top of page

# DataWeave programming challenge #6: Using tail-recursion to get the factorial of a number

Â

Other posts from this series:

Â

In this post:

Â

Try to solve this challenge on your own to maximize learning. We recommend you refer to the DataWeave documentation only. Try to avoid using Google or asking others so you can learn on your own and become a DataWeave expert!

## Input

Consider the following input payload (can be of txt format):

```3
5
-6
300```

## Explanation of the problem

1. Create a tail-recursive function to get the factorial of each positive number from the payload.

2. Sum the results.

3. Retrieve the digits/characters located at positions 20-25.

4. Make sure to return a number and not a string.

5. Numbers 0 or less won't be calculated.

For example:

• The factorial of 3 is 6 (3 x 2 x 1)

• The factorial of 5 is 120 (5 x 4 x 3 x 2 x 1)

• The factorial of -6 won't be calculated so the result will be 0

• And so on

• Once the results are summed (6 + 120 + 0...), extract the digits at indexes 20 to 25.

• Return this last 6-digit number.

## Expected output

In this case, the expected output would be:

`537046`

## Clues

If you're stuck with your solution, feel free to check out some of these clues to give you ideas on how to solve it!

Clue #1

Clue #2

Clue #3

Clue #4

Clue #5

If you haven't solved this challenge yet, we encourage you to keep trying! It's ok if it's taking longer than you thought. We all have to start somewhere âœ¨ Check out the clues and read the docs before giving up. You got this!! ðŸ’™

There are many ways to solve this challenge, but you can find here one of my solutions!

Solution #1

Feel free to comment your code below for others to see! ðŸ˜„

Subscribe to receive notifications as soon as new content is published âœ¨

Tags:

For those using dw::core::Arrays slice it excludes the last index, so select 20-26

Like

%dw 2.0 output application/json fun solve(x)= x reduce(item, acc) -> ( (if(item > 0) acc + factorial(item) else acc)) fun factorial(x)= 2 to x reduce(item, acc) -> (acc*item) --- (solve(payload splitBy('\n')) splitBy(''))[20 to 25] joinBy('')

Like

%dw 2.0 fun factorial(num)= num to 1 reduce ((item, accumulator) -> accumulator * item) fun pos(val) = val[20 to 25] --- pos(sum((payload splitBy "\n" ) map factorial(\$))) as Number

Like
Alex Martinez
May 10, 2023

Another person using reduce :D haha nice job!

Like

Disha Bhar
May 10, 2023

Tried this out. :)

Like
Alex Martinez
May 10, 2023

awesome!!

Like

Felix Schnabel
May 09, 2023