A lot of solutions don't work with something like n=5 and m=0
Not at all...
For n-by-n matrix, m=1 if n is odd number.
ur problems are really interesting
This problem has some issues because some pairs (n,m) do not have a solution. For instance, it is impossible to find a matrix such that m = n^2 -1 since a rotation pivot does not move. Moreover, there are floor((n^2)/4) cycles of numbers, and when we match the cycle beginning with its end, it creates two matches. It is possible to do some manipulations to have an m greater than n^2 - floor((n^2)/4), but not always.
My code fails for some of the cases I've mentioned. The current leading solution does better (finding some solutions my code can't), yet sometimes it enters into an infinite loop.
i had no idea this was gonna work :p
Add more zeros
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