Impossible, It would almost completely depends on the size of the pieces, which you don’t have.

Imagine this is a rectangle. You could have two large pieces on each side, making a total of 8 pieces around the perimeter. Then the remaining 100 pieces would be much smaller in the center of the puzzle.

You could reduce it to 4 pieces around the side, or increase to 16 pieces around the side. My principle is still correct.

There is not enough information in the question to be able to make an educated estimate.

The bandit, ^^^ is absolutely right.
The only calculation would be to guess that is would be less than half of the total.
However, there are now round puzzles, so another conundrum.

You can only do that, if the puzzle is a square, then it is easy: n=4√(108)-4.
But that does not work, because the result is not a natural number.
You would have to make a table with all divisors of 108 that are natural numbers.
So the possible solutions are:
1*108, an edge case, with 108 pieces at the edge
2*54, 112 pieces at the edge, BUT because the corners can not count double (because they are themselves areas, not lines), you have to subtract 4, so still 108 pieces at the edge.
The same must be done to all other results.
3*36, 74 pieces
4*27, 58 pieces
6*18, 44 pieces
9*12, 38 pieces

Assuming of course that all pieces are the same size.

This is a meaningless riddle.

I have a jigsaw puzzle shaped like a jaguar (cat, not car), tail and all. No computing that one.

I also have jigsaw puzzles with a wide variety of sizes of pieces. Not all puzzles are made in strips of little squares with innie and outie lobes. That’s just one design.

Anyway, if this is about taunting and irritating, why play along with it? Just ignore it.

I’ll bite…

I’ll say a typical puzzle is in the shape of 3 to 4 ratio. Let’s say the pieces on average are square. so what ratio is in 3 to 4 and multiplies out to 108. 9×12 = 108

2×9 = 18 2×12 = 24 18 + 24 = 36 edge pieces.

By adding one piece at a time.