# Scala algorithm: Find the longest palindrome within a string

Published

## Algorithm goal

The palindrome in String `XYZdZeXdZdZdX`is XdZdZdX. Compute the longest sub-string palindrome in a given string.

Note: there are many solutions to this problem that may be wrong.

## Test cases in Scala

``````assert(longestPalindrome("X") == "X".length)
assert(longestPalindrome("XYZdZ") == "ZdZ".length)
assert(longestPalindrome("XYZdZeX") == "ZdZ".length)
assert(longestPalindrome("XYZdZeXdZdZd") == "dZdZd".length)
assert(longestPalindrome("XYZdZeXdZdZdX") == "XdZdZdX".length)
assert(longestPalindrome("XYZdZeXdZZdX") == "XdZZdX".length)
assert(longestPalindromeBrute("XYZdZ") == "ZdZ")
assert(longestPalindromeBrute("XYZdZeX") == "ZdZ")
assert(longestPalindromeBrute("XYZdZeXdZdZd") == "dZdZd")
assert(longestPalindromeBrute("XYZdZeXdZdZdX") == "XdZdZdX")
assert(longestPalindromeBrute("XYZdZeXdZZdX") == "XdZZdX")
``````

## Algorithm in Scala

43 lines of Scala (compatible versions 2.13 & 3.0).

## Explanation

We use a dynamic programming style solution.

Notice that if 'aBa' is a palindrome, then '?aBa!' is also a palindrome if '?' and '!' are equal. (this is Â© from www.scala-algorithms.com)

By defining Palindrome(i, j) to be 'true' if there is a palindrome starting at i, and ending at j, we have the following:

1. Palindrome(i, i) = true
2. Palindrome(i, i + 1) = char at i == char at i + 1
3. Palindrome(i, j) = Palindrome (i + 1, j - 1) && char at i == char at i + 1

## Scala concepts & Hints

1. ### Drop, Take, dropRight, takeRight

Scala's `drop` and `take` methods typically remove or select `n` items from a collection.

``````assert(List(1, 2, 3).drop(2) == List(3))

assert(List(1, 2, 3).take(2) == List(1, 2))

assert(List(1, 2, 3).dropRight(2) == List(1))

assert(List(1, 2, 3).takeRight(2) == List(2, 3))

assert((1 to 5).take(2) == (1 to 2))
``````
2. ### For-comprehension

The for-comprehension is highly important syntatic enhancement in functional programming languages.

``````val Multiplier = 10

val result: List[Int] = for {
num <- List(1, 2, 3)
anotherNum <-
List(num * Multiplier - 1, num * Multiplier, num * Multiplier + 1)
} yield anotherNum + 1

assert(result == List(10, 11, 12, 20, 21, 22, 30, 31, 32))
``````
3. ### Range

The `(1 to n)` syntax produces a "Range" which is a representation of a sequence of numbers.

``````assert((1 to 5).toString == "Range 1 to 5")

assert((1 to 5).reverse.toString() == "Range 5 to 1 by -1")

assert((1 to 5).toList == List(1, 2, 3, 4, 5))
``````

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