9/18 每日一題(顧客購買價值5元的產品,並指會使用5,10,20付款,檢查是否可以找零)

 At a lemonade stand, each lemonade costs $5. Customers are standing in a queue to buy from you and order one at a time (in the order specified by bills). Each customer will only buy one lemonade and pay with either a $5, $10, or $20 bill. You must provide the correct change to each customer so that the net transaction is that the customer pays $5.

Note that you do not have any change in hand at first.

Given an integer array bills where bills[i] is the bill the ith customer pays, return true if you can provide every customer with the correct change, or false otherwise.

 

Example 1:

Input: bills = [5,5,5,10,20]
Output: true
Explanation: 
From the first 3 customers, we collect three $5 bills in order.
From the fourth customer, we collect a $10 bill and give back a $5.
From the fifth customer, we give a $10 bill and a $5 bill.
Since all customers got correct change, we output true.

Example 2:

Input: bills = [5,5,10,10,20]
Output: false
Explanation: 
From the first two customers in order, we collect two $5 bills.
For the next two customers in order, we collect a $10 bill and give back a $5 bill.
For the last customer, we can not give the change of $15 back because we only have two $10 bills.
Since not every customer received the correct change, the answer is false.

 

Constraints:

• 1 <= bills.length <= 105
• bills[i] is either 5, 10, or 20.

class Solution:

    def lemonadeChange(self, bills: List[int]) -> bool:

        My_dict={5:0,10:0,20:0}

        for i in bills:

            if i ==5:

                My_dict[i]+=1

           

            elif i ==10:

                if My_dict[5]>0:

                    My_dict[5] -= 1

                    My_dict[10] += 1

                else:

                    print(f'{i}找不開,只有{My_dict}')

                    return False

            else:

                if My_dict[5]>0 and My_dict[10]>0:

                    My_dict[5] -= 1

                    My_dict[10] -= 1

                    My_dict[20] += 1

                elif My_dict[5] > 2:

                    My_dict[5] -= 3

                    My_dict[20] += 1

                else:

                    print(f'{i}找不開,只有{My_dict}')

                    return False

        return True