How do you solve Find Winner on a Tic Tac Toe Game on LeetCode?

Find Winner on a Tic Tac Toe Game (LeetCode #1275) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding how to track game states efficiently can significantly reduce time complexity.

What is the brute force solution for Find Winner on a Tic Tac Toe Game?

The brute force approach for Find Winner on a Tic Tac Toe Game is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible winning condition after each move. It verifies if either player has filled a row, column, or diagonal with their marks. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Winner on a Tic Tac Toe Game?

Find Winner on a Tic Tac Toe Game uses the following concepts: Array, Hash Table, Matrix, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find Winner on a Tic Tac Toe Game?

Always clarify the rules and constraints before diving into coding. Think about edge cases, such as the last move leading to a win or a draw. Practice explaining your thought process as you code, as this helps in interviews.

What are common mistakes when solving Find Winner on a Tic Tac Toe Game?

Not checking all winning conditions after every move. Confusing the player turns and not updating the board correctly.

#1275

Find Winner on a Tic Tac Toe Game

Easy

ArrayHash TableMatrixSimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution uses a more efficient approach by tracking the counts of marks in rows, columns, and diagonals. This allows us to determine the winner without checking the entire board after each move.

⚙️

Algorithm

5 steps

1Step 1: Initialize counters for rows, columns, and two diagonals.
2Step 2: Iterate through the moves and update the respective counters based on the player's turn.
3Step 3: After each move, check if any row, column, or diagonal has reached 3 for either player.
4Step 4: If a player wins, return their character ('A' or 'B').
5Step 5: If all moves are played and no winner, return 'Draw'. If moves remain, return 'Pending'.

solution.py14 lines

1# Full working Python code
2class Solution:
3    def tictactoe(self, moves):
4        rows, cols = [0]*3, [0]*3
5        diag1, diag2 = 0, 0
6        for i, move in enumerate(moves):
7            player = 1 if i % 2 == 0 else -1
8            rows[move[0]] += player
9            cols[move[1]] += player
10            if move[0] == move[1]: diag1 += player
11            if move[0] + move[1] == 2: diag2 += player
12            if abs(rows[move[0]]) == 3 or abs(cols[move[1]]) == 3 or abs(diag1) == 3 or abs(diag2) == 3:
13                return 'A' if player == 1 else 'B'
14        return 'Draw' if len(moves) == 9 else 'Pending'

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the moves once, updating counters and checking for a winner in constant time.

1Understanding how to track game states efficiently can significantly reduce time complexity.
2Recognizing patterns in winning conditions can help optimize checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.