Cracking the Coding Interview, 4 Edition - 150 Programming Interview Questions and Solutions

Solutions to Chapter 19 | Moderate
19.2 Design an algorithm to figure out if someone has won in a game of tic-tac-toe.
SOLUTION
pg 89
The first thing to ask your interviewer is whether the hasWon function will be called just
once, or multiple times. If it will be called multiple times, you can get a very fast algorithm
by amortizing the cost (especially if you can design your own data storage system for the
tic-tac-toe board).
Approach #1: If hasWon is called many times
There are only 3^9, or about twenty thousand tic-tac-toe boards. We can thus represent our
tic-tac-toe board as an int, with each digit representing a piece (0 means Empty, 1 means
Red, 2 means Blue). We set up a hashtable or array in advance with all possible boards as
keys, and the values are 0, 1, and 2. Our function then is simply this:
Easy!
int hasWon(int board) {
return winnerHashtable[board];
}
Approach #2: If hasWon is only called once
1 enum Piece { Empty, Red, Blue };
2 enum Check { Row, Column, Diagonal, ReverseDiagonal }
3
4 Piece getIthColor(Piece[][] board, int index, int var, Check check) {
5 if (check == Check.Row) return board[index][var];
6 else if (check == Check.Column) return board[var][index];
7 else if (check == Check.Diagonal) return board[var][var];
8 else if (check == Check.ReverseDiagonal)
9 return board[board.length - 1 - var][var];
10 return Piece.Empty;
11 }
12
13 Piece getWinner(Piece[][] board, int fixed_index, Check check) {
14 Piece color = getIthColor(board, fixed_index, 0, check);
15 if (color == Piece.Empty) return Piece.Empty;
16 for (int var = 1; var < board.length; var++) {
17 if (color != getIthColor(board, fixed_index, var, check)) {
18 return Piece.Empty;
19 }
20 }
21 return color;
22 }
23
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