Count number of changes (manipulations) needed to make an anagram in immutable/pure functional Scala with foldLeft and a MultiSet

Algorithm goal

Count number of changes needed to make. This problem is similar to:

  • On HackerRank: You must split it [a string] into two contiguous substrings, then determine the minimum number of characters to change to make the two substrings into anagrams of one another.

One string is an anagram of another if they have the same number of each character, not necessarily in the same order - for example abcc is an anagram of accb and vice-versa.

ab is composed of a and b, and exchanging a to b is enough to create an anagram - ie 1 change of letter is enough.

abc is not possible to create anagrams of because it cannot be split.

poeq requires 2 changes, to create an anagram, for example to pqpq, popo, eoeo and so forth.

Explanation

Once we split a string into 2 parts, we are left with 2 strings. Number of changes required is effectively the number of characters that are not in common.

If we're counting a "difference" between two "sets", then we need a set differentiation operation. But this set is of a particular type - we have counters, so in fact it is a Map-like type. There is a name for it and it is a MultiSet (I discovered this while working out a solution for this problem). (this is © from www.scala-algorithms.com)

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Scala Concepts & Hints

Collect

'collect' allows you to use Pattern Matching, to filter and map items.

assert("Hello World".collect {
  case character if Character.isUpperCase(character) => character.toLower
} == "hw")

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Pattern Matching

Pattern matching in Scala lets you quickly identify what you are looking for in a data, and also extract it.

assert("Hello World".collect {
  case character if Character.isUpperCase(character) => character.toLower
} == "hw")

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foldLeft and foldRight

A 'fold' allows you to perform the equivalent of a for-loop, but with a lot less code.

def foldMutable[I, O](initialState: O)(items: List[I])(f: (O, I) => O): O =
  items.foldLeft(initialState)(f)

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Stack Safety

Stack safety is present where a function cannot crash due to overflowing the limit of number of recursive calls.

This function will work for n = 5, but will not work for n = 2000 (crash with java.lang.StackOverflowError) - however there is a way to fix it :-)

In Scala Algorithms, we try to write the algorithms in a stack-safe way, where possible, so that when you use the algorithms, they will not crash on large inputs. However, stack-safe implementations are often more complex, and in some cases, overly complex, for the task at hand.

def sum(from: Int, until: Int): Int =
  if (from == until) until else from + sum(from + 1, until)

def thisWillSucceed: Int = sum(1, 5)

def thisWillFail: Int = sum(1, 300)

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Option Type

The 'Option' type is used to describe a computation that either has a result or does not. In Scala, you can 'chain' Option processing, combine with lists and other data structures. For example, you can also turn a pattern-match into a function that return an Option, and vice-versa!

assert(Option(1).flatMap(x => Option(x + 2)) == Option(3))

assert(Option(1).flatMap(x => None) == None)

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View

The .view syntax creates a structure that mirrors another structure, until "forced" by an eager operation like .toList, .foreach, .forall, .count.

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Algorithm in Scala

44 lines of Scala (version 2.13).

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Test cases in Scala

assert(anagramChanges("aaabbb") == Some(3))
assert(anagramChanges("ab") == Some(1))
assert(anagramChanges("mnop") == Some(2))
assert(anagramChanges("xyyx") == Some(0))
assert(anagramChanges("xaxbbbxx") == Some(1))
assert(anagramChanges("abc") == None)
def anagramChanges(input: String): Option[Int] = ???