Interview Question

Software Engineer In Test Interview Kirkland, WA

How would you determine if someone has won a game of

  tic-tac-toe on a board of any size?
brain teaser

Interview Answer

13 Answers


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 Apr 29, 2009

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

Anonymous on Apr 30, 2009

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 May 5, 2009

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 May 29, 2009

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 Jun 3, 2009

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 Sep 7, 2009

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')
                            if (i == n-1 && columnSums[j] == n) return true
                            else if (i == j)
                                     if (i == n-1 && diagonalSums[0] == n) return true
                            else if (i = n-1-j)
                                     if (j == 0 && diagonalSums[i] == n) return true
            if (rowSum == n) return true

ellemeno on Nov 15, 2009

(sorry - forgot last line, put this after if (rowSum == n)....)

            else rowSum = 0 // Need to reset the row counter

ellemeno on Nov 15, 2009


You could just ask. on Jun 3, 2010

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

count on Dec 9, 2010

there can be more than one winner

veeru on Dec 17, 2011

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

D on Mar 4, 2012

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 Jan 22, 2015

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