How to resolve the algorithm Convert seconds to compound duration step by step in the OCaml programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Convert seconds to compound duration step by step in the OCaml programming language
Table of Contents
Problem Statement
Write a function or program which: This is detailed below (e.g., "2 hr, 59 sec").
Demonstrate that it passes the following three test-cases: Test CasesDetailsThe following five units should be used: However, only include quantities with non-zero values in the output (e.g., return "1 d" and not "0 wk, 1 d, 0 hr, 0 min, 0 sec"). Give larger units precedence over smaller ones as much as possible (e.g., return 2 min, 10 sec and not 1 min, 70 sec or 130 sec) Mimic the formatting shown in the test-cases (quantities sorted from largest unit to smallest and separated by comma+space; value and unit of each quantity separated by space).
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Convert seconds to compound duration step by step in the OCaml programming language
Source code in the ocaml programming language
let divisors = [
(max_int, "wk"); (* many wk = many wk *)
(7, "d"); (* 7 d = 1 wk *)
(24, "hr"); (* 24 hr = 1 d *)
(60, "min"); (* 60 min = 1 hr *)
(60, "sec") (* 60 sec = 1 min *)
]
(* Convert a number of seconds into a list of values for weeks, days, hours,
* minutes and seconds, by dividing the number of seconds 'secs' successively by
* the values contained in the list 'divisors' (taking them in reverse order).
* Ex:
* compute_duration 7259
* returns
* [(0, "wk"); (0, "d"); (2, "hr") (0, "min"); (59, "sec")]
*)
let compute_duration secs =
let rec doloop remain res = function
| [] -> res
| (n, s) :: ds -> doloop (remain / n) ((remain mod n, s) :: res) ds
in
doloop secs [] (List.rev divisors)
(* Format nicely the list of values.
* Ex:
* pretty_print [(0, "wk"); (0, "d"); (2, "hr") (0, "min"); (59, "sec")]
* returns
* "2 hr, 59 sec"
*
* Intermediate steps:
* 1. Keep only the pairs where duration is not 0
* [(2, "hr"); (59, "sec")]
* 2. Format each pair as a string
* ["2 hr"; "59 sec"]
* 3. Concatenate the strings separating them by a comma+space
* "2 hr, 59 sec"
*)
let pretty_print dur =
List.filter (fun (d, _) -> d <> 0) dur
|> List.map (fun (d, l) -> Printf.sprintf "%d %s" d l)
|> String.concat ", "
(* Transform a number of seconds into the corresponding compound duration
* string.
* Not sure what to do with 0... *)
let compound = function
| n when n > 0 -> compute_duration n |> pretty_print
| n when n = 0 -> string_of_int 0 ^ "..."
| _ -> invalid_arg "Number of seconds must be positive"
(* Some testing... *)
let () =
let test_cases = [
(7259, "2 hr, 59 sec");
(86400, "1 d");
(6000000, "9 wk, 6 d, 10 hr, 40 min");
(0, "0...");
(3599, "59 min, 59 sec");
(3600, "1 hr");
(3601, "1 hr, 1 sec")
] in
let testit (n, s) =
let calc = compound n in
Printf.printf "[%s] %d seconds -> %s; expected: %s\n"
(if calc = s then "PASS" else "FAIL")
n calc s
in
List.iter testit test_cases
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