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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.
++++++++++++++++++++++++++++++++++++++++++++++++++
++++++++++++++++++++++++++++++++++++++++++++++++++
This question is number 17.12 in "Elements of Programming Interviews" by Adnan Aziz, Tsung-Hsien Lee, and Amit Prakash.
🏃 Run Code As You Learn
💾 Save Progress
❓New Unseen Questions
🔎 Get All Solutions
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.
++++++++++++++++++++++++++++++++++++++++++++++++++
++++++++++++++++++++++++++++++++++++++++++++++++++
This question is number 17.12 in "Elements of Programming Interviews" by Adnan Aziz, Tsung-Hsien Lee, and Amit Prakash.
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