I am trying to write a recursive function in Scala that takes a positive integer and prints out its prime factors. For example, 200 = 2, 2, 2, 5, 5. This is the code I have written.
println(calculatePrimeFactors(200,2))
def isPrime( x:Int ):Boolean = {
def recIsP( n:Int, i:Int ) : Boolean = (i == n) || (n % i != 0) && recIsP(n, i + 1)
recIsP(x,2)
}
def calculatePrimeFactors(n:Int,i:Int):String= {
if (i==n) {
""
}else if(isPrime(i) && n%i==0){
i.toString + ", " + calculatePrimeFactors(n,i+1)
}else{
calculatePrimeFactors(n,i+1)
}
}
My code outputs 2,5, .
How can I make this so that the code works correctly and outputs 2,2,2,5,5,. I tried using a loop with this statement n = n / i, but Scala returns an error saying that reassignment to val can't be done.
A clean version of this algorithm might look like this:
def calculatePrimeFactors(n: Int): List[Int] = {
def loop(p: Int, rem: Int, res: List[Int]): List[Int] =
if (rem % p == 0) {
loop(p, rem / p, res :+ p)
} else if (p > rem) {
res
} else {
loop(p + 1, rem, res)
}
loop(2, n, Nil)
}
Use mkString(", ") to convert the result to a String if required.
Note that there is no need to check that the divisor is prime since any non-prime divisor will have smaller prime factors which will have already been removed.
Well, you can debug your code, and see that once you passed a number, you won't print it again.
In order to print all primes, you need to divide in the prime you found, and start over with the divided number. Please consider the following implantation for calculatePrimeFactors:
def calculatePrimeFactors(n:Int, i:Int):String= {
if (i==n) {
i.toString
} else if(n%i==0) {
i.toString + ", " + calculatePrimeFactors(n/i,i)
} else {
calculatePrimeFactors(n,i+1)
}
}
Now you get the output: 2, 2, 2, 5, 5. Code run can be found at Scastie.
Related
I a teaching myself Scala, and have been dealing with the algorithmic problem of finding all balanced pairs of n parenthesis.
So, the following solution works:
object GetAllValidParenthesis extends App {
def generateParenthesis(n: Int): Unit = {
parenthesis("", 0, 0, n)
}
private def parenthesis(output: String, open: Int, close: Int, pairs: Int): Unit = {
if ((open == pairs) && (close == pairs)) {
println(output)
}
else {
if (open < pairs) parenthesis(output + "(", open + 1, close, pairs)
if (open > close) parenthesis(output + ")", open, close + 1, pairs)
}
}
println(generateParenthesis(3))
}
But, if I wanted to return a String instead of Unit (so as to later keep all strings in some data structure, like an array buffer), i.e., I did this:
private def parenthesis(output: String, open: Int, close: Int, pairs: Int): String = {
if ((open == pairs) && (close == pairs)) {
output
}
else {
if (open < pairs) parenthesis(output + "(", open + 1, close, pairs)
if (open > close) parenthesis(output + ")", open, close + 1, pairs)
}
}
The compiler tells me:
type mismatch;
found : Unit
required: String
if (open > close) parenthesis(output + ")", open, close + 1, pairs)
But it is impossible for the function to return anything but a String, since the only "return" is in the line that says "output", the rest are recursive calls.
Why does this happen? Is there any way in which I could fix this?
Thanks in advance for your help.
In Scala, every expression has a value.
That's to say
// this is not a Boolean value
val exp = if (2 > 1) true
// in fact, is the same as following. The `Unit` and `Boolean`'s common type is `AnyVal`
val exp: AnyVal = if (2 > 1) true else {}
In your problem, your function should generate a string of parenthesis and append it to result set rather than just return the value.
I have just write solution of this problem.
