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This is a Statue View puzzle, an original invention combining two logic puzzle genres: Statue Park and Canal View.

Rules of Statue View:

• Shade some cells of the grid to form the given set of pieces. Pieces may be rotated or reflected.


• Pieces cannot be adjacent (though they can touch at a corner).


• All unshaded cells must be (orthogonally) connected.


• Any cells with numbers in them must be unshaded. These numbers give the total lengths of the runs of shaded cells starting immediately adjacent to the clue, and extending outwards from the clue.

Some initial observations:

The 4 can have at most two cells shaded to the north, so the two cells east of it must be filled in.


Then, since the unshaded cells (including the numbers) must be connected, the cell north of the 4 is also unshaded, and the four cells to the east of the 4 must be the I-tetromino.

Also,

The 5 can have at most two cells to the north and two to the west, so the cell south of it must be shaded.

Then,

The lone unshaded square at the east end of the "I" doesn't have any way of connecting with the other unshaded cells just by going farther east, so the cell north of it must also be unshaded.

Then,

The shaded cell south of the 5 could be fit as part of the L-, S, or T-tetromino (the O won't fit). However, if the "T" were there, only one cell to the west and one cell to the south of the 5 would be shaded, and at most two can fit to the north, which is not enough.


Similarly, an upright "L" would give two cells connected to the 5 and block the square to its north. So, it must be the "S" or the upside-down "L" in that spot. The "S" would fill in one cell west of the five and two cells south, and the "L" would shade three cells to the south and block the cell west of the 5. A flipped "Z" would shade the two cells between the 3 and the 5, and just one cell south of the 5.


In any of those cases, the 5 would have three shades cells from the shape south of it, and both cells north of the 5 need to be shaded, as well as the cell immediately southwest of the 5.


(I previously misread the rule about reflection, so I'm backing up to this point because it is possible for a Z-tetromino to fill both spaces between the 5 and the 3, or for a J-tetromino to sit below the 2 without shading the cell northwest of the 5).

Now, a slightly reductionist approach.

Let's assume for a moment that the O-tetromino is not in the top corner, south of the 2. That would mean the L/J-tetromino is in that corner, extending three squares to the west. The "O" thus cannot occupy any of the squares north of the central 3, or it would be directly adjacent.
If the "O" were between the two 3s, each 3 would also need exactly one more adjacent cell to be shaded. For the westernmost 3, there is no way to do that without a shape that would extend east and end up next to the "O" as we have placed it. So, that is also not a possibility.
Similarly, placing the "O" in the northwest corner, or one space to the east of that, does not leave enough room for other shapes to fill in the needs of the 3s.
Finally, if the "O" occupied the two cells north of the western 3, the cell to its south would need to be shaded but there is no shape that can do that without ending up adjacent to the "O," blocking off the 3, or both.
Thus, by process of elimination, the "O" must indeed be in the northeast corner, south of the 2.


There are probably "nicer" ways to prove this without so much mental trial-and-error.

From there,

It's obvious that the "L" must be in the southeast corner.

Finally,

With just two pieces left, a visual approach is simple enough (I may come back and fill in logical steps later). This is where the reflection becomes necessary, and the Z- and T-tetrominoes fit to complete the puzzle: