## Interview Question

Software Engineer In Test Interview

-Kirkland, WA

Google## How would you determine if someone has won a game of tic-tac-toe on a board of any size?

Answer

## Interview Answers

15 Answers

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I think maybe this question is worded a bit wrong, because given a tic-tac-toe board you would need to read in at least some of the values on the board to figure out if someone has won, and this would be impossible to do in constant time (the larger the board, the more values you would have to read). I think they must mean how can you determine if someone has won during a game in real time, as in checking after every move. This can be solved with a strategy in constant time. My solution would be: Create an array of size 2n+2 at the beginning of the game and fill it with zeros. Each spot in the array will be a sum of X's or O's horizontally (the first n places in the array), vertically (the second n places in the array) and diagonally (the last 2 places). Then with every move, you add 1 to the 2 places (or 3 if on a diagnol) of the array if X, and subtract 1 if its an O. After adding you check and see if the value of the array is equal to n or -n, if it is, n mean X has won and -n means O has won. I would bet there is a more elegant solution than creating a large array, but since this isn't my job interview I can't be bothered trying to figure one out. :)

woctaog on

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Happier player doesn't always mean the winner. A father teaching his son how to play tic tac toe for instance could be happier if his son actually beat him at the game. Your "simplest answer" is wrong.

Anonymous on

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Assume that you are handed a board with no prior knowledge of what has happened in the game. Assume that, to win on a board of size NxN, the player must have N 'X' characters or 'O' characters in the same row, column, or diagonal. Assume that, for our problem, we are only checking if the winner is 'X'. We have to make at least one pass through the game board, but we should be able to solve the problem in one pass without checking any cell twice. Target running time O(N^2) for a board of size NxN. boolean checkXWinner(int[][] a, int n){ int[] diagonalSums = new int[2]; int[] columnSums = new int[n]; initialize diagonalSums and columnSums with zeroes; int rowSum = 0; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) if (a[i][j] = 'X') rowSum++; columnSums[j]++; if (i == n-1 && columnSums[j] == n) return true else if (i == j) diagonalSums[0]++; if (i == n-1 && diagonalSums[0] == n) return true else if (i = n-1-j) diagonalSums[1]++; if (j == 0 && diagonalSums[i] == n) return true if (rowSum == n) return true

ellemeno on

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I came across this question on a Google search for something else related to tic-tac-toe. I was just moments ago thinking of this exact problem (fastest way to determine if game is in "won" state) and I'm surprised I have not seen it here... Why check every square??? Assuming an NxN square board, ANY winning arrangement MUST include a square on the diagonal. Iterate over the diagonals, and recurse both vertically and horizontally for matches, breaking when a non-match is found. On the center square check both diagonals in addition to the vertical and horizontal. The only scenario that requires all squares to be checked is if one edge is a winning edge and even then there's a few constraints required to force it.

Cam on

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@count : The game can be tied even though one has greater count.

D on

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iterate the board and every time you find a players symbol peek forward if the board contains other two symbols at correct indices (there are four feasible patterns). this is in constant time.

Anonymous on

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Constant Time Solution Keep a counter for every row and column, plus 2 counters for each diagonal. When X makes a move, increment the counter for that row, column, the left-right horizontal (if the row == col), and the right-left horizontal (if the row == 2 - col). When Y makes a move, decrement the corresponding columns. If a counter reaches 3, X has won. If a counter reaches -3, Y has won. 0 | 0 | 0 0 <- right-left horizontal counter | 0 | 0 | 0 0 <- left-right horizontal counter

Alex on

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there can be more than one winner

veeru on

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(sorry - forgot last line, put this after if (rowSum == n)....) else rowSum = 0 // Need to reset the row counter

ellemeno on

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Ask.

You could just ask. on

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There's a way to do this in constant time...

Anonymous on

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Count the number of times X and O appear on the board. Whichever has the greater count is the winner.

count on

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Check all rows, check all columns, check two diagonals. If there exists a diagonal, row or column of 'X' or 'O' someone has won the game. For a board of size N^2 runtime is N.

B on

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You would determine the winner by identifying the first player to string together three consecutive X's or O's. The size of the board is irrelevant.

Anonymous on

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You are looking for the winner. The easy and obvious answer is to check the happier player, not the board. While this may ignore the "engineering" side of the question, it does demonstrate the logic of searching for the simplest answer becuase that solution will be the same regardless of the board configuration.

AtlCreditGuru on

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