May be it's more appropriate like following
import scala.collection.mutable.ArrayBuffer
object GenerateParenthesis {
def generateParenthesis(n: Int): List[String] = {
val res = ArrayBuffer.empty[String]
def travel(l: Int, r: Int, curString: String): Unit = {
if (l == 0 && r == 0) res += curString
if (l > 0) travel(l - 1, r, curString + "(")
if (r > l) travel(l, r - 1, curString + ")")
}
travel(n, n, "")
res.toList
}
}
The test code can be found here:
https://github.com/counter2015/LeetCodeScala/blob/6a3ff3d28a7b378490105684d080418c287680d8/src/test/scala/algorithms/medium/implementation/GenerateParenthesisTest.scala
I am writing a recursive function to check if a non-negative Integer is Prime in SCALA. My function takes two inputs to do so. Here is my code:
import io.StdIn._
def isPrime (x:Int, i:Int): Boolean = {
if (i < 3) {
true
}else if (x % i == 0) {
false
}else{
isPrime(x,i-1)
}
}
print("Enter a number to check if it is a prime number: " )
val num1 = readInt()
println(isPrime(num1,num1-1))
My question is that is there a way for me to take one input as the parameter for the function? The code has to use a recursive function. I know that there are more effective ways to check if a number is prime (using iteration perhaps), but I'm just doing this as a practice problem.
It is perfectly fine, I'd say even standard practice, to wrap recursive functions with a calling function. For example:
def isPrime( x:Int ) = {
def recIsP( n:Int, i:Int ) : Boolean = i == n || n % i != 0 && recIsP(n, i + 1)
recIsP(x,2)
}
isPrime(3301)
One option is to make an inner method
def isPrime(x:Int): Boolean = {
def loop(i: Int): Boolean =
if (i<3 ) {
true
} else if (x % i==0) {
false
} else {
isPrime(x,i-1)
}
loop(x, x - 1)
}
another option is to make a default parameter. Those can't refer back to the previous parameter, so you'll have to use a workaround. Here I pass some j, and make i = x - j, so I can just increment j starting from 1
def isPrime(x: Int, j: Int = 1): Boolean = {
val i = x - j
if (i < 3) {
true
} else if (x % i==0) {
false
} else {
isPrime(x, j + 1)
}
}
Unrelatedly, there is a bug in your code: 4 should not be prime.
Since I liked programming in Scala, for my Google interview, I asked them to give me a Scala / functional programming style question. The Scala functional style question that I got was as follows:
You have two strings consisting of alphabetic characters as well as a special character representing the backspace symbol. Let's call this backspace character '/'. When you get to the keyboard, you type this sequence of characters, including the backspace/delete character. The solution you are to implement must check if the two sequences of characters produce the same output. For example, "abc", "aa/bc". "abb/c", "abcc/", "/abc", and "//abc" all produce the same output, "abc". Because this is a Scala / functional programming question, you must implement your solution in idiomatic Scala style.
I wrote the following code (it might not be exactly what I wrote, I'm just going off memory). Basically I just go linearly through the string, prepending characters to a list, and then I compare the lists.
def processString(string: String): List[Char] = {
string.foldLeft(List[Char]()){ case(accumulator: List[Char], char: Char) =>
accumulator match {
case head :: tail => if(char != '/') { char :: head :: tail } else { tail }
case emptyList => if(char != '/') { char :: emptyList } else { emptyList }
}
}
}
def solution(string1: String, string2: String): Boolean = {
processString(string1) == processString(string2)
}
So far so good? He then asked for the time complexity and I responded linear time (because you have to process each character once) and linear space (because you have to copy each element into a list). Then he asked me to do it in linear time, but with constant space. I couldn't think of a way to do it that was purely functional. He said to try using a function in the Scala collections library like "zip" or "map" (I explicitly remember him saying the word "zip").
Here's the thing. I think that it's physically impossible to do it in constant space without having any mutable state or side effects. Like I think that he messed up the question. What do you think?
Can you solve it in linear time, but with constant space?
This code takes O(N) time and needs only three integers of extra space:
def solution(a: String, b: String): Boolean = {
def findNext(str: String, pos: Int): Int = {
#annotation.tailrec
def rec(pos: Int, backspaces: Int): Int = {
if (pos == 0) -1
else {
val c = str(pos - 1)
if (c == '/') rec(pos - 1, backspaces + 1)
else if (backspaces > 0) rec(pos - 1, backspaces - 1)
else pos - 1
}
}
rec(pos, 0)
}
#annotation.tailrec
def rec(aPos: Int, bPos: Int): Boolean = {
val ap = findNext(a, aPos)
val bp = findNext(b, bPos)
(ap < 0 && bp < 0) ||
(ap >= 0 && bp >= 0 && (a(ap) == b(bp)) && rec(ap, bp))
}
rec(a.size, b.size)
}
The problem can be solved in linear time with constant extra space: if you scan from right to left, then you can be sure that the /-symbols to the left of the current position cannot influence the already processed symbols (to the right of the current position) in any way, so there is no need to store them.
