Design a parking system for a parking lot. The parking lot has three kinds of parking spaces: big, medium, and small, with a fixed number of slots for each size.

Implement the `ParkingSystem` class:

*   `ParkingSystem(int big, int medium, int small)` Initializes object of the `ParkingSystem` class. The number of slots for each parking space are given as part of the constructor.
*   `bool addCar(int carType)` Checks whether there is a parking space of `carType` for the car that wants to get into the parking lot. `carType` can be of three kinds: big, medium, or small, which are represented by `1`, `2`, and `3` respectively. **A car can only park in a parking space of its** `carType`. If there is no space available, return `false`, else park the car in that size space and return `true`.

**Example 1:**

**Input**
\[ "ParkingSystem ",  "addCar ",  "addCar ",  "addCar ",  "addCar "\]
\[\[1, 1, 0\], \[1\], \[2\], \[3\], \[1\]\]
**Output**
\[null, true, true, false, false\]

**Explanation**
ParkingSystem parkingSystem = new ParkingSystem(1, 1, 0);
parkingSystem.addCar(1); // return true because there is 1 available slot for a big car
parkingSystem.addCar(2); // return true because there is 1 available slot for a medium car
parkingSystem.addCar(3); // return false because there is no available slot for a small car
parkingSystem.addCar(1); // return false because there is no available slot for a big car. It is already occupied.

**Constraints:**

*   `0 <= big, medium, small <= 1000`
*   `carType` is `1`, `2`, or `3`
*   At most `1000` calls will be made to `addCar`


def runningSum(nums):
    for i in range(1, len(nums)):
        nums[i] += nums[i - 1]
    return nums
