Yea, your solution will work as well. It's just that it will take additional O(n2) space as we are creating a new matrix. The pattern that you mentioned is indeed what the algorithm mentioned by me does. But, since it tries to do in-place, we rotate one number of row/column at a time.
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This is an interesting (and practical) problem.
I have trouble following the solution, though.
Here's the pattern I noticed:
1) Row #1 becomes Column #3
2) Row #2 becomes Column #2
3) Row #3 becomes Column #1
Basically, if the matrix is
nxn, Row #ibecomes Column #n-i+1.But we can make things easier for ourselves by switching up our perspective:
Algorithm:
For
col = 0throughn - 1:Iterate through the rows in reverse:
row = n - 1 through 0and fill the new matrix withoriginal[row][col]Yea, your solution will work as well. It's just that it will take additional O(n2) space as we are creating a new matrix. The pattern that you mentioned is indeed what the algorithm mentioned by me does. But, since it tries to do in-place, we rotate one number of row/column at a time.