1275. Find Winner on a Tic Tac Toe Game

🟩 Easy

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 i^th 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

tic_tac_toe

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

Example 2

tic_tac_toe

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

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

Hint-1

It's straightforward to check if A or B won or not, check for each row/column/diag if all the three are the same.

Hint-2

Then if no one wins, the game is a draw if the board is full, i.e. moves.length = 9 otherwise is pending.

Solution

My Solution

func tictactoe(moves [][]int) string {
    grid := [3][3]int{}
    for i, move := range moves {
        grid[move[0]][move[1]] = i&1 + 1
    }

    for i := 0; i < 3; i++ {
        if player := grid[i][i]; player != 0 {
            if (grid[i][0] == grid[i][1] && grid[i][0] == grid[i][2]) ||
                (grid[0][i] == grid[1][i] && grid[0][i] == grid[2][i]) {
                return string(byte('@' + player))
            }
        }
    }

    if player := grid[1][1]; player != 0 {
        if (grid[0][0] == grid[2][2] && grid[0][0] == player) ||
            (grid[2][0] == grid[0][2] && grid[2][0] == player) {
            return string(byte('@' + player))
        }
    }

    if len(moves) != 9 {
        return "Pending"
    }

    return "Draw"
}

Optimal solution

func tictactoe(moves [][]int) string {
    // Initialize score arrays
    rows, cols := [3]int{}, [3]int{}
    diagonal, antiDiagonal := 0, 0

    // Iterate over moves
    for i, move := range moves {
        player := 1 // Player A
        if i%2 == 1 {
            player = -1 // Player B
        }
        row, col := move[0], move[1]

        // Update scores
        rows[row] += player
        cols[col] += player
        if row == col {
            diagonal += player
        }
        if row+col == 2 {
            antiDiagonal += player
        }

        // Check for a winner
        if rows[row] == 3 || cols[col] == 3 || diagonal == 3 || antiDiagonal == 3 {
            return "A"
        }
        if rows[row] == -3 || cols[col] == -3 || diagonal == -3 || antiDiagonal == -3 {
            return "B"
        }
    }

    // Determine game state
    if len(moves) == 9 {
        return "Draw"
    }
    return "Pending"
}

result

Leetcode: link