At every point, you need to know only two things:
Where are you in the string?
How many symbols do you have to throw away because of the backspaces
That makes two integers for storing the positions, and one additional integer for temporary storing the number of accumulated backspaces during the findNext invocation. That's a total of three integers of space overhead.
Intuition
Here is my attempt to formulate why the right-to-left scan gives you a O(1) algorithm:
The future cannot influence the past, therefore there is no need to remember the future.
The "natural time" in this problem flows from left to right. Therefore, if you scan from right to left, you are moving "from the future into the past", and therefore you don't need to remember the characters to the right of your current position.
Tests
Here is a randomized test, which makes me pretty sure that the solution is actually correct:
val rng = new util.Random(0)
def insertBackspaces(s: String): String = {
val n = s.size
val insPos = rng.nextInt(n)
val (pref, suff) = s.splitAt(insPos)
val c = ('a' + rng.nextInt(26)).toChar
pref + c + "/" + suff
}
def prependBackspaces(s: String): String = {
"/" * rng.nextInt(4) + s
}
def addBackspaces(s: String): String = {
var res = s
for (i <- 0 until 8)
res = insertBackspaces(res)
prependBackspaces(res)
}
for (i <- 1 until 1000) {
val s = "hello, world"
val t = "another string"
val s1 = addBackspaces(s)
val s2 = addBackspaces(s)
val t1 = addBackspaces(t)
val t2 = addBackspaces(t)
assert(solution(s1, s2))
assert(solution(t1, t2))
assert(!solution(s1, t1))
assert(!solution(s1, t2))
assert(!solution(s2, t1))
assert(!solution(s2, t2))
if (i % 100 == 0) {
println(s"Examples:\n$s1\n$s2\n$t1\n$t2")
}
}
A few examples that the test generates:
Examples:
/helly/t/oj/m/, wd/oi/g/x/rld
///e/helx/lc/rg//f/o, wosq//rld
/anotl/p/hhm//ere/t/ strih/nc/g
anotx/hb/er sw/p/tw/l/rip/j/ng
Examples:
//o/a/hellom/, i/wh/oe/q/b/rld
///hpj//est//ldb//y/lok/, world
///q/gd/h//anothi/k/eq/rk/ string
///ac/notherli// stri/ig//ina/n/g
Examples:
//hnn//ello, t/wl/oxnh///o/rld
//helfo//u/le/o, wna//ova//rld
//anolq/l//twl//her n/strinhx//g
/anol/tj/hq/er swi//trrq//d/ing
Examples:
//hy/epe//lx/lo, wr/v/t/orlc/d
f/hk/elv/jj//lz/o,wr// world
/anoto/ho/mfh///eg/r strinbm//g
///ap/b/notk/l/her sm/tq/w/rio/ng
Examples:
///hsm/y//eu/llof/n/, worlq/j/d
///gx//helf/i/lo, wt/g/orn/lq/d
///az/e/notm/hkh//er sm/tb/rio/ng
//b/aen//nother v/sthg/m//riv/ng
Seems to work just fine. So, I'd say that the Google-guy did not mess up, looks like a perfectly valid question.
You don't have to create the output to find the answer. You can iterate the two sequences at the same time and stop on the first difference. If you find no difference and both sequences terminate at the same time, they're equal, otherwise they're different.
But now consider sequences such as this one: aaaa/// to compare with a. You need to consume 6 elements from the left sequence and one element from the right sequence before you can assert that they're equal. That means that you would need to keep at least 5 elements in memory until you can verify that they're all deleted. But what if you iterated elements from the end? You would then just need to count the number of backspaces and then just ignoring as many elements as necessary in the left sequence without requiring to keep them in memory since you know they won't be present in the final output. You can achieve O(1) memory using these two tips.
