Funky chess board problem

#1

can anyone suggest me approach to this problem

#2

This problem can be solved using BFS graph traversal.
Lets see how.
you first need to create 2D array for chess board having 0 entry where knight can’t move. and a 2D array for visited cells(we usually do in graph traversal)
we will find how many maximum cells can knight cover. then your answer will be (total cells - max covered cells).
your recursion would be like; let f(x) be no of cells covered starting from position x.
f(r,c) = 1 + max{ f(r-2,c-1), f(r-2,c+1), f(r-1,c-2), f(r-1,c+2), f(r+1,c-2), f(r+1,c+2), f(r+2,c-1), f(r+2,c+1) } ;
and f(r,c) = 0 if r,c does not exists (i,e it contains 0 in input matrix ) or it is already visited.
here starting r,c will be top left corner.

you need to carefully update visited matrix(it should be local to path).
Thanks

#3

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