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This approach involves using a greedy strategy where we maintain counters for $5 and $10 bills. When a $5 bill is received, simply increase the count. For a $10 bill, check if you have a $5 bill to give as change. For a $20 bill, prefer giving one $10 and one $5 bill as change if available, otherwise give three $5 bills.
Time Complexity: O(n), where n is the length of the bills array.
Space Complexity: O(1), since only a fixed amount of additional space is used for counters.
1#include <vector>
2using namespace std;
3
4class Solution {
5public:
6 bool lemonadeChange(vector<int>& bills) {
7 int five = 0, ten = 0;
8 for (int bill : bills) {
9 if (bill == 5) five++;
10 else if (bill == 10) {
11 if (five == 0) return false;
12 five--; ten++;
13 } else {
14 if (ten > 0 && five > 0) {
15 ten--; five--;
16 } else if (five >= 3) {
17 five -= 3;
18 } else {
19 return false;
20 }
21 }
22 }
23 return true;
24 }
25};The C++ solution uses a similar approach with vector iteration to check each bill and determine if the proper change can be given, updating the count of $5 and $10 bills accordingly.
This approach focuses on prioritizing the use of $10 bills when giving change to customers with $20 bills. By simulating the process, we ensure that change is handled with a clear order of preference, utilizing available higher denominations first.
Time Complexity: O(n), where n is the size of the bills array.
Space Complexity: O(1), using only fixed space for counters.
1 public bool LemonadeChange(int[] bills) {
int five = 0, ten = 0;
foreach (int bill in bills) {
if (bill == 5) five++;
else if (bill == 10) {
if (five == 0) return false;
five--; ten++;
} else {
if (ten > 0 && five > 0) {
ten--; five--;
} else if (five >= 3) {
five -= 3;
} else {
return false;
}
}
}
return true;
}
}This C# solution tracks the quantity of bills in hand, preferring to give $10 as part of change if it simplifies future transactions, ensuring a smooth customer experience.