# Longest increasing sub-sequence length

## Algorithm goal

In a sequence, find the length of the longest increasing sub-sequence.

Example: $$[11, 10, 3, 6, 4, 9, 101, 18]$$ has 2 longest increasing sub-sequences: $$[3, 4, 9, 101]$$, and $$[3, 6, 9, 101]$$.

The length of these sub-sequences is $$4$$.

Solution is expected to be efficient as brute-force is computationally expensive.

## Brute-force solution

To put things in perspective, we sometimes need to consider a brute-force solution. In this solution, for every item in the array, we select all the items ahead of it that are greater than it, and from that, we can run the algorithm again. However, this is highly inefficient (some variants will be $$O(2^n)$$ which is not feasible at all!).

## $$O(n^2)$$ solution

Let's consider $$[11, 10, 3, 6, 4, 9, 101, 18]$$. Often problems like this can be solved by not searching, but by assuming. What a strange idea, right? (The concept is very similar to MaximumSubArraySum) Well, if we assumed that our LIS (longest-increasing subsequence) ends at $$18$$, then the LIS ending at that is $$1 +$$ the LIS that ends at any of the numbers before it, that are under $$18$$. Likewise, we can do the same check for $$101$$. We can turn this understanding into the following formula (might take some time to wrap your head around it!): (this is © from www.scala-algorithms.com)

$$L(i) = 1 + \max_{\substack { j < i \\ v_j < v_i } }L(j)$$

Here, thus we end up with a simple relationship between LIS for $$i$$ and LIS for $$i+1$$. What we can do here now for optimisation is to compute the value forward (it is called Dynamic Programming, because you need to remember all the previous values that you had computed - to benefit from the programming optimisation). Our solution, then, is $$\max_i{L(i)}$$.

## Scala Concepts & Hints

Collect

'collect' allows you to use Pattern Matching, to filter and map items.

assert("Hello World".collect {
case character if Character.isUpperCase(character) => character.toLower
} == "hw")


Drop, Take, dropRight, takeRight

Scala's drop and take methods typically remove or select n items from a collection.

assert(List(1, 2, 3).drop(2) == List(3))

assert(List(1, 2, 3).take(2) == List(1, 2))

assert(List(1, 2, 3).dropRight(2) == List(1))

assert(List(1, 2, 3).takeRight(2) == List(2, 3))

assert((1 to 5).take(2) == (1 to 2))


foldLeft and foldRight

A 'fold' allows you to perform the equivalent of a for-loop, but with a lot less code.

def foldMutable[I, O](initialState: O)(items: List[I])(f: (O, I) => O): O =
items.foldLeft(initialState)(f)


For-comprehension

The for-comprehension is highly important syntatic enhancement in functional programming languages.

val Multiplier = 10

val result: List[Int] = for {
num <- List(1, 2, 3)
anotherNum <-
List(num * Multiplier - 1, num * Multiplier, num * Multiplier + 1)
} yield anotherNum + 1

assert(result == List(10, 11, 12, 20, 21, 22, 30, 31, 32))


Option Type

The 'Option' type is used to describe a computation that either has a result or does not. In Scala, you can 'chain' Option processing, combine with lists and other data structures. For example, you can also turn a pattern-match into a function that return an Option, and vice-versa!

assert(Option(1).flatMap(x => Option(x + 2)) == Option(3))

assert(Option(1).flatMap(x => None) == None)


Pattern Matching

Pattern matching in Scala lets you quickly identify what you are looking for in a data, and also extract it.

assert("Hello World".collect {
case character if Character.isUpperCase(character) => character.toLower
} == "hw")


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.

Zip

'zip' allows you to combine two lists pair-wise (meaning turn a pair of lists, into a list of pairs)

It can be used over Arrays, Lists, Views, Iterators and other collections.

assert(List(1, 2, 3).zip(List(5, 6, 7)) == List(1 -> 5, 2 -> 6, 3 -> 7))

assert(List(1, 2).zip(List(5, 6, 7)) == List(1 -> 5, 2 -> 6))

assert(List(5, 6).zipWithIndex == List(5 -> 0, 6 -> 1))


## Algorithm in Scala

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

## Test cases in Scala

assert(
lengthOfLongestIncreasingSubSequence(
Array(
0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15
)
).contains(6),
"Wikipedia example has LIS of 6"
)
assert(
lengthOfLongestIncreasingSubSequence(Array(11, 10, 3, 6, 4, 9, 101, 18))
.contains(4),
"Subsequence is length of 4, [3, 4 (or 6), 9, 101]"
)
assert(
lengthOfLongestIncreasingSubSequence(
Array(
0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15
)
).contains(6),
"Subsequence is length of 6, [0, 2, 6, 9, 11, 15]"
)
assert(
lengthOfLongestIncreasingSubSequence(Array.empty).isEmpty,
"Empty subsequence is empty"
)
assert(
lengthOfLongestIncreasingSubSequence(Array(5)).contains(1),
"Sequence of 1 is just 1"
)
assert(
lengthOfLongestIncreasingSubSequence(Array(-2, -1)).contains(2),
"Negative numbers work too"
)

def lengthOfLongestIncreasingSubSequence(array: Array[Int]): Option[Int] = ???