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TCO21 Jan 2008
TCO08 Round 1

BoxFilling

Problem statement, definition, constraints, and public examples.

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

  1. 4
    5
    1
    2
    4
    1
    Returns: 13
    This is the box from the first example in the problem statement.
  2. 4
    3
    3
    2
    2
    2
    Returns: 25
    This is the box from the second example in the problem statement.
  3. 4
    3
    3
    4
    2
    1
    Returns: 9
    Same box, different cube coordinates.
  4. 2
    2
    2
    1
    1
    1
    Returns: 1
  5. 43633
    35453
    34533
    2344
    32442
    34221
    Returns: 9416237809215
  6. 3
    1
    10
    3
    1
    2
    Returns: 14
  7. 4
    2
    4
    3
    2
    3
    Returns: 28
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