I tried it and it seems to work:
def areEqual(s1: String, s2: String) = {
def charAt(s: String, index: Int) = if (index < 0) '#' else s(index)
#tailrec
def recSol(i1: Int, backspaces1: Int, i2: Int, backspaces2: Int): Boolean = (charAt(s1, i1), charAt(s2, i2)) match {
case ('/', _) => recSol(i1 - 1, backspaces1 + 1, i2, backspaces2)
case (_, '/') => recSol(i1, backspaces1, i2 - 1, backspaces2 + 1)
case ('#' , '#') => true
case (ch1, ch2) =>
if (backspaces1 > 0) recSol(i1 - 1, backspaces1 - 1, i2 , backspaces2 )
else if (backspaces2 > 0) recSol(i1 , backspaces1 , i2 - 1, backspaces2 - 1)
else ch1 == ch2 && recSol(i1 - 1, backspaces1 , i2 - 1, backspaces2 )
}
recSol(s1.length - 1, 0, s2.length - 1, 0)
}
Some tests (all pass, let me know if you have more edge cases in mind):
// examples from the question
val inputs = Array("abc", "aa/bc", "abb/c", "abcc/", "/abc", "//abc")
for (i <- 0 until inputs.length; j <- 0 until inputs.length) {
assert(areEqual(inputs(i), inputs(j)))
}
// more deletions than required
assert(areEqual("a///////b/c/d/e/b/b", "b"))
assert(areEqual("aa/a/a//a//a///b", "b"))
assert(areEqual("a/aa///a/b", "b"))
// not enough deletions
assert(!areEqual("aa/a/a//a//ab", "b"))
// too many deletions
assert(!areEqual("a", "a/"))
PS: just a few notes on the code itself:
Scala type inference is good enough so that you can drop types in the partial function inside your foldLeft
Nil is the idiomatic way to refer to the empty list case
Bonus:
I had something like Tim's soltion in mind before implementing my idea, but I started early with pattern matching on characters only and it didn't fit well because some cases require the number of backspaces. In the end, I think a neater way to write it is a mix of pattern matching and if conditions. Below is my longer original solution, the one I gave above was refactored laater:
def areEqual(s1: String, s2: String) = {
#tailrec
def recSol(c1: Cursor, c2: Cursor): Boolean = (c1.char, c2.char) match {
case ('/', '/') => recSol(c1.next, c2.next)
case ('/' , _) => recSol(c1.next, c2 )
case (_ , '/') => recSol(c1 , c2.next)
case ('#' , '#') => true
case (a , b) if (a == b) => recSol(c1.next, c2.next)
case _ => false
}
recSol(Cursor(s1, s1.length - 1), Cursor(s2, s2.length - 1))
}
private case class Cursor(s: String, index: Int) {
val char = if (index < 0) '#' else s(index)
def next = {
#tailrec
def recSol(index: Int, backspaces: Int): Cursor = {
if (index < 0 ) Cursor(s, index)
else if (s(index) == '/') recSol(index - 1, backspaces + 1)
else if (backspaces > 1) recSol(index - 1, backspaces - 1)
else Cursor(s, index - 1)
}
recSol(index, 0)
}
}
If the goal is minimal memory footprint, it's hard to argue against iterators.
def areSame(a :String, b :String) :Boolean = {
def getNext(ci :Iterator[Char], ignore :Int = 0) : Option[Char] =
if (ci.hasNext) {
val c = ci.next()
if (c == '/') getNext(ci, ignore+1)
else if (ignore > 0) getNext(ci, ignore-1)
else Some(c)
} else None
val ari = a.reverseIterator
val bri = b.reverseIterator
1 to a.length.max(b.length) forall(_ => getNext(ari) == getNext(bri))
}
On the other hand, when arguing FP principals it's hard to defend iterators, since they're all about maintaining state.
Here is a version with a single recursive function and no additional classes or libraries. This is linear time and constant memory.
