“We can just do bruteforce by putting ‘X’ in each of the empty place of grid and check whether the grid has KK consecutive ‘X’'s or not.”
I think, this will not work for a test case like-
1
3 2
. . .
. . .
. . .
Here the actual answer is ‘NO’ but using this algorithm it will give ‘YES’.
My code is here-
https://www.codechef.com/viewsolution/9499155
@debjitdj, I think the editorialist meant, putting X at given position and checking if a solution exists, then removing the X placed for next iteration. Putting it simply, as the chef can play only one move, putting X at all possible locations is not what the editorialist meant to say.
The links to the Problem in the Contest section and Practice section are incorrect. Please fix them.
Could someone point out what’s wrong in my solution here or give a set of n=k=3 tests?
https://www.codechef.com/viewsolution/9503691
Thanks
@likecs I have a doubt. Is this a valid test case?-
1
4 3
. X O .
X O O X
O X X O
X O X O
Because in this case that logic will also no work.
Can anyone help me find out test cases for 3rd subtask for which my solln. does not work?
https://www.codechef.com/viewsolution/9502837
how can we access the minor diagonal elements?
Can anyone, give some hints or algorithm to be used to solve it in O(n^2)?
the question was poorly expressed. I thought that we had to check all possible diagonals for win
You can add only one X.
And what about this test case?
1
4 3
. X O .
X O O X
O X X O
X O X O
In that case also, it will work. You can put X in two places and check for those cases. Answer will be NO.
But the answer should be ‘YES’…because there are already 3 consecutive ‘X’.
That input is invalid according to problem. The board cannot already have some winner.
Invalid, read the last line of the problem saying that it will be always chef’s chance and the board will not have any winner already.
Oh! I missed the last line. Thankz 
Sorry. I missed the last line.
Oh, I didn’t see that. Yes it’s invalid input.
