Please explain the question more clearly

#1

please explain the problem and its approach

#2

Imagine you have a floor of size n × m and you are given tiles of size 1× m.
Now you can either place the tiles vertically. Where a tile can be either place vertically,meaning it will only occupy a cell of width 1 and a complete height of M

there are 3 possibilities:

  1. n>m - here you can place a tile vertically(left with n-m,m) as well as horizontally(left with n-1,m).
  2. n==m then there are always 2 ways - horizontally or vertically.
  3. n<m here you have only 1 choice,that is to place tile horizontally, so 1 way only.
    Also mere recursion won’t help in passing all test cases, you have to use DP

Tile placed horizontally(one row is covered now recurse for remaining rows):-
countWays(n - 1, m)

Tile placed vertically if and only if :-
if(n - m >= 0) {
countWays(n - m, m)
}
Now where is the issue, try to visualize the case for 4*3.
there is a 4 * 3 floor, then you have a 1 * 3 tile.

Now if you place the tile horizontally then you have 3 * 3 floor left to tile.
So you make a call for say tiling(n - 1) i.e n - 1 rows left to tile.
Now one vertical call is made by you in the code. So your work is to add the horizontal call.