Find The Longest Increasing Subsequence - Dynamic Programming Fundamentals

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Question: Given an array of all integers that may or may not be sorted, determine the length of the longest subsequence that is strictly non-decreasing.

What Is A Subsequence?

A subsequence of an array is a subset of the array that maintains temporal order.

It does not have to be contiguous but it might turn out to be contiguous by chance.
Problem Name / Variants

This problem also comes in the form of asking for the longest strictly decreasing subsequence.

This is longest non-decreasing subsequence meaning that we will have a non-strictly increasing subsequence (aka we can have deltas of 0 between contiguous elements in the subsequence).
Approach 1 (Brute Force)

We can enumerate all 2^n subsets of the original array and then test them for the non-decreasing property.

The answer will be the longest subset that has the property.

This is too expensive.
Approach 2 (Dynamic Programming)

Do you see the potential for a subproblem here?

If you do, then we have the opportunity to use dynamic programming.

Example
[ -1, 3, 4, 5, 2, 8 ]

At the index 0 I always know that I can have a subsequence of length 1.

In fact, at all positions the longest non-decreasing subsequence can be at least length 1.

We then look at index 1, I need to ask myself if the item at index 1 can lengthen the longest subseqence found at index 0.

We check if 3 is greater than or equal to 1...it is. Great. index 1 can be tacked on but...should I?

The LNDS (longest non-decreasing subsequence) at index 1 is 0.

The LNDS at index 0 is 1.

Yeah...it makes sense because if I tack 3 onto the LNDS I found for the subproblem of just [ -1 ] then at index 1 I will also have a LDNS.

So what we basically do is build a table and ask ourselves these questions all along the way.

EACH CELL REPRESENTS THE ANSWER TO THE SUBPROBLEM ASKED AGAINST the subsequence from index 0 to index i (including the element at index i).
Complexities:

Time: O( n^2 )

n is the length of the array.

For each of the n elements we will solve the LNDS problem which takes O(n) time, therefore we yield a O( n^2 ) runtime complexity.

Space: O( n )

We will store our answers for each of the n LNDS subproblems.
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This question is number 17.12 in "Elements of Programming Interviews" by Adnan Aziz, Tsung-Hsien Lee, and Amit Prakash.

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