def compare(a: String, b: String): Boolean = {
#tailrec
def loop(aIndex: Int, aDeletes: Int, bIndex: Int, bDeletes: Int): Boolean = {
val aVal = if (aIndex < 0) None else Some(a(aIndex))
val bVal = if (bIndex < 0) None else Some(b(bIndex))
if (aVal.contains('/')) {
loop(aIndex - 1, aDeletes + 1, bIndex, bDeletes)
} else if (aDeletes > 0) {
loop(aIndex - 1, aDeletes - 1, bIndex, bDeletes)
} else if (bVal.contains('/')) {
loop(aIndex, 0, bIndex - 1, bDeletes + 1)
} else if (bDeletes > 0) {
loop(aIndex, 0, bIndex - 1, bDeletes - 1)
} else {
aVal == bVal && (aVal.isEmpty || loop(aIndex - 1, 0, bIndex - 1, 0))
}
}
loop(a.length - 1, 0, b.length - 1, 0)
}
I wrote sum code in scala to find the majority element(the element which appears more than n/2 times where 'n' is the no.of elements in an array.I want to know where there is functional / scala native style of version(which includes match cases and transformations like "map/"flatmap" etc..) for the below imperative style of scala code which includes looping. The code which i used in:
object MajorityElement{
def printMajority(arr:Array[Int]) ={
var cand:Int=findCandidate(arr);
if(isMajority(arr,cand))
println(cand);
else
println("No Majority Element");
}
def isMajority(arr:Array[Int],Cand:Int):Boolean ={
var count=0;
for(i <- 0 until arr.length){
if(arr(i)== Cand)
count+=1;
}
if (count > arr.size/2)
return true;
else false
}
def findCandidate(arr:Array[Int]):Int ={
val s = arr.size
var majIndex:Int = 0;
var count = 1;
for(i <- 0 until arr.length){
if(arr(majIndex) == arr(i))
count+=1;
else
count-=1;
if(count==0)
{
majIndex = i;
count =1
}
}
return arr(majIndex);
}
}
please let me know, whether it is possible to write/ convert imperative style to functional version in scala(which uses match cases) for any scenario.
If you're only interested in the final result (and so you don't need isMajority etc), it's very simple
def findMajority(xs: List[Int]) = {
val mostFrequent = xs.groupBy(identity).values.maxBy(_.length)
if (mostFrequent.length >= xs.length / 2) Some(mostFrequent.head) else None
}
findMajority(List(1, 2, 2, 2, 3, 3, 3, 3, 3, 4))
//Option[Int] = Some(3)
Group equal elements into lists (the values of the Map returned by GroupBy). Pick the longest list. If its length is more than half the list, then it's a majority, return Some(the head) (any element will do, they're all the same value). Otherwise, return None
The transition from imperative thinking to functional thinking takes time and study. One approach is to find code examples here on SO and, with the help of the Standard Library, break it down until you understand what's going on.
Here's a little something to get you started.
def isMajority(arr:Array[Int],cand:Int):Boolean =
arr.count(_ == cand) > arr.size/2
Threr is no Native Scala Style, but code can be Functional Style(value oriented)
(No var, No Side-Effect, Pure Function)
object MajorityElement {
case class Candidate(idx: Int, count: Int)
def solve(arr: Array[Int]): Option[Int] = {
val candidate = findCandidate(arr)
if (isMajority(arr, candidate)) Option(arr(candidate.idx))
else None
}
def isMajority(arr: Array[Int], candidate: Candidate) =
arr.count(_ == arr(candidate.idx)) > arr.size / 2
def findCandidate(arr: Array[Int]): Candidate =
arr.indices.foldLeft(Candidate(0, 1)) { (acc, idx) =>
val newAcc =
if (arr(acc.idx) == arr(idx)) acc.copy(count = acc.count + 1)
else acc.copy(count = acc.count - 1)
if (newAcc.count == 0) Candidate(idx, 1)
else newAcc
}
}
val arr = Array(1, 1, 1, 2, 3, 4, 1)
val ret = MajorityElement.solve(arr)
ret match {
case Some(n) => println(s"Found Majority Element: $n")
case None => println("No Majority Element")
}
I came across this problem from CodeChef. The problem states the following:
A positive integer is called a palindrome if its representation in the
decimal system is the same when read from left to right and from right
to left. For a given positive integer K of not more than 1000000
digits, write the value of the smallest palindrome larger than K to
output.
I can define a isPalindrome method as follows:
def isPalindrome(someNumber:String):Boolean = someNumber.reverse.mkString == someNumber
The problem that I am facing is how do I loop from the initial given number and break and return the first palindrome when the integer satisfies the isPalindrome method? Also, is there a better(efficient) way to write the isPalindrome method?
It will be great to get some guidance here
If you have a number like 123xxx you know, that either xxx has to be below 321 - then the next palindrom is 123321.
Or xxx is above, then the 3 can't be kept, and 124421 has to be the next one.
