學習

1. 拆解問題的能力：刷了不少題目後，發現現在更能把問題拆解成小項目，然後一個個用程式碼實現出來，像這次寫題目前就先列出下面：
   • 分兩個 count： counter_25, counter_50 當其中一個小於0就是 NO，不然就是 YES
   • 收 25 時，counter_25 +1
   • 收 50 時，counter_25 -1
   • 收 100 時，counter_25 -3 || counter_25 -1 && counter_50 -1
2. 刷題前先簡單看個討論：刷題網站在 Codewars 上分成 test 與 attempt，前者只給 1, 2 個測試，後者就會給到 20 個左右。今天刷題不管怎麼看就是會有一個測試過不了，先是學到了原來刷題網站還是可以用「p」印出每次測試 input 的值，來檢查到底什麼情況下噴錯。結果一看，明明是測試檔有問題，後來在討論區發現非常多人反應。

題目：

The new "Avengers" movie has just been released! There are a lot of people at the cinema box office standing in a huge line. Each of them has a single 100, 50 or 25 dollar bill. An "Avengers" ticket costs 25 dollars.
Vasya is currently working as a clerk. He wants to sell a ticket to every single person in this line.
Can Vasya sell a ticket to every person and give change if he initially has no money and sells the tickets strictly in the order people queue?
Return YES, if Vasya can sell a ticket to every person and give change with the bills he has at hand at that moment. Otherwise return NO.

# 翻譯：某家電影院，賣一張門票是 25 元，只接受客戶給 25 元、50 元、100 元三種面額，假設從開店就沒有任何錢的情況下，如果店員能在服務完所有客人後中間都順利找錢就回傳 'YES' 否則 'NO'

def tickets(people)

end

答案需要過以下測試：

RSpec.describe "Vasya - Clerk" do
  it "ticket" do
    expect(tickets([25, 25, 50])).to eq('YES')
    expect(tickets([25, 100])).to eq('NO')
    expect(tickets([50, 100, 100])).to eq('NO')
  end
end

我的答案

def tickets(people)
  counter_25, counter_50 = 0, 0
  people.each do |pay|
    counter_25 += 1 if pay == 25
    
    if pay == 50
      counter_25 -= 1
      counter_50 += 1
    end
    
    if pay == 100
      if counter_25 >= 3
        counter_25 -= 3 
      else
        counter_25 -= 1
        counter_50 -= 1
      end
    end
  end

  if counter_25 < 0 || counter_50 < 0
    'NO'
  else
    'YES'
  end
end

思路：

1. 分兩個 count： counter_25, counter_50 當其中一個小於0就是 NO，不然就是 YES
2. 收 25 時，counter_25 +1，收 50 時，counter_25 -1
3. 收 100 時，counter_25 -3 或是 counter_25 -1 && counter_50 -1，總要選擇一個做，所以先判斷 4. counter_25 是否 ≥ 3，是就選前者，否就選後者