Convert seconds to compound duration

Questions / Help
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Challenge

I have recently changed to CodeWars to make practice my Elixir skills. To complete an exercise I need to convert seconds to compound duration. This looks complex, but in reality is something most developers did at one time or another:

Write a function or program which:

  • takes a positive integer representing a duration in seconds as input (e.g., 100 ), and
  • returns a string which shows the same duration decomposed into hours, minutes, and seconds like: “01|20|34” ( 1 hour, 20 minutes, 34 seconds ).

Each string needs to have 2 digits.

Question

I noticed that some languages like Julia already have a function that does this. I wonder if Elixir / Erlang, with so many years going on, also has one.

Does any one know of an native function that helps me accomplish this without having to reinvent the wheel all over again?

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I may be wrong but I thought the point of these challenges was to implement such functions. But answering your question, I would create a zero NaiveDateTime and add seconds using the add function from this module.

Edit: In fact you can use Time and Time.add

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The idea of exercise is not to make a compound date. That is merely a necessary step to make the exercise, like knowing how to add to numbers in order to calculate integrals in math.

So I was hoping I could skip this with a native function and actually focus on the core of the exercise itself.

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I haven’t found a native implementation of this. But I had to do the same for my website just now.

I think you can accomplish the behavior by using rem and div.

hours = div seconds, 3600
minutes = div seconds, 60
seconds = rem seconds, 60

I haven’t comprehensively tested this to make sure there are no edge cases, but I think it should work :slight_smile:

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If you take 3661 seconds, that’s 1 hour, 1 minute, 1 second.

But div seconds, 60 is equal to 61.

I would write :

  def seconds_to_hours_minutes_seconds(seconds) do
    { div(seconds, 3600), rem(seconds, 3600) |> div(60),  rem(seconds, 3600) |> rem(60)}
  end

Also, I don’t think you can use NaiveDateTime easily, because you will soon get days, months and years. And if you use Time, it won’t work for durations above 24h, as you will get back to 0.