Here is some code without guarantees, not very elegant, but the case of (multiple) Nines in the middle is a bit hairy (19992):
object Palindrome extends App {
def nextPalindrome (inNumber: String): String = {
val len = inNumber.length ()
if (len == 1 && inNumber (0) != '9')
"" + (inNumber.toInt + 1) else {
val head = inNumber.substring (0, len/2)
val tail = inNumber.reverse.substring (0, len/2)
val h = if (head.length > 0) BigInt (head) else BigInt (0)
val t = if (tail.length > 0) BigInt (tail) else BigInt (0)
if (t < h) {
if (len % 2 == 0) head + (head.reverse)
else inNumber.substring (0, len/2 + 1) + (head.reverse)
} else {
if (len % 2 == 1) {
val s2 = inNumber.substring (0, len/2 + 1) // 4=> 4
val h2 = BigInt (s2) + 1 // 5
nextPalindrome (h2 + (List.fill (len/2) ('0').mkString)) // 5 + ""
} else {
val h = BigInt (head) + 1
h.toString + (h.toString.reverse)
}
}
}
}
def check (in: String, expected: String) = {
if (nextPalindrome (in) == expected)
println ("ok: " + in) else
println (" - fail: " + nextPalindrome (in) + " != " + expected + " for: " + in)
}
//
val nums = List (("12345", "12421"), // f
("123456", "124421"),
("54321", "54345"),
("654321", "654456"),
("19992", "20002"),
("29991", "29992"),
("999", "1001"),
("31", "33"),
("13", "22"),
("9", "11"),
("99", "101"),
("131", "141"),
("3", "4")
)
nums.foreach (n => check (n._1, n._2))
println (nextPalindrome ("123456678901234564579898989891254392051039410809512345667890123456457989898989125439205103941080951234566789012345645798989898912543920510394108095"))
}
I guess it will handle the case of a one-million-digit-Int too.
Doing reverse is not the greatest idea. It's better to start at the beginning and end of the string and iterate and compare element by element. You're wasting time copying the entire String and reversing it even in cases where the first and last element don't match. On something with a million digits, that's going to be a huge waste.
This is a few orders of magnitude faster than reverse for bigger numbers:
def isPalindrome2(someNumber:String):Boolean = {
val len = someNumber.length;
for(i <- 0 until len/2) {
if(someNumber(i) != someNumber(len-i-1)) return false;
}
return true;
}
There's probably even a faster method, based on mirroring the first half of the string. I'll see if I can get that now...
update So this should find the next palindrome in almost constant time. No loops. I just sort of scratched it out, so I'm sure it can be cleaned up.
def nextPalindrome(someNumber:String):String = {
val len = someNumber.length;
if(len==1) return "11";
val half = scala.math.floor(len/2).toInt;
var firstHalf = someNumber.substring(0,half);
var secondHalf = if(len % 2 == 1) {
someNumber.substring(half+1,len);
} else {
someNumber.substring(half,len);
}
if(BigInt(secondHalf) > BigInt(firstHalf.reverse)) {
if(len % 2 == 1) {
firstHalf += someNumber.substring(half, half+1);
firstHalf = (BigInt(firstHalf)+1).toString;
firstHalf + firstHalf.substring(0,firstHalf.length-1).reverse
} else {
firstHalf = (BigInt(firstHalf)+1).toString;
firstHalf + firstHalf.reverse;
}
} else {
if(len % 2 == 1) {
firstHalf + someNumber.substring(half,half+1) + firstHalf.reverse;
} else {
firstHalf + firstHalf.reverse;
}
}
}
This is most general and clear solution that I can achieve:
Edit: got rid of BigInt's, now it takes less than a second to calculate million digits number.
def incStr(num: String) = { // helper method to increment number as String
val idx = num.lastIndexWhere('9'!=, num.length-1)
num.take(idx) + (num.charAt(idx)+1).toChar + "0"*(num.length-idx-1)
}
def palindromeAfter(num: String) = {
val lengthIsOdd = num.length % 2
val halfLength = num.length / 2 + lengthIsOdd
val leftHalf = num.take(halfLength) // first half of number (including central digit)
val rightHalf = num.drop(halfLength - lengthIsOdd) // second half of number (also including central digit)
val (newLeftHalf, newLengthIsOdd) = // we need to calculate first half of new palindrome and whether it's length is odd or even
if (rightHalf.compareTo(leftHalf.reverse) < 0) // simplest case - input number is like 123xxx and xxx < 321
(leftHalf, lengthIsOdd)
else if (leftHalf forall ('9'==)) // special case - if number is like '999...', then next palindrome will be like '10...01' and one digit longer
("1" + "0" * (halfLength - lengthIsOdd), 1 - lengthIsOdd)
else // other cases - increment first half of input number before making palindrome
(incStr(leftHalf), lengthIsOdd)
// now we can create palindrome itself
newLeftHalf + newLeftHalf.dropRight(newLengthIsOdd).reverse
}
According to your range-less proposal: the same thing but using Stream:
def isPalindrome(n:Int):Boolean = n.toString.reverse == n.toString
def ints(n: Int): Stream[Int] = Stream.cons(n, ints(n+1))
val result = ints(100).find(isPalindrome)
And with iterator (and different call method, the same thing you can do with Stream, actually):
val result = Iterator.from(100).find(isPalindrome)
But as #user unknown stated, it is direct bruteforce and not practical with large numbers.
