If you want to learn how to code, you need to learn algorithms. Learning algorithms improves your problem solving skills by revealing design patterns in programming. In this tutorial, you will learn how to code the longest increasing subsequence algorithm in JavaScript and Python.

This article originally published at jarednielsen.com

How to Code the Longest Increasing Subsequence Algorithm

Programming is problem solving. There are four steps we need to take to solve any programming problem:

1. Understand the problem

2. Make a plan

3. Execute the plan

4. Evaluate the plan

Understand the Problem

To understand our problem, we first need to define it. Let’s reframe the problem as acceptance criteria:

GIVEN an unsorted array of numbers 
WHEN I calculate the longest increasing subsequence
THEN I am returned the length of that sequence 

Let's use the first 16 digits following the decimal in Pi for an example.

1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 

Let's manually find the longest increasing subsequence. We'll place an X under the values in the sequence. The first value is obviously 1.

1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 
X

The second value is 2.

1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 
X         X

The third value is 3.

1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 
X         X     X

The fourth value is 5.

1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 
X         X     X X

The fifth value is 7.

1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 
X         X     X X     X

The sixth value is 9.

1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 
X         X     X X     X X

The length of the longest increasing subsequence of the first 16 digits of Pi is 6.

That’s our general outline. We know our input conditions, an unsorted array of postiive integers, and our output requirements, the length of the sequence which is a value greater than or equal to 1, and our goal is to find the longest increasing subsequence of values in the array.

Let’s make a plan!

Make a Plan

Let’s revisit our computational thinking heuristics as they will aid and guide is in making a plan. They are:

• Decomposition

• Pattern recognition

• Abstraction

• Algorithm design

The first step is decomposition, or breaking our problem down into smaller problems. Continuing with the first 16 Pi decimals, what's the smallest problem we can solve?

1

What's the longest subsequence?

Also 1.

What's the next smallest?

1 4

And what's the longest subsequence?

2

How did we calculate that?

We can see that there are two values and the last value is greater than the first, so the LIS is equal to 2. In other words, we made a comparison and tallied up the increasing values.

So what's the next smallest problem?

1 4 1

Ah! Now it gets interesting.

But the longest subsequence is still 2.

How do we solve this problem?

Again, we start a tally of increasing values. We know that the LIS is at least 1. We then compare 4 to 1, and, because 4 is greater than 1, we add 1 to our LIS tally. We compare the next value, 1, to 4, and, because 1 is less than 4, we do not add 1 to our LIS tally.

What's the next smallest problem?

1 4 1 5

Just when we thought we found the LIS we add a larger value!

How do we solve this problem?

We need to compare the current LIS, which is 3, with the previous LIS, which is 2.

This is starting to get complicated. We need a way to keep track of all these values. What if we keep a tally? There are a two approaches we can take to creating our tally:

1. Generate a new array of n length assigning each element a value of 1. On each iteration, reassign the corresponding value with the tally.

2. Initialize an array with only one element assigned a value of 1. On each iteration, add a new element containing the corresponding value of the tally.

Let's take the second approach. We can implement it without needing to use any constructors or an additional loop to generate the array.

If we start sketching out our pseudocode:

INPUT n

SET result TO 1
SET tally TO [1]

WORK SOME MAGIC! 

RETURN result

Now we need to work some magic.

We know we're going to need to iterate, and, if we need to calculate a result for every value in the array, we're going to need nested iteration.

Let's visualize this. Here's our array of four elements and our tally.

tally = [1]

array = [1, 4, 1, 5]

Following convention, we'll use the variables i and j for our outer and inner loops, respectively.

Let's initialize i with a value of 1 and j with a value of 0.

tally = [1]

            i
array = [1, 4, 1, 5]
         j

Why?

If we initialize i with 0, there's nowhere for j to go. We only need to iterate up to i. If we iterate beyond i in our nested loop, we won't get an accurate result.

In each iteration, we compare the value indexed by i and the value indexed by j. In this iteration, we see that 1 is less than 4. We take the value of our previous LIS, add 1, and update our tally. The LIS is now 2.

