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lemonade-change.py
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lemonade-change.py
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# Time: O(n)
# Space: O(1)
# 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 don't have any change in hand at first.
#
# Return true if and only if you can provide every customer
# with correct change.
#
# Example 1:
#
# Input: [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: [5,5,10]
# Output: true
# Example 3:
#
# Input: [10,10]
# Output: false
# Example 4:
#
# Input: [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't give change of $15 back
# because we only have two $10 bills.
# Since not every customer received correct change, the answer is false.
#
# Note:
# - 0 <= bills.length <= 10000
# - bills[i] will be either 5, 10, or 20.
class Solution(object):
def lemonadeChange(self, bills):
"""
:type bills: List[int]
:rtype: bool
"""
five, ten = 0, 0
for bill in bills:
if bill == 5:
five += 1
elif bill == 10:
if not five:
return False
five -= 1
ten += 1
else:
if ten and five:
ten -= 1
five -= 1
elif five >= 3:
five -= 3
else:
return False
return True