To check if List of Any Type is palindrome using slice, without any Loops
def palindrome[T](list: List[T]): Boolean = {
if(list.length==1 || list.length==0 ){
false
}else {
val leftSlice: List[T] = list.slice(0, list.length / 2)
var rightSlice :List[T]=Nil
if (list.length % 2 != 0) {
rightSlice = list.slice(list.length / 2 + 1, list.length).reverse
} else {
rightSlice = list.slice(list.length / 2, list.length).reverse
}
leftSlice ==rightSlice
}
}
Though the simplest solution would be
def palindrome[T](list: List[T]): Boolean = {
list == list.reverse
}
You can simply use the find method on collections to find the first element matching a given predicate:
def isPalindrome(n:Int):Boolean = n.toString.reverse == n.toString
val (start, end) = (100, 1000)
val result: Option[Int] = (start to end).find(isPalindrome)
result foreach println
>Some(101)
Solution to verify if a String is a palindrome
This solution doesn't reverse the String. However I am not sure that it will be faster.
def isPalindrome(s:String):Boolean = {
s.isEmpty ||
((s.last == s.head) && isPalindrome(s.tail.dropRight(1)))
}
Solution to find next palindrome given a String
This solution is not the best for scala (pretty the same as Java solution) but it only deals with Strings and is suitable for large numbers
You just have to mirror the first half of the number you want, check if it is higher than the begin number, otherwise, increase by one the last digit of the first half and mirror it again.
First, a function to increment the string representation of an int by 1:
def incrementString(s:String):String = {
if(s.nonEmpty){
if (s.last == '9')
incrementString(s.dropRight(1))+'0'
else
s.dropRight(1) + (s.last.toInt +1).toChar
}else
"1"
}
Then, a function to compare to string representation of ints: (the function 'compare' doesn't work for that case)
/* is less that 0 if x<y, is more than 0 if x<y, is equal to 0 if x==y */
def compareInts(x:String, y:String):Int = {
if (x.length !=y.length)
(x.length).compare(y.length)
else
x.compare(y)
}
Now the function to compute the next palindrome:
def nextPalindrome(origin_ :String):String = {
/*Comment if you want to have a strictly bigger number, even if you already have a palindrome as input */
val origin = origin_
/* Uncomment if you want to have a strictly bigger number, even if you already have a palindrome as input */
//val origin = incrementString(origin_)
val length = origin.length
if(length % 2 == 0){
val (first, last) = origin.splitAt(length/2);
val reversed = first.reverse
if (compareInts(reversed,last) > -1)
first ++ reversed
else{
val firstIncr = incrementString(first)
firstIncr ++ firstIncr.reverse
}
} else {
val (first,last) = origin.splitAt((length+1)/2)
val reversed = first.dropRight(1).reverse
if (compareInts(reversed,last) != -1)
first ++ reversed
else{
val firstIncr = incrementString(first)
firstIncr ++ firstIncr.dropRight(1).reverse
}
}
}
You could try something like this, I'm using basic recursive:
object Palindromo {
def main(args: Array[String]): Unit = {
var s: String = "arara"
println(verificaPalindromo(s))
}
def verificaPalindromo(s: String): String = {
if (s.length == 0 || s.length == 1)
"true"
else
if (s.charAt(0).toLower == s.charAt(s.length - 1).toLower)
verificaPalindromo(s.substring(1, s.length - 1))
else
"false"
}
}
#tailrec
def palindrome(str: String, start: Int, end: Int): Boolean = {
if (start == end)
true
else if (str(start) != str(end))
false
else
pali(str, start + 1, end - 1)
}
println(palindrome("arora", 0, str.length - 1))