CODEWITHSUNDEEP
scalaimmutability
sfwh
sfwh

Reputation: 37

functional approach suggestions instead of using mutablelist

I was trying to solve breaktherope challenge of codility in scala but i could not come up with a functional solution. The snippet below is an example of where i'm stuck. Both list elements have to change throughtout the program . Can we solve this with map or sth like that?

var A = MutableList(5, 3, 6, 3, 3)
var B = MutableList(2, 3, 1, 1, 2)

for (i <- 0 to N){
    A(i) -= B(i)
}

Upvotes: 0

Views: 59

Answers (2)

Maximiliano Felice
Maximiliano Felice

Reputation: 367

One little enhancement to Maxim's answer would be the use of scala's for comprehensions, just to get a more declarative solution:

val result = for {
  (a, b) <- A zip B
} yield a - b

Upvotes: 0

Maxim
Maxim

Reputation: 7348

A functional solution will return a new list and not a mutation of A.

val result = A.zip(B).map { case (a, b) => a - b}

if A, B are immutable the list will be immutable as well. If A, B are mutable you can convert the result to immutable list using .toList

Upvotes: 3

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