Solution: I Replicate

The 5 in the orange i can be duplicated into R1C1 (row 1 column 1). Also, we have some easy clues in rows 2 and 6, and columns 2 and 8.

The data in the blue Is can be copied. In addition, the 4-clue to the left of row 8 forces the 10 in that row.

In row 1, the 4-clue on the left forces the 10 of that row to be in the 5th column or farther to the right, since the tile with a 1 in it can’t be visible. Similarly, the 5 clue on the right forces the 10 onto the 5th column or farther to the left, since the tile with a 2 can’t be visible. This forces the 10 into R1C5 exactly. Following this, the top row can be filled out uniquely, as the left side must have 5-6-7-10 visible due to the 8/9 already placed in column 4, and the right side needs to take the remaining numbers in order.

Let’s try to place the remaining 10s–we have three left, one each in rows 3, 5, and 7. In column 3, the 10 can’t be in row 3, as it makes the 5-clue on the top of the column impossible. It can’t be in row 5, as duplicating that square into the corresponding places of the blue Is yields a contradiction. We don’t have quite enough info yet to place the remaining two 10s–we’ll mark the possible squares with "10?".

In column 3, the 5-clue on top requires the buildings 6-7-8-9-10 to be visible. The blue cell in R2C3 can’t be a 7, due to it being duplicated in R7C7 and the 7 already in R7C2. Also, the blue cell in R5C3 can’t be bigger than 5, due to it being duplicated as a clue below column 1 and the 10 in R6C1. Thus, 7, 8, and 9 must fit in the three other cells between the 6 and 10 of that column.

Let’s start looking at what numbers could go in the blue I. We know that the bottom left one must be at most 5, and the bottom right one must be at most 4, due to being clues below columns and the known positions of 10s in those columns. From that, plus eliminating candidates present in the same row or column, we can limit them to the following candidates, written in the cells.

In the bottom right corner, the 2-clue at the bottom of column 9 implies that R10C9 > R9C9. The 3-clue at the right of row 9 implies that R9C9 > R9C10. Thus, R10C9 > R9C9 > R9C10, so R10C9 must be at least 3. Since it is currently constrained to be 2 or 3, we conclude that it is equal to 3, and R9C9 = 2, R9C10 = 1.

In row 7, the 5-clue to the right needs the two squares to the right of the 8 to be visible, in addition to 8-9-10. To get this, we need R7C9 > R7C10, so we can’t have R7C9=1 and conclude that R7C9=5. Additionally, we need the 9 to be in between the 10 and 8 of that row, so it must go in R7C5.

Let’s try to place the remaining 9s. There are three of them remaining in rows 4, 5, and 10, and columns 7, 9, and 10. In row 10, the only allowed column for the 9 is row 10. In row 5, placing a 9 in column 9 (or 10) would contradict the 4-clue on the right , so the 9 must be in column 7. We conclude that the 9s must go in R10C10, R5C7, R4C9.

The remaining 10s can now be placed. In addition, the clue to the right of row 4 has to be a 3 (and can be replicated).

The remaining orange square has to be at most 4 and can’t be 1, due to it being a clue to the right of row 3 and the placement of the 10 in that row. It also can’t be 3 or 4, due to numbers in the same row, so it must be 2.

In column 3, there is exactly one row available for a 1, and exactly one row available for a 3, so we can place both of those. In the same column, the remaining 2-4 pair can be disambiguated by the 3-clue below the column. We can fill in the whole column, and all the remaining blue squares.

In row 5, the square in the first column has to be a 4 to satisfy the 3-clue to the left. The remaining 7-8 pair can only be filled in one way to satisfy the 4-clue to the right. Now looking at column 1, we have 6, 7, and 8 remaining. To satisfy the 2-clue on the bottom, 8 must go in the bottom. Furthermore, the cell in R7C1 must be 6, so we can fill out the entire column.

Column 9 can now be filled in uniquely due to latin square rules. This puts a 6 in R3C9, also forcing an 8 in R3C10 to satisfy the 2-clue to the right of row 3. After this, the rest of row 3 can be filled out.

From here, the skyscraper clues have all been used, and this can be solved as a straight latin square (plus the constraint from the purple cells). For an example of a next step, the remaining 5s can be uniquely placed, first in row 10, then in row 9, then in row 6.

Let’s fill out candidates for row 2, row 10, and the purple cells (being sure to use the duplicate constraint).

In column 4, 6 can only go in row 9. Propagating this to R8C6, we can immediately fill in some squares.

From here, we can fill out the puzzle as a Latin square in a few steps.