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1275. Find Winner on a Tic Tac Toe Game

SourceWeekly Contest 165 Q1DifficultyEasyRating1336

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

Thinking

Moves are legal and play stops at a win, so we only test whether the last player already has three in a row. Walking backward every other move, we tally that player's rows, columns, and diagonals; a \(3\) is a win. Otherwise nine moves is a draw, fewer is pending.

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 \(\textit{cnt}\) of length \(8\) to record the number of moves in rows, columns, and diagonals. Where \(\textit{cnt}[0, 1, 2]\) represent the number of moves in the \(0, 1, 2\) rows respectively, and \(\textit{cnt}[3, 4, 5]\) represent the number of moves in the \(0, 1, 2\) columns respectively. Additionally, \(\textit{cnt}[6]\) and \(\textit{cnt}[7]\) 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 \(\textit{moves}\).

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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"

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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";
    }
}

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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";
    }
};

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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"
}

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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';
}

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