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Problem solution · Java

Design Tic-Tac-Toe

Design Tic-Tac-Toe: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
53 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Design Tic-Tac-Toe, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 53 lines of Java from the credited upstream file 348.java.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeDesign Tic-Tac-Toe · JavaJava
Use this to learn the idea, then write your own version.
class TicTacToe {  public TicTacToe(int n) {    this.n = n;    rows = new int[n];    cols = new int[n];  }   /**   * Player {player} makes a move at ({row}, {col}).   *   * @param row    The row of the board.   * @param col    The column of the board.   * @param player The player, can be either 1 or 2.   * @return The current winning condition, can be either:   *         0: No one wins.   *         1: Player 1 wins.   *         2: Player 2 wins.   */  public int move(int row, int col, int player) {    final int toAdd = player == 1 ? 1 : -1;    final int target = player == 1 ? n : -n;     if (row == col) {      diag += toAdd;      if (diag == target)        return player;    }     if (row + col == n - 1) {      antiDiag += toAdd;      if (antiDiag == target)        return player;    }     rows[row] += toAdd;    if (rows[row] == target)      return player;     cols[col] += toAdd;    if (cols[col] == target)      return player;     return 0;  }   private final int n;  // Record count('X') - count('O').  private int[] rows;  private int[] cols;  private int diag = 0;  private int antiDiag = 0;}

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