Unable to come up with a logic for this proble using recursions,
kindly dry run it for me and explain the logic
Tiling problem 2
Hello @mkidwai74 In this problem you have figured the base cases correct that if n<m then return 1 and if n==m return 2. The recurrsion relation will be that if n>m then add the two recursions . One with same function but n as n-1 and the other will be same function but with n= n-m . Now this will the two cases one in which tile are placed vertically will be where n=n-1 and the one where tiles are placed horizontally will be where n=n-m . And also i will recommend that rather than forming a recurrence relation try to take a array of size n+1 and fill it from index i= 1 where size of n =1 till index i= n and where each index has the total possible way of that case on that index i.
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:
n>m - here you can place a tile vertically(left with n-m,m) as well as horizontally(left with n-1,m).
n==m then there are always 2 ways - horizontally or vertically.
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.
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