Formatted question description: https://leetcode.ca/all/1275.html

# 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.

Here are the rules of Tic-Tac-Toe:

• 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 3 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 an array moves where each element is another array of size 2 corresponding to the row and column of the grid where they mark their respective character in the order in which A and B play.

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 (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, he always plays first.
"X  "    "X  "    "X  "    "X  "    "X  "
"   " -> "   " -> " X " -> " X " -> " X "
"   "    "O  "    "O  "    "OO "    "OOX"


Example 2:

Input: moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]
Output: "B"
Explanation: "B" wins.
"X  "    "X  "    "XX "    "XXO"    "XXO"    "XXO"
"   " -> " O " -> " O " -> " O " -> "XO " -> "XO "
"   "    "   "    "   "    "   "    "   "    "O  "


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.
"XXO"
"OOX"
"XOX"


Example 4:

Input: moves = [[0,0],[1,1]]
Output: "Pending"
Explanation: The game has not finished yet.
"X  "
" O "
"   "


Constraints:

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

Related Topics:
Array

## Solution 1.

// OJ: https://leetcode.com/problems/find-winner-on-a-tic-tac-toe-game/
// Time: O(1)
// Space: O(1)
class Solution {
string check(int A[3][3]) {
for (int i = 0; i < 3; ++i) {
int j = 1;
while (j < 3 && A[i][0] != 0 && A[i][j] == A[i][0]) ++j;
if (j == 3) return A[i][0] == 1 ? "A" : "B";
j = 1;
while (j < 3 && A[0][i] != 0 && A[j][i] == A[0][i]) ++j;
if (j == 3) return A[0][i] == 1 ? "A" : "B";
}
int i = 1;
while (i < 3 && A[0][0] != 0 && A[i][i] == A[0][0]) ++i;
if (i == 3) return A[0][0] == 1 ? "A" : "B";
i = 1;
while (i < 3 && A[0][2] != 0 && A[i][2 - i] == A[0][2]) ++i;
if (i == 3) return A[0][2] == 1 ? "A" : "B";
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 3; ++j) {
if (A[i][j] == 0) return "Pending";
}
}
return "Draw";
}
public:
string tictactoe(vector<vector<int>>& moves) {
int A[3][3] = {}, ch = 1;
for (auto &m : moves) {
A[m[0]][m[1]] = ch;
ch = ch == 1 ? 2 : 1;
}
return check(A);
}
};


## Solution 2.

// OJ: https://leetcode.com/problems/find-winner-on-a-tic-tac-toe-game/
// Time: O(1)
// Space: O(1)
class Solution {
public:
string tictactoe(vector<vector<int>>& moves) {
int A[8] = {}, B[8] = {};
for (int i = 0; i < moves.size(); ++i) {
int r = moves[i][0], c = moves[i][1];
if (i % 2 == 0 && (++A[r] == 3 || ++A[3 + c] == 3 || (r == c && ++A[6] == 3) || (r == 2 - c && ++A[7] == 3))) return "A";
if (i % 2 && (++B[r] == 3 || ++B[3 + c] == 3 || (r == c && ++B[6] == 3) || (r == 2 - c && ++B[7] == 3))) return "B";
}
return moves.size() == 9 ? "Draw" : "Pending";
}
};


Java

• class Solution {
public String tictactoe(int[][] moves) {
final char[] SIGNALS = {'X', 'O'};
final int SIDE = 3;
char[][] board = new char[SIDE][SIDE];
int[][] countRows = new int[2][SIDE];
int[][] countColumns = new int[2][SIDE];
int[][] countDiagonals = new int[2][2];
int length = moves.length;
for (int i = 0; i < length; i++) {
int player = i % 2;
int[] move = moves[i];
int row = move[0], column = move[1];
board[row][column] = SIGNALS[player];
int countRow = countRows[player][row];
countRow++;
countRows[player][row] = countRow;
if (countRow == SIDE)
return player == 0 ? "A" : "B";
int countColumn = countColumns[player][column];
countColumn++;
countColumns[player][column] = countColumn;
if (countColumn == SIDE)
return player == 0 ? "A" : "B";
if (row == column) {
int countDiagonal = countDiagonals[player][0];
countDiagonal++;
countDiagonals[player][0] = countDiagonal;
if (countDiagonal == SIDE)
return player == 0 ? "A" : "B";
}
if (row + column == SIDE - 1) {
int countDiagonal = countDiagonals[player][1];
countDiagonal++;
countDiagonals[player][1] = countDiagonal;
if (countDiagonal == SIDE)
return player == 0 ? "A" : "B";
}
}
if (length == 9)
return "Draw";
else
return "Pending";
}
}

• // OJ: https://leetcode.com/problems/find-winner-on-a-tic-tac-toe-game/
// Time: O(1)
// Space: O(1)
class Solution {
string check(int A[3][3]) {
for (int i = 0; i < 3; ++i) {
int j = 1;
while (j < 3 && A[i][0] != 0 && A[i][j] == A[i][0]) ++j;
if (j == 3) return A[i][0] == 1 ? "A" : "B";
j = 1;
while (j < 3 && A[0][i] != 0 && A[j][i] == A[0][i]) ++j;
if (j == 3) return A[0][i] == 1 ? "A" : "B";
}
int i = 1;
while (i < 3 && A[0][0] != 0 && A[i][i] == A[0][0]) ++i;
if (i == 3) return A[0][0] == 1 ? "A" : "B";
i = 1;
while (i < 3 && A[0][2] != 0 && A[i][2 - i] == A[0][2]) ++i;
if (i == 3) return A[0][2] == 1 ? "A" : "B";
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 3; ++j) {
if (A[i][j] == 0) return "Pending";
}
}
return "Draw";
}
public:
string tictactoe(vector<vector<int>>& moves) {
int A[3][3] = {}, ch = 1;
for (auto &m : moves) {
A[m[0]][m[1]] = ch;
ch = ch == 1 ? 2 : 1;
}
return check(A);
}
};

• # 1275. Find Winner on a Tic Tac Toe Game
# https://leetcode.com/problems/find-winner-on-a-tic-tac-toe-game/

class Solution:
def tictactoe(self, moves: List[List[int]]) -> str:
grid = [[""]*3 for _ in range(3)]

player = 0
for r,c in moves:
grid[r][c] = "X" if player % 2 == 0 else "O"
player += 1

for r in range(3):
if all(b == "X" for b in grid[r]): return "A"
elif all(b == "O" for b in grid[r]): return "B"

for c in range(3):
if all(grid[i][j] == "X" for i in range(3) for j in range(3) if j == c): return "A"
elif all(grid[i][j] == "O" for i in range(3) for j in range(3) if j == c): return "B"

if all(grid[r][c] == "X" for r in range(3) for c in range(3) if r == c): return "A"
if all(grid[r][c] == "O" for r in range(3) for c in range(3) if r == c): return "B"

if all(grid[r][c] == "X" for r in range(2,-1,-1) for c in range(2,-1,-1) if r + c == 2): return "A"
if all(grid[r][c] == "O" for r in range(2,-1,-1) for c in range(2,-1,-1) if r + c == 2): return "B"

return "Draw" if player == 9 else "Pending"