@sunny:
This test:
if(! ( (countx == counto + 1) || (countx == counto) ) )
cout no endl;
prints no if countx counto
and this one
if(o x)
cout no endl;
else
cout yes endl;
prints no if both have won or else
no i didn't mean that
in first test u checking if count of X should be either equal of one more
than that of O
and in last u r checking if both are winning or only only one
but what i meant is if O has already won but no of moves of X are greater
than O the answer should be No but your solution
@sunny: why the answer for the case u mentioned is no.. those are
possible set of moves according to me and hence my program outputs
yes
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as you can see in this case no of moves of X are 4 and that of O are 3
as X starts first, after both players has played 3 moves each, O would have
already won the game so next move of X is invalid
i got your solution AC after adding this condition :)
On Fri, Jun 17, 2011 at 2:48 PM, KK
oops !! :) i'll look into that.. thx
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For
i think we can do this if the last move is given and that we have processed
the previous moves before, so only O(n) time is required if the last move's
column row or diagonal is filled or not
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What do you mean by N term? Is it a size of the matrix N*N?
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For each row, column and both diagonals keep track amount of equal marks.
Update these counts in O(1) when player makes move.
To determine winner, iterater over each column, row and diagonals to check
whether there is N equal marks.
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