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 either5
,10
, or20
.
Approach 01:
-
C++
-
Python
class Solution { public: bool lemonadeChange(vector<int>& bills) { int fiveDollarCount = 0; int tenDollarCount = 0; for (const int bill : bills) { if (bill == 5) { ++fiveDollarCount; } else if (bill == 10) { --fiveDollarCount; ++tenDollarCount; } else { // bill == 20 if (tenDollarCount > 0) { --tenDollarCount; --fiveDollarCount; } else { fiveDollarCount -= 3; } } if (fiveDollarCount < 0) return false; } return true; } };
class Solution: def lemonadeChange(self, bills: List[int]) -> bool: fiveDollarCount = 0 tenDollarCount = 0 for bill in bills: if bill == 5: fiveDollarCount += 1 elif bill == 10: fiveDollarCount -= 1 tenDollarCount += 1 else: # bill == 20 if tenDollarCount > 0: tenDollarCount -= 1 fiveDollarCount -= 1 else: fiveDollarCount -= 3 if fiveDollarCount < 0: return False return True
Time Complexity
- Iterating Through Bills:
The function iterates through each bill in the
bills
vector exactly once. Each iteration involves a constant amount of work (checking conditions and updating counters), making the time complexity \( O(n) \), where \( n \) is the number of bills.
Space Complexity
- Space Usage:
The function uses a fixed amount of additional space for storing two integer counters (
fiveDollarCount
andtenDollarCount
). This results in a space complexity of \( O(1) \), indicating constant space usage.