Coin Vending Machine
Problem
Design a vending machine class. It is built with a price per slot code and a stock count per slot, and it keeps the credit inserted so far. insert(coin) accepts only 5, 10, 25 and 100 and answers "credit N", or "rejected N" for any other coin, which drops straight back out. select(code) answers "sold out" for an empty slot, "need N" when the credit is N short, and otherwise vends, returns the change and resets the credit: "vend CODE, change N". cancel() returns the whole credit as "refund N". The machine must never vend without payment or keep money after a vend or a cancel.
Examples
Input: prices = A1 65, B2 40; stock = A1 1, B2 0
insert(25), insert(3), select("A1")
Output: ['credit 25', 'rejected 3', 'need 40']Input: then insert(100), select("A1"), select("A1")
Output: ['credit 125', 'vend A1, change 60', 'sold out']
Why: the only A1 is gone after the first vendInput: then select("B2"), insert(10), cancel()
Output: ['sold out', 'credit 10', 'refund 10']
Why: edge case, a slot that starts empty never takes money for itselfHints
0 / 3
List what the machine has to remember between calls. It is less than it looks: the tables it was built with, and one number.
Each method checks its failure cases first and returns early, and only the success path changes state. That ordering is what keeps money from going missing.
insert adds to credit only for an allowed coin. select checks stock, then credit, and on success works out the change, sets credit to 0 and takes one item from stock. cancel hands back the credit and sets it to 0.
Solution
The machine's whole state is its price table, its stock and one credit counter, and every method guards its failure cases before touching any of them, so a rejected coin, a sold-out slot or a short credit leaves the state exactly as it was. Stock is checked before credit so that an empty slot is reported as sold out rather than as a price to be paid. On a vend the change and the reset credit are computed together, which makes it impossible to hand out change and still hold the credit. A small run helper replays a list of calls so each example is one line. Every call is O(1).
class VendingMachine:
COINS = {5, 10, 25, 100}
def __init__(self, prices, stock):
self.prices, self.stock, self.credit = prices, stock, 0
def insert(self, coin):
if coin not in self.COINS:
return "rejected " + str(coin) # the coin drops straight back out
self.credit += coin
return "credit " + str(self.credit)
def select(self, code):
if self.stock.get(code, 0) == 0:
return "sold out"
if self.credit < self.prices[code]:
return "need " + str(self.prices[code] - self.credit)
change, self.credit = self.credit - self.prices[code], 0
self.stock[code] -= 1
return "vend " + code + ", change " + str(change)
def cancel(self):
refund, self.credit = self.credit, 0
return "refund " + str(refund)
def run(machine, calls):
return [getattr(machine, name)(*args) for name, *args in calls]
vm = VendingMachine({"A1": 65, "B2": 40}, {"A1": 1, "B2": 0})
print(run(vm, [("insert", 25), ("insert", 3), ("select", "A1")])) # -> ['credit 25', 'rejected 3', 'need 40']
print(run(vm, [("insert", 100), ("select", "A1"), ("select", "A1")])) # -> ['credit 125', 'vend A1, change 60', 'sold out']
print(run(vm, [("select", "B2"), ("insert", 10), ("cancel",)])) # -> ['sold out', 'credit 10', 'refund 10']Stuck on the idea rather than the code? Vending Machine covers it.