# Compute maximum sum of subarray (Kadane's algorithm) in purely functional immutable Scala

## 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]'.

## 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 © 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

- 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)`

- View
The

`.view`

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

## Algorithm in Scala

9 lines of Scala (version 2.13), showing how concise Scala can be!

## 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)
```

```
def computeForArrayClearKadane(array: Array[Int]): Int = ???
```