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Write a function rotate(m) that takes a list representation m of a square matrix as input, and returns the matrix obtained by rotating the original matrix clockwize by 90 degrees. For instance, if we rotate the matrix above, we get

the matrix is n*n

>>> rotate([[1,2,3],[4,5,6],[7,8,9]])
[[7, 4, 1], [8, 5, 2], [9, 6, 3]]

>>> rotate([[1,1,1],[2,2,2],[3,3,3]])
[[3, 2, 1], [3, 2, 1], [3, 2, 1]]

What I have tried:

Python
def rotate(m):
    l1=m[::]
    n=len(m)-1 
    count=n
    for i in range(n+1):
        for j in range(n+1):
            l1[j][count]=l1[i][j]
        count-=1
    return l1
Posted
Updated 12-Aug-19 22:11pm
v2
Comments
Pranav Sharma 13-Aug-19 2:53am    
all the loops have proper indentations, they are not showing up in the code

So ... imagine you were the matrix. What do you think happens to you when you run that code on yourself?

Let's make it easier: you are a one dimensional array, and you want to reverse yourself:
1, 2, 3, 4

When you do this:
Python
for i in range(4)
   me[3-i] = me[i]
What do you get?
is it
4, 3, 2, 1
Or
1, 2, 2, 1

To reverse the array, don't you need to swap the elements?
Python
for i in range(4)
   temp = me[3-i]
   me[3-i] = me[i]
   me[i] = temp
 
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Comments
Richard Deeming 13-Aug-19 11:32am    
Except your final code will reverse the array twice, leaving you with the original array.

Or was than intended as an "exercise for the reader"? :)
OriginalGriff 13-Aug-19 11:45am    
You gotta leave 'em something to do! :laugh:
Your code replace elements of matrix on itself:
Python
l1[j][count]=l1[i][j]

this means that
123 121 121 121
456 456 452 452
789 789 789 781

So first you kill the 3, then you kill the 6, and you kill the 9 with the 1 that was 3 at beginning.
Try something like
Python
l1[j][count]=m[i][j]

at least, it will not kill some elements.
 
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v2

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