0961EasyArrayHash TableMatrixSimulation

Who Comes Out Ahead on the Three-by-Three

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Trains the technique from

LeetCode 1275Find Winner on a Tic Tac Toe Game

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

Two players take turns marking cells of a three-by-three board, the first player writing "A" and the second "B". The plays are given in order as moves, each [row, column], with the first entry being the first player's opening mark.

A player wins the moment they hold all three cells of any row, any column, or either of the two long diagonals, and the game stops there.

Return "A" or "B" for the winner, "Draw" when the board fills with no winner, or "Pending" when the game is unfinished.

Examples

Example 1

Input: moves = [[0, 0], [1, 0], [0, 1], [1, 1], [0, 2]]
Output: "A"

The first player takes the whole top row on their third mark, which wins.

Example 2

Input: moves = [[0, 0], [0, 1], [1, 1], [0, 2], [2, 2]]
Output: "A"

The first player's three marks run down the long diagonal from the top left.

Example 3

Input: moves = [[0, 0], [1, 1]]
Output: "Pending"

Only two marks have been made, so the game is still going.

Constraints

1 <= moves.length <= 9
moves[i].length == 2
0 <= moves[i][0] <= 2
0 <= moves[i][1] <= 2
No cell is played twice
The plays follow the rules of the game

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def tictactoe(moves: list[list[int]]) -> str:

Java

public String tictactoe(int[][] moves)

The intended approach

Time O(1): a fixed board and eight lines.Space O(1).

Lay the plays onto a board, remembering that the players alternate and the first entry belongs to `"A"`. Then read off the eight lines that can win: three rows, three columns and the two diagonals. A line held entirely by one player names the winner, and there can only be one such player because the game stops the moment it appears. With no winner, the answer turns on how many plays were made: all nine means the board is full and the game is a draw, and fewer means it is unfinished.

Pythonentry point: tictactoe

def tictactoe(moves):
    board = [["", "", ""], ["", "", ""], ["", "", ""]]
    for i in range(len(moves)):
        r, c = moves[i]
        board[r][c] = "A" if i % 2 == 0 else "B"
    lines = []
    for k in range(3):
        lines.append([board[k][0], board[k][1], board[k][2]])
        lines.append([board[0][k], board[1][k], board[2][k]])
    lines.append([board[0][0], board[1][1], board[2][2]])
    lines.append([board[0][2], board[1][1], board[2][0]])
    for line in lines:
        if line[0] != "" and line[0] == line[1] and line[1] == line[2]:
            return line[0]
    if len(moves) == 9:
        return "Draw"
    return "Pending"

The obvious version

Keep the cells each player holds as two sets and, after every play, check each of the eight winning lines against the set of the player who has just moved. Rebuilds the check per play rather than reading the finished board once.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_tictactoe(moves):
    lines = [
        [(0, 0), (0, 1), (0, 2)],
        [(1, 0), (1, 1), (1, 2)],
        [(2, 0), (2, 1), (2, 2)],
        [(0, 0), (1, 0), (2, 0)],
        [(0, 1), (1, 1), (2, 1)],
        [(0, 2), (1, 2), (2, 2)],
        [(0, 0), (1, 1), (2, 2)],
        [(0, 2), (1, 1), (2, 0)],
    ]
    held = {"A": set(), "B": set()}
    for i in range(len(moves)):
        who = "A" if i % 2 == 0 else "B"
        held[who].add((moves[i][0], moves[i][1]))
        for line in lines:
            full = True
            for cell in line:
                if cell not in held[who]:
                    full = False
                    break
            if full:
                return who
    if len(moves) == 9:
        return "Draw"
    return "Pending"

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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