Might there be a better way to achieve this? ``` val elementChanges = listOf("a", "b", "b").zipWith...

# codereview

[Daniel]

Might there be a better way to achieve this?

val elementChanges = listOf("a", "b", "b").zipWithNext { a, b ->
            if(b != a) 1 else 0
        }.sum()

I try to get the count of "changes" in a list in relation to one element to the next:

Here for example I see that there is a change from "a" to "b", but after that there are no changes anymore "b" to "b"

[Pavlo Liapota]

Looks good to me.
Other ways to write it would be:

listOf("a", "b", "b")
    .zipWithNext { a, b -> a != b }
    .count { it }

listOf("a", "b", "b")
    .zipWithNext()
    .count { (a, b) -> a != b }

[Daniel]

Thank you for your suggestions! They are also good possibilities

[Milan Hruban]

I would probably do

listOf("a", "b", "b")
    .windowed(size = 2, step = 1)
    .count { it.distinct().size == 2 }

because then the "windowSize" (2 in this case) could be extracted to parameter

[Daniel]

Also a good option, thank you!