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This approach leverages an array to manage the count of available parking slots for each type of car. The array indices correspond to the car types: index 0 for big cars, index 1 for medium, and index 2 for small. The addCar function decrements the corresponding index if space is available.
Time Complexity: O(1) for each addCar operation as we are directly accessing an array element.
Space Complexity: O(1) as only a fixed-size array is used.
1class ParkingSystem {
2 private int[] slots;
3
4 public ParkingSystem(int big, int medium, int small) {
5 slots = new int[]{big, medium, small};
6 }
7
8 public boolean addCar(int carType) {
9 if (slots[carType - 1] > 0) {
10 slots[carType - 1]--;
11 return true;
12 }
13 return false;
14 }
15}In Java, an array is utilized to store the available slots for different car types. The addCar method checks availability and decrements the slot count for the specified carType.
This method uses distinct variables to handle each type of car's parking slots. This approach makes the code very clear for small data sets. While this isn't necessarily more efficient than the array method for this particular problem, it offers an alternative for simple systems where explicit clarity is beneficial.
Time Complexity: O(1) for checking and updating.
Space Complexity: O(1) as three variables track the state.
1 int big, medium, small;
public:
ParkingSystem(int big, int medium, int small) : big(big), medium(medium), small(small) {}
bool addCar(int carType) {
if (carType == 1) {
if (big > 0) { big--; return true; }
} else if (carType == 2) {
if (medium > 0) { medium--; return true; }
} else if (carType == 3) {
if (small > 0) { small--; return true; }
}
return false;
}
};This code uses dedicated variables for each parking slot type. The addCar function updates the correct variable similar to the array-based solution but directly checks carType through conditionals.