Question

1275. Find Winner on a Tic Tac Toe Game

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Description

Tic-tac-toe is played by two players A and B on a 3 x 3 grid. The rules of Tic-Tac-Toe are:

• Players take turns placing characters into empty squares ' '.
• The first player A always places 'X' characters, while the second player B always places 'O' characters.
• 'X' and 'O' characters are always placed into empty squares, never on filled ones.
• The game ends when there are three of the same (non-empty) character filling any row, column, or diagonal.
• The game also ends if all squares are non-empty.
• No more moves can be played if the game is over.

Given a 2D integer array moves where moves[i] = [rowi, coli] indicates that the ith move will be played on grid[rowi][coli]. return _the winner of the game if it exists_ (A or B). In case the game ends in a draw return "Draw". If there are still movements to play return "Pending".

You can assume that moves is valid (i.e., it follows the rules of Tic-Tac-Toe), the grid is initially empty, and A will play first.

 

Example 1:

Input: moves = [[0,0],[2,0],[1,1],[2,1],[2,2]]
Output: "A"
Explanation: A wins, they always play first.

Example 2:

Input: moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]
Output: "B"
Explanation: B wins.

Example 3:

Input: moves = [[0,0],[1,1],[2,0],[1,0],[1,2],[2,1],[0,1],[0,2],[2,2]]
Output: "Draw"
Explanation: The game ends in a draw since there are no moves to make.

 

Constraints:

• 1 <= moves.length <= 9
• moves[i].length == 2
• 0 <= rowi, coli <= 2
• There are no repeated elements on moves.
• moves follow the rules of tic tac toe.

Solutions

Solution 1: Determine if the last player to move can win

Since all moves are valid, that is, there is no situation where a person continues to play after someone has won. Therefore, we only need to determine whether the last player to move can win.

We use an array cnt of length $8$ to record the number of moves in rows, columns, and diagonals. Where $cnt[0, 1, 2]$ represent the number of moves in the $0, 1, 2$ rows respectively, and $cnt[3, 4, 5]$ represent the number of moves in the $0, 1, 2$ columns respectively. Additionally, $cnt[6]$ and $cnt[7]$ represent the number of moves on the two diagonals respectively. During the game, if a player makes $3$ moves in a row, column, or diagonal, that player wins.

If the last player to move does not win, then we determine whether the board is full. If it is full, it is a draw; otherwise, the game is not over yet.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of moves.

class Solution:
  def tictactoe(self, moves: List[List[int]]) -> str:
    n = len(moves)
    cnt = [0] * 8
    for k in range(n - 1, -1, -2):
      i, j = moves[k]
      cnt[i] += 1
      cnt[j + 3] += 1
      if i == j:
        cnt[6] += 1
      if i + j == 2:
        cnt[7] += 1
      if any(v == 3 for v in cnt):
        return "B" if k & 1 else "A"
    return "Draw" if n == 9 else "Pending"

class Solution {
  public String tictactoe(int[][] moves) {
    int n = moves.length;
    int[] cnt = new int[8];
    for (int k = n - 1; k >= 0; k -= 2) {
      int i = moves[k][0], j = moves[k][1];
      cnt[i]++;
      cnt[j + 3]++;
      if (i == j) {
        cnt[6]++;
      }
      if (i + j == 2) {
        cnt[7]++;
      }
      if (cnt[i] == 3 || cnt[j + 3] == 3 || cnt[6] == 3 || cnt[7] == 3) {
        return k % 2 == 0 ? "A" : "B";
      }
    }
    return n == 9 ? "Draw" : "Pending";
  }
}

class Solution {
public:
  string tictactoe(vector<vector<int>>& moves) {
    int n = moves.size();
    int cnt[8]{};
    for (int k = n - 1; k >= 0; k -= 2) {
      int i = moves[k][0], j = moves[k][1];
      cnt[i]++;
      cnt[j + 3]++;
      if (i == j) {
        cnt[6]++;
      }
      if (i + j == 2) {
        cnt[7]++;
      }
      if (cnt[i] == 3 || cnt[j + 3] == 3 || cnt[6] == 3 || cnt[7] == 3) {
        return k % 2 == 0 ? "A" : "B";
      }
    }
    return n == 9 ? "Draw" : "Pending";
  }
};

func tictactoe(moves [][]int) string {
  n := len(moves)
  cnt := [8]int{}
  for k := n - 1; k >= 0; k -= 2 {
    i, j := moves[k][0], moves[k][1]
    cnt[i]++
    cnt[j+3]++
    if i == j {
      cnt[6]++
    }
    if i+j == 2 {
      cnt[7]++
    }
    if cnt[i] == 3 || cnt[j+3] == 3 || cnt[6] == 3 || cnt[7] == 3 {
      if k%2 == 0 {
        return "A"
      }
      return "B"
    }
  }
  if n == 9 {
    return "Draw"
  }
  return "Pending"
}

function tictactoe(moves: number[][]): string {
  const n = moves.length;
  const cnt = new Array(8).fill(0);
  for (let k = n - 1; k >= 0; k -= 2) {
    const [i, j] = moves[k];
    cnt[i]++;
    cnt[j + 3]++;
    if (i == j) {
      cnt[6]++;
    }
    if (i + j == 2) {
      cnt[7]++;
    }
    if (cnt[i] == 3 || cnt[j + 3] == 3 || cnt[6] == 3 || cnt[7] == 3) {
      return k % 2 == 0 ? 'A' : 'B';
    }
  }
  return n == 9 ? 'Draw' : 'Pending';
}