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 <= 9moves[i].length == 20 <= moves[i][j] <= 2- There are no repeated elements on
moves. movesfollow 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"