Daily Challenge #69 - Going to the Cinema

Today's challenge is to help John make an important financial decision!

John likes to go to the cinema, but he wants to find the most cost-effective way to go. He can buy a ticket for $15, or he can buy a membership card for $500. Every time he uses the membership card, the ticket will be 0.9 times the price he paid for the last one.

Ex. If John goes to the cinema three times:
A : 15 * 3 = 45
B : 500 + 15 * 0.90 + (15 * 0.90) * 0.90 + (15 * 0.90 * 0.90) * 0.90 ( = 536.5849999999999, no rounding for each ticket)

Create a function movie with three parameters: card (price of the card), ticket(normal price of a ticket), perc(fraction of what he paid for the previous ticket) and returns the first n such that ceil(price of B) < price of A

More examples:
movie(500, 15, 0.9)should return43
(with card the total price is 634, with tickets 645)

movie(100, 10, 0.95) should return 24
(with card the total price is 235, with tickets 240)

Good luck!

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Today's challenge comes from CodeWars user g964. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge idea for a future post? Email [e-mail] with your suggestions!

Comments

[TadPol]

Here is my try:

function movie(card, ticket,perc){
    let i = 0
    let ticketFullPrice = 0
    let lowedTicket = ticket

    while(card>ticketFullPrice){
        ticketFullPrice += ticket
        lowedTicket *= perc 
        card += lowedTicket 
        i++
    }
    return i
}
movie(500,15,0.9)
movie(100, 10, 0.95)

[Mihail Malo]

"first" lmao

const movie = (card, ticket, part) => {
  let i = 0
  let a = 0
  let b = card
  do {
    a += ticket
    b += ticket * part ** i
    i++
  } while (Math.ceil(b) >= a)
  return i - 1
}

an "optimization" (no power)

const movie = (card, ticket, part) => {
  let i = 0
  let a = 0
  let b = card
  let cheap = ticket
  while (true) {
    a += ticket
    b += cheap
    cheap *= part
    if (Math.ceil(b) < a) return i
    i++
  }
}

"optimization II" (no Math.ceil, comparison with 0 instead of 1)

const movie = (card, ticket, part) => {
  let i = 0
  let dif = -card - 1
  let cheap = ticket
  while (true) {
    dif += ticket - cheap
    cheap *= part
    if (dif > 0) return i
    i++
  }
}

idiocy.

const movie = (card, ticket, part) => {
  let i = 1
  let dif = -card - 1
  let cheap = ticket
  while ((dif += ticket - (cheap *= part)) <= 0) i++
  return i
}

[Yasser Beyer]

denied on first lol

[Mihail Malo]

Check timestamp by hovering the date lol

[Amin]

Elm

module Movie exposing (movie)


getReduction : Int -> Float -> Float
getReduction times percentage =
    percentage ^ toFloat times


profitableNumberOfMovies : Int -> Int -> Float -> Int -> Int
profitableNumberOfMovies card ticket percentage times =
    let
        reduction : Float
        reduction =
            getReduction times percentage

        increasedCard : Int
        increasedCard =
            ceiling <| toFloat card + toFloat ticket * reduction

        increasedTicket : Int
        increasedTicket =
            ticket * times
    in
    if increasedCard < increasedTicket then
        times - 1

    else
        profitableNumberOfMovies increasedCard ticket percentage <| times + 1


movie : Int -> Int -> Float -> Int
movie card ticket percentage =
    profitableNumberOfMovies card ticket percentage 1

Tests

module MovieTest exposing (suite)

import Expect
import Movie exposing (movie)
import Test exposing (Test)


suite : Test
suite =
    Test.describe "Movie"
        [ Test.test "It should return 43 for movie 500 15 0.9" <|
            \_ ->
                Expect.equal 43 <| movie 500 15 0.9
        , Test.test "It should return 24 for movie 100 10 0.95" <|
            \_ ->
                Expect.equal 24 <| movie 100 10 0.95
        ]

[E. Choroba]

#!/usr/bin/perl
use warnings;
use strict;

sub movie {
    my ($card, $ticket, $perc) = @_;
    my $reduced = my $sum = $ticket * $perc;
    my $count = 0;
    $sum += ($reduced *= $perc)
        while $ticket * ++$count < $card + int $sum;
    return $count
}

use Test::More tests => 3;

is movie(100, 200, 0.5),   1;
is movie(500,  15, 0.9),  43;
is movie(100,  10, 0.95), 24;

[Yasser Beyer]

JavaScript

const movie = (card, ticket, perc) => {
    let n = 1;
    let frac = ticket * perc;
    let a = ticket * n;
    let b = card + frac;
    while(a < Math.ceil(b)) {
        n++;
        frac = frac * perc;
        b = b + frac;
        a = ticket * n;
    }
    return n;
}