Create n x 2n "mirror" matrix of this type:
For n = 2
m = [ 1 2 2 1
1 2 2 1 ]
For n = 3
m = [ 1 2 3 3 2 1
1 2 3 3 2 1
1 2 3 3 2 1 ]
This problem was fun. I was able to find a better way to solve beat my own solution. Not the best way though. But lot better.
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Colon operator of two vectors
find the maximum element of the matrix
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Sum of series II
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