CODEWITHSUNDEEP
scalafunctional-programmingkadanes-algorithm
S0rin
S0rin

Reputation: 1293

Kadane's Algorithm in Scala

Does anyone have a Scala implementation of Kadane's algorithm done in a functional style?

Edit Note: The definition on the link has changed in a way that invalidated answers to this question -- which goes to show why questions (and answers) should be self-contained instead of relying on external links. Here's the original definition:

In computer science, the maximum subarray problem is the task of finding the contiguous subarray within a one-dimensional array of numbers (containing at least one positive number) which has the largest sum. For example, for the sequence of values −2, 1, −3, 4, −1, 2, 1, −5, 4; the contiguous subarray with the largest sum is 4, −1, 2, 1, with sum 6.

Upvotes: 11

Views: 1390

Answers (3)

Heiko Seeberger
Heiko Seeberger

Reputation: 3722

What about this, if an empty subarray is allowed or the input array cannot be all negative:

numbers.scanLeft(0)((acc, n) => math.max(0, acc + n)).max

Or, failing the conditions above this (which assumes the input is non-empty):

numbers.tail.scanLeft(numbers.head)((acc, n) => (acc + n).max(n)).max

Upvotes: 19

Nader Ghanbari
Nader Ghanbari

Reputation: 4310

The following code returns the start and end index as well as the sum:


import scala.math.Numeric.Implicits.infixNumericOps
import scala.math.Ordering.Implicits.infixOrderingOps

case class Sub[T: Numeric](start: Index, end: Index, sum: T)

def maxSubSeq[T](arr: collection.IndexedSeq[T])(implicit n: Numeric[T]) =
  arr
    .view
    .zipWithIndex
    .scanLeft(Sub(-1, -1, n.zero)) {
      case (p, (x, i)) if p.sum > n.zero => Sub(p.start, i, p.sum + x)
      case (_, (x, i))                   => Sub(i, i, x)
    }
    .drop(1)
    .maxByOption(_.sum)

Upvotes: 1

Daniel C. Sobral
Daniel C. Sobral

Reputation: 297320

I prefer the folding solution to the scan solution -- though there's certainly elegance to the latter. Anyway,

numbers.foldLeft(0 -> 0) {
  case ((maxUpToHere, maxSoFar), n) =>
    val maxEndingHere = 0 max maxUpToHere + n
    maxEndingHere -> (maxEndingHere max maxSoFar)
}._2

Upvotes: 6

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