Any algorithm to fill the gap with the least number of boxes

Question

Let a 3D grid, like a chessboard, have an extra dimension. Now let's say that I have a certain number of cubes in this grid, each cube occupies 1x1x1 cells. Let them say that each of these cubes is an element.

What I would like to do is replace / merge these cubes into larger boxes, occupying any number of cells along the X, Y and Z axes, so that the resulting number of boxes is as small as possible, while preserving the overall appearance. "

This is probably unclear, so I will give a 2D example. Say I have a 2D grid containing several squares occupying 1x1 cells. The letter represents the cells occupied by this element, and each element has a different letter from the others. In the first example, we have 10 different elements, each of which occupies 1x1x1 cells.

+---+---+---+---+---+---+
|   |   |   |   |   |   |
+---+---+---+---+---+---+
|   | A | B | C | D |   |
+---+---+---+---+---+---+
|   | E | F | G | H |   |
+---+---+---+---+---+---+
|   |   | K | L |   |   |
+---+---+---+---+---+---+
|   |   |   |   |   |   |
+---+---+---+---+---+---+

This is my input. Now I can optimize it, i.e. Reduce the number of elements while still occupying the same cells in several possible ways, one of which might be:

+---+---+---+---+---+---+
|   |   |   |   |   |   |
+---+---+---+---+---+---+
|   | A | B | B | C |   |
+---+---+---+---+---+---+
|   | A | B | B | C |   |
+---+---+---+---+---+---+
|   |   | B | B |   |   |
+---+---+---+---+---+---+
|   |   |   |   |   |   |
+---+---+---+---+---+---+

Here, instead of 10 elements, I have only 3 (i, A, B and C). However, it can be optimized even more:

+---+---+---+---+---+---+
|   |   |   |   |   |   |
+---+---+---+---+---+---+
|   | A | A | A | A |   |
+---+---+---+---+---+---+
|   | A | A | A | A |   |
+---+---+---+---+---+---+
|   |   | B | B |   |   |
+---+---+---+---+---+---+
|   |   |   |   |   |   |
+---+---+---+---+---+---+

Here I have only two elements: A and B. It is as optimized as possible.

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Answer 1

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Min       x1 + x2 + x3 + ... //where x1 is 1 if the 1th partition is chosen, 0 otherwise
such that x1 +    + x3 = 1// if 1st and 3rd partition contain 1st item
               x2 + x3 = 1//if 2nd and 3rd partition contain 2nd item and so on.

          x1, x2, x3,... are binary

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Answer 2

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