Problem Statement
We have a box that consists of (sizeX x sizeY x sizeZ) = N unit cubes. These cubes have coordinates ranging from (1,1,1) to (sizeX,sizeY,sizeZ).
We want to number the unit cubes, using integers from 1 to N. We will do this algorithmically.
We will call a box "1-dimensional (1D)" if at least two of its dimensions are 1, "2-dimensional (2D)" if exactly one of its dimensions is 1, and "3-dimensional (3D)" otherwise.
The algorithm used to number a 1-dimensional box is simple: order the cubes according to the sums of their coordinates (in ascending order), and number them in this order.
To number a 2-dimensional box, we follow this algorithm:
- If X size is greater than 1, consider all cubes with both Y and Z coordinates minimal, number them as a 1D box, and throw them away.
- If we still have a 2D box, if Y size is greater than 1, consider all cubes with both X and Z coordinates minimal, number them as a 1D box, and throw them away.
- If we still have a 2D box, if Z size is greater than 1, consider all cubes with both X and Y coordinates minimal, number them as a 1D box, and throw them away.
- Recursively number the rest of the box.
For example, a 4x5x1 box filled using this algorithm looks as follows:
z=1 x:1 2 3 4 y:+--+--+--+--+ 1| 1| 2| 3| 4| +--+--+--+--+ 2| 5| 9|10|11| +--+--+--+--+ 3| 6|12|15|16| +--+--+--+--+ 4| 7|13|17|19| +--+--+--+--+ 5| 8|14|18|20| +--+--+--+--+
To number a 3-dimensional box, we follow this algorithm:
- Consider all cubes with the Z coordinate minimal, number them as a 2D box, and throw them away.
- If we still have a 3D box, consider all cubes with the Y coordinate minimal, number them as a 2D box, and throw them away.
- If we still have a 3D box, consider all cubes with the X coordinate minimal, number them as a 2D box, and throw them away.
- Recursively number the rest of the box.
For example, a 4x3x3 box filled using this algorithm looks as follows:
z=1 x:1 2 3 4 y:+--+--+--+--+ 1| 1| 2| 3| 4| +--+--+--+--+ 2| 5| 7| 8| 9| +--+--+--+--+ 3| 6|10|11|12| +--+--+--+--+ z=2 x:1 2 3 4 y:+--+--+--+--+ 1|13|14|15|16| +--+--+--+--+ 2|21|25|26|27| +--+--+--+--+ 3|22|28|29|30| +--+--+--+--+ z=3 x:1 2 3 4 y:+--+--+--+--+ 1|17|18|19|20| +--+--+--+--+ 2|23|31|32|33| +--+--+--+--+ 3|24|34|35|36| +--+--+--+--+
You will be given the box dimensions sizeX, sizeY, and sizeZ, and the coordinates of a single cube (cubeX,cubeY,cubeZ). Write a method that will compute the number assigned to the cube at the given coordinates, when using the algorithm described above.
Definition
- Class:
- BoxFilling
- Method:
- getNumber
- Parameters:
- int, int, int, int, int, int
- Returns:
- long
- Method signature:
- long getNumber(int sizeX, int sizeY, int sizeZ, int cubeX, int cubeY, int cubeZ)
- (be sure your method is public)
Notes
- Note that the box described by sizeX, sizeY, and sizeZ is not necessarily a 3D box.
Constraints
- sizeX, sizeY and sizeZ will be between 1 and 109, inclusive.
- The volume of the box will not exceed 1018.
- cubeX will be between 1 and sizeX, inclusive.
- cubeY will be between 1 and sizeY, inclusive.
- cubeZ will be between 1 and sizeZ, inclusive.
Examples
4
5
1
2
4
1
Returns: 13
This is the box from the first example in the problem statement.4
3
3
2
2
2
Returns: 25
This is the box from the second example in the problem statement.4
3
3
4
2
1
Returns: 9
Same box, different cube coordinates.2
2
2
1
1
1
Returns: 1
43633
35453
34533
2344
32442
34221
Returns: 9416237809215
3
1
10
3
1
2
Returns: 14
4
2
4
3
2
3
Returns: 28