# Scala algorithm: Compute maximum sum of subarray (Kadane's algorithm)

## Algorithm goal

Find the maximum sum of a sub-sequence of an array. This problem is known as 'MaxSliceSum' on Codility.

If we have an array like '[1, 2, -3, 4, -5]', the maximum sum of a subarray is 4, that of '[4]'.

## Test cases in Scala

assert(
computeForArrayClearKadane(Array[Int](-2, 1, -3, 4, -1, 2, 1, -5, 4)) == 6
)
assert(computeForArrayClearKadane(Array(1, 2, -3, 4, -5)) == 4)


## Algorithm in Scala

10 lines of Scala (compatible versions 2.13 & 3.0), showing how concise Scala can be!

## Explanation

### The mathematics

We need to break the problem down into parts: if we have a sequence $$[a,b,c]$$, by brute-force, we would need to look through $$[a]$$, $$[b]$$, $$[c]$$, $$[a,b]$$, $$[b,c]$$, and $$[a,b,c]$$. However, if we had computed $$[a,b]$$, computing $$[a,b,c]$$ from scratch would be redundant (as we have to do many summations more than once) and here we see a possibility of optimisation.

How do we split it into parts? If we define the result $$M_e$$ to be the maximum sum of subarray ending at position $$e$$, then $$M_{e+1}$$ is either that, plus the value $$V_{e+1}$$ at position $$e+1$$, or it is the new value $$V_{e+1}$$. (this is Â© from www.scala-algorithms.com)

This leads to a formula $$M_{e+1} = \max\{ M_e, M_e + V_{e+1}\}$$, and now that we have the list of maximum subarrays ending at position $$e$$, we can find the maximum value from those items.

### The code

Once you understand the mathematics, the solution becomes quite straightforward.

## Scala concepts & Hints

1. ### scanLeft and scanRight

Scala's scan functions enable you to do folds like foldLeft and foldRight, while collecting the intermediate results

assert(List(1, 2, 3, 4, 5).scanLeft(0)(_ + _) == List(0, 1, 3, 6, 10, 15))

2. ### 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)

3. ### 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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