We start the next iteration, making the same comparisons as above...

tally = [1, 2]

               i
array = [1, 4, 1, 5]
         j

...until we reach the condition where we compare the value indexed by i and the value indexed by j and see that 4 is not less than 1, meaning our subsequence did not increase, so our LIS is unchanged.

tally = [1, 2]

               i
array = [1, 4, 1, 5]
            j

We still update our tally with this value and start the next iteration of the outer loop.

tally = [1, 2, 2]

                  i
array = [1, 4, 1, 5]
         j

Our nested loop iterates, making the same comparisons as above...

tally = [1, 2, 2]

                  i
array = [1, 4, 1, 5]
            j

...until we reach the condition where the value indexed by j is less than the value indexed by i, meaning our subsequence is increasing. We update the value in tally and exit our loops.

tally = [1, 2, 2, 3]

                  i
array = [1, 4, 1, 5]
               j

Let's update our pseudocode:

INPUT n

SET result TO 1
SET tally TO [1]

FOR EACH VALUE, i, BETWEEN 1 AND THE LENGTH OF n
    SET tally[i] TO 1
    FOR EACH VALUE, j, BETWEEN 0 AND i
        SET lis TO THE VALUE STORED IN tally[j] PLUS 1

        IF THE VALUE STORED IN n[j] IS LESS THAN THE VALUE STORED IN n[i] AND lis IS GREATER THAN THE VALUE STORED IN tally[i]
            SET tally[i] TO THE VALUE STORED IN lis
            IF lis IS GREATER THAN result
                SET result TO THE VALUE STORED IN lis

RETURN result 

Let's walk through this. We pass our LIS function an unsorted array, n.

We first initialzie a result variable and give it a value of 1 because we know that the result of our LIS calculation will be at least one.

We next initialize an array, tally, with one element assigned a value of 1. We do this for two reasons:

1. We know that the longest increasing subsequence is at least 1. It can't be 0.

2. We need to keep a record of which iteration contained the longest increasing subsquence.

We initialize our outer for loop, beginning the iteration at 1 and iterating up to the length of n. We start iterating at 1 because we use i as the condition in the nested for loop. If we started at 0, the nested loop would not execute its first iteration.

With each iteration of our outer loop, we add another element to our tally array with a value of 1.

We then initialize our nested for loop, beginning the iteration at 0. As above, note that we are iterating up to i. We are only iterating up to i to count the subsequence.

Within the nested loop, we initialize a lis variable.

If the value of n[j]is less than n[i] and the value of lis is greater than the value stored in lengths[i], we set lengths[i] to lis. This is how we store our count and increase it with each iteration.

Before we exit this condition our loops, we check if lis is greater than result. If so, we need to update result with the value stored in lis. Finally, when our iterations are complete, we return result.

Let's just use the first 8 values, [1 4 1 5 9 2 6 5]. The length of the longest increasing subsequence is 4.

Table time!
| i | j | lis | lengths | result |
| --- | --- | --- | --- | --- |
| 1 | 0 | 2 | [ 1, 2, 1, 1, 1, 1, 1, 1 ] | 2 |
| 2 | 0 | 2 | [ 1, 2, 1, 1, 1, 1, 1, 1 ] | 2 |
| 2 | 1 | 3 | [ 1, 2, 1, 1, 1, 1, 1, 1 ] | 2 |
| 3 | 0 | 2 | [1, 2, 1, 2, 1, 1, 1, 1] | 2 |
| 3 | 1 | 3 | [1, 2, 1, 3, 1, 1, 1, 1] | 3 |
| 3 | 2 | 2 | [1, 2, 1, 3, 1, 1, 1, 1] | 3 |
| 4 | 0 | 2 | [1, 2, 1, 3, 2, 1, 1, 1] | 3 |
| 4 | 1 | 3 | [1, 2, 1, 3, 3, 1, 1, 1] | 3 |
| 4 | 2 | 2 | [1, 2, 1, 3, 3, 1, 1, 1] | 3 |
| 4 | 3 | 4 | [1, 2, 1, 3, 4, 1, 1, 1] | 4 |

And so on...

Execute the Plan

Now it's simply a matter of translating our pseudocode into the syntax of our programming language.

How to Code the Longest Increasing Subsequence Algorithm in JavaScript

Let's start with JavaScript...

const longestIncreasingSubsequence = (n) => {
    let result = 1;
    const tally = [1];

    for (let i = 1; i < n.length; i++) {
        tally[i] = 1;
        for (let j = 0; j < i; j++) {
            let lis = tally[j] + 1;

            if (n[j] < n[i] && lis > tally[i]) {
                tally[i] = lis
                if (lis > result) {
                    result = lis;
                }
            }
        }
    }
    return result; 
}

How to Code the Longest Increasing Subsequence Algorithm in Python

Now let's see it in Python...

def longest_increasing_subsequence(n):
    result = 1
    tally = [1]

    for i in range(1, len(n)):
        tally += [1]
        for j in range(i):
            lis = tally[j] + 1

            if (n[j] < n[i] and lis > tally[i]):
                tally[i] = lis

                if lis > result:
                    result = lis

    return result

Evaluate the Plan

Can we do better?

The nested iteration isn't great for performance, but there's no getting around it. There are some refactors we could make. For example, we could use the max methods in our Math modules to find the lis in place of the result reassignment. I prefer this approach as it's more legible.

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