Find Winner on a Tic Tac Toe Game - Array - Easy - LeetCode
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Find Winner on a Tic Tac Toe Game - Array - Easy - LeetCode

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  • 1Tic-tac-toe is played on a 3x3 grid by two players, A and B, who take turns placing 'X' and 'O'.
  • 2The game ends when a player has three of their characters in a row, column, or diagonal, or if all squares are filled.
  • 3The function returns the winner, 'Draw', or 'Pending' based on the game's state after processing the moves.

AI-generated summary · May not capture all nuances

Key Insight
AskGif

"Tic-tac-toe is played on a 3x3 grid by two players, A and B, who take turns placing 'X' and 'O'."

Find Winner on a Tic Tac Toe Game - Array - Easy - LeetCode

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.

Solution:

using System;
using System.Collections.Generic;
using System.Text;

namespace LeetCode.AskGif.Easy.Array
{
 public class TictactoeSoln
 {
 public string Tictactoe(int[][] moves)
 {
 var matrix = new char[3, 3];

 var playerA = true;
 var input = 'X';
 for (int i = 0; i < moves.Length; i++)
 {
 if (!playerA)
 {
 input = 'O'; 
 }
 else
 {
 input = 'X';
 }
 playerA = !playerA;

 //considering inputs are valid
 matrix[moves[i][0], moves[i][1]] = input;
 var check = ValidateMatrix(matrix);
 switch (check)
 {
 case "O":
 return "B"; 
 case "X":
 return "A";
 case "Draw":
 return "Draw";
 case "Pending":
 break;
 }
 }

 return "Pending";
 }

 private string ValidateMatrix(char[,] matrix)
 {
 char defaultValue = '\0';
 //First row
 if(matrix[0, 0] != defaultValue && matrix[0, 0] == matrix[0, 1] && matrix[0, 0] == matrix[0, 2])
 {
 return matrix[0, 0].ToString();
 }

 //Second row
 if (matrix[1, 0] != defaultValue && matrix[1, 0] == matrix[1, 1] && matrix[1, 0] == matrix[1, 2])
 {
 return matrix[1, 0].ToString();
 }

 //Third row
 if (matrix[2, 0] != defaultValue && matrix[2, 0] == matrix[2, 1] && matrix[2, 0] == matrix[2, 2])
 {
 return matrix[2, 0].ToString();
 }

 //First Column
 if (matrix[0, 0] != defaultValue && matrix[0, 0] == matrix[1, 0] && matrix[0, 0] == matrix[2, 0])
 {
 return matrix[0, 0].ToString();
 }

 //Second Column
 if (matrix[0, 1] != defaultValue && matrix[0, 1] == matrix[1, 1] && matrix[0, 1] == matrix[2, 1])
 {
 return matrix[0, 1].ToString();
 }

 //Third Column
 if (matrix[0, 2] != defaultValue && matrix[0, 2] == matrix[1, 2] && matrix[0, 2] == matrix[2, 2])
 {
 return matrix[0, 2].ToString();
 }

 //Diagonal
 if (matrix[0, 0] != defaultValue && matrix[0, 0] == matrix[1, 1] && matrix[0, 0] == matrix[2, 2])
 {
 return matrix[0, 0].ToString();
 }

 //Diagonal
 if (matrix[0, 2] != defaultValue && matrix[0, 2] == matrix[1, 1] && matrix[0, 2] == matrix[2, 0])
 {
 return matrix[0, 2].ToString();
 }

 //Check if any block is available to play
 for (int i = 0; i < 3; i++)
 {
 for (int j = 0; j < 3; j++)
 {
 if(matrix[i,j] == defaultValue)
 {
 return "Pending";
 }
 }
 }

 return "Draw";
 }
 }
}

Time Complexity: O(n)

Space Complexity: O(n) To Create Matrix

Unit Tests:

using LeetCode.AskGif.Easy.Array;
using Microsoft.VisualStudio.TestTools.UnitTesting;
using System;
using System.Collections.Generic;
using System.Text;

namespace CodingUnitTest.Easy.Array
{
 [TestClass]
 public class TictactoeSolnTests
 {
 [TestMethod]
 public void TictactoeSoln_First()
 {
 var moves = new int[,] {
 { 0, 0 },
 { 2, 0 },
 { 1, 1 },
 { 2, 1 },
 { 2, 2 }
 };

 var output = "A";
 var res = new TictactoeSoln().Tictactoe(ArrayMapper(moves));

 Assert.AreEqual(res, output);
 }

 [TestMethod]
 public void TictactoeSoln_Second()
 {
 var moves = new int[,] {
 { 0, 0 },
 { 1, 1 },
 { 0, 1 },
 { 0, 2 },
 { 1, 0 },
 { 2, 0 }
 };

 var output = "B";
 var res = new TictactoeSoln().Tictactoe(ArrayMapper(moves));

 Assert.AreEqual(res, output);
 }

 [TestMethod]
 public void TictactoeSoln_Third()
 {
 var moves = new int[,] {
 { 0, 0 },
 { 1, 1 },
 { 2, 0 },
 { 1, 0 },
 { 1, 2 },
 { 2, 1 },
 { 0, 1 },
 { 0, 2 },
 { 2, 2 }
 };

 var output = "Draw";
 var res = new TictactoeSoln().Tictactoe(ArrayMapper(moves));

 Assert.AreEqual(res, output);
 }

 [TestMethod]
 public void TictactoeSoln_Fourth()
 {
 var moves = new int[,] {
 { 0, 0 },
 { 1, 1 }
 };

 var output = "Pending";
 var res = new TictactoeSoln().Tictactoe(ArrayMapper(moves));

 Assert.AreEqual(res, output);
 }

 private int[][] ArrayMapper(int[,] matrix)
 {
 var arr = new int[matrix.GetLength(0)][];
 for (int i = 0; i < matrix.GetLength(0); i++)
 {
 arr[i] = new int[matrix.GetLength(1)];
 for (int j = 0; j < matrix.GetLength(1); j++)
 {
 arr[i][j] = matrix[i, j];
 }
 }

 return arr;
 }
 }
}

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AskGif

Published on 10 June 2020 · 5 min read · 962 words

Part of AskGif Blog · coding

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