digit. Throws an exception if this value is not in the range 0..9.
If this value is in 0..9, the decimal digit Char with code '0'.code + this is returned.
Example
println(5.digitToChar()) // 5
println(3.digitToChar(radix = 8)) // 3
println(10.digitToChar(radix = 16)) // A
println(20.digitToChar(radix = 36)) // K
Like you said, probably the easiest way to convert an Int to the Char representation of that same digit is to add 48 and call toChar():
val number = 3
val character = (number + 48).toChar()
println(character) // prints 3
If you don't want to have the magic 48 number in your program, you could first parse the number to a String and then use toCharArray()[0] to get the Char representation:
val number = 3
val character = number.toString().toCharArray()[0]
println(character) // prints 3
Edit: in the spirit of the attempt in your question, you can do math with '0'.toInt() and get the result you were expecting:
val number = 7
val character = (number + '0'.toInt()).toChar()
println(number) // prints 7
How about 0.toString() instead of 0.toChar() ? If you are specifically after single digits, then 0.toString()[0] will give you a Char type
You can use an extension like this:
fun Int.toReadableChar(): Char {
return ('0'.toInt() + this).toChar()
}
You can apply this to any other class you want :)
Example:
println(7.toReadableChar())
>> 7
I wanted to sum the digits of Long variable and add it to the variable it self, I came with the next working code:
private fun Long.sumDigits(): Long {
var n = this
this.toString().forEach { n += it.toString().toLong() }
return n
}
Usage: assert(48.toLong() == 42.toLong().sumDigits())
I had to use it.toString() in order to get it work, so I came with the next test and I don't get it's results:
#Test
fun toLongEquality() {
println("'4' as Long = " + '4'.toLong())
println("\"4\" as Long = " + "4".toLong())
println("\"42\" as Long = " + "42".toLong())
assert('4'.toString().toLong() == 4.toLong())
}
Output:
'4' as Long = 52
"4" as Long = 4
"42" as Long = 42
Is it a good practice to use char.toString().toLong() or there is a better way to convert char to Long?
Does "4" represented by chars? Why it is not equal to it char representation?
From the documentation:
class Char : Comparable (source) Represents a 16-bit Unicode
character. On the JVM, non-nullable values of this type are
represented as values of the primitive type char.
fun toLong(): Long
Returns the value of this character as a Long.
When you use '4' as Long you actually get the Unicode (ASCII) code of the char '4'
As mTak says, Char represents a Unicode value. If you are using Kotlin on the JVM, you can define your function as follows:
private fun Long.sumDigits() = this.toString().map(Character::getNumericValue).sum().toLong()
There's no reason to return Long rather than Int, but I've kept it the same as in your question.
Non-JVM versions of Kotlin don't have the Character class; use map {it - '0'} instead.
I was solving a problem on codeforces in which I had to sum up the digits of a big number (it can have up to 100k digits) and I'd have to repeat that process until there is only one digit left and count the number of times I did that and I came up with a working solution, however I'd like to know if some things could have been done in a more "Kotlin-ish like way", so given:
fun main(args: Array<String>) {
println(transform(readLine()!!))
}
fun transform(n: String): Int {
var count = 0
var sum : Int
var s = n
while(s.length > 1) {
sum = (0 until s.length).sumBy { s[it].toInt() - '0'.toInt() }
s = sum.toString()
count++
}
return count
}
sum = (0 until s.length).sumBy { s[it].toInt() - '0'.toInt() } is there a way to I guess map the sum of digits in the string to the sum variable, or in general a better approach than the one I used?
When converting a Char to an Int it converts it to the ASCII value so I had to add "-'0'.toInt()" is there a faster way (not that it's too much to write, asking out of curiosity)?
How to make the String n mutable without creating a new String s and manipulating it? Or is that the desired (and only) way?
P.S. I'm a beginner with Kotlin.
When converting a Char to an Int it converts it to the ASCII value so I had to add "-'0'.toInt()" is there a faster way (not that it's too much to write, asking out of curiosity)?
You can simply write s[it] - '0', because subtracting Chars in Kotlin already gives you an Int:
public class Char ... {
...
/** Subtracts the other Char value from this value resulting an Int. */
public operator fun minus(other: Char): Int
...
}
But why are looping over the indexes when you could loop over the Chars directly?
sum = s.sumBy { it - '0' }
This is a functional (and recursive) style to solve it:
private fun sum(num: String, count: Int) : Int {
return num
//digit to int
.map { "$it".toInt() }
//sum digits
.sum()
//sum to string
.toString()
//if sum's length is more than one, do it again with incremented count. Otherwise, return the current count
.let { if (it.length > 1) sum(it, count + 1) else count }
}
And you call it like this:
val number = "2937649827364918308623946..." //and so on
val count = sum(number, 0)
Hope it helps!
val listNumbers = generateSequence(1) { it + 1 }
val listNumber1to100 = listNumbers.takeWhile { it < 100 }
val secNum:Unit = listNumber1to100.forEach {it}
println(listNumber1to100.asSequence().filter { it%(listNumber1to100.forEach { it })!=0 }.toList())
I have an error in reminder sign!
This is Error: None of the following functions can be called with the arguments supplied
In your first approach, the error appears in this line:
it%(listNumber1to100.forEach { it })
A Byte, Double, Float, Int, Long or Short is prefered right after the % operator, however, forEach is a function which the return type is Unit.
In your second approach, you have the correct expression in isPrime(Int). Here are some suggestions for you:
listNumber1to100 is excluding 100 in your code, if you want to include 100 in listNumber1to100, the lambda you pass to takeWhile should be changed like this:
val listNumber1to100 = listNumbers.takeWhile { it <= 100 }
listNumber1to100.asSequence() is redundant here since listNumber1too100 is itself a TakeWhileSequence which implements Sequence.
isPrime(Int) is a bit confusing since it is check for isComposite and it does not work for every input it takes(it works for 1 to 99 only). I will rewrite it in this way:
fun isPrime(num: Int): Boolean = if (num <= 1) false else !(2..num/2).any { num % it == 0 }
Since prime number must be positive and 1 is a special case(neither a prime nor composite number), it just return false if the input is smaller or equal to 1. If not, it checks if the input is divisible by a range of number from 2 to (input/2). The range ends before (input/2) is because if it is true for num % (num/2) == 0, it is also true for num % 2 == 0, vise versa. Finally, I add a ! operator before that because a prime number should not be divisible by any of those numbers.
Finally, you can filter a list by isPrime(Int) like this:
println(listNumber1to100.filter(::isPrime).toList())
PS. It is just for reference and there must be a better implementation than this.
To answer your question about it, it represents the only lambda parameter inside a lambda expression. It is always used for function literal which has only one parameter.
The error is because the expression: listNumber1to100.forEach { it } - is not a number, it is a Unit (ref).
The compiler try to match the modulo operator to the given function signatures, e.g.: mod(Byte) / mod(Int) / mod(Long) - etc.
val listNumbers = generateSequence(1) { it + 1 }
val listNumber1to100 = listNumbers.takeWhile { it < 100 }
fun isPrime(num: Int): Boolean = listNumber1to100.asSequence().any { num%it==0 && it!=num && it!=1 }
println(listNumber1to100.asSequence().filter { !isPrime(it)}.toList())
I found this solution and worked
But why can I have a non-number here in the right side of reminder