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# 1275. Find Winner on a Tic Tac Toe Game

### Description

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**.

### 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.

### Approach

### Links

* Binarysearch
* GeeksforGeeks
* [Leetcode](https://leetcode.com/problems/find-winner-on-a-tic-tac-toe-game/)
* ProgramCreek
* YouTube

### **Examples**

{% tabs %}
{% tab title="Example 1" %}
**Input:** moves = \[\[0, 0], \[2, 0], \[1, 1], \[2, 1], \[2, 2]]

**Output:** "A"

**Explanation:** "A" wins, he always plays first.

<div align="left"><img src="/files/-Mk1eoiNeatnWQXvsPRs" alt=""></div>
{% endtab %}

{% tab title="Example 2" %}
**Input:** moves = \[\[0,0],\[1,1],\[0,1],\[0,2],\[1,0],\[2,0]]

**Output:** "B"

**Explanation:** "B" wins.

<div align="left"><img src="/files/-Mk1fGbH-pJJLuD3MP8t" alt=""></div>
{% endtab %}

{% tab title="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"
{% endtab %}

{% tab title="Example 4" %}
**Input:** moves = \[\[0,0],\[1,1]]

**Output:** "Pending"

**Explanation:** The game has not finished yet.

"X "

" O "

" "
{% endtab %}
{% endtabs %}

### **Solutions**

{% tabs %}
{% tab title="Solution 1" %}

```java
/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    
    public String tictactoe(int[][] moves) {
        if(moves.length > 4) {
            int n = 3;
            char[] players = {'A', 'B'};
            char[][] board = new char[n][n];
            for(int i = 0; i < moves.length; i++) {
                int[] move = moves[i];
                board[move[0]][move[1]] = players[i%2];
            }
        
            for(int i = 0; i < n; i++) {
                for(char player: players) {
                    if(rowCheck(board, i, player) || colCheck(board, i, player)) {
                        return player + "";
                    }
                }
            }

            for(char player: players) {
                if(diagonalCheck(board, player)) {
                    return player + "";
                }
            }
        }
        
        return moves.length == 9? "Draw": "Pending";
    }
    
    private boolean rowCheck(char[][] board, int row, char player) {
        return board[row][0] == player && board[row][1] == player && board[row][2] == player;
    }
    
    private boolean colCheck(char[][] board, int col, char player) {
        return board[0][col] == player && board[1][col] == player && board[2][col] == player;
    }
    
    private boolean diagonalCheck(char[][] board, char player) {
        return (board[0][0] == player && board[1][1] == player && board[2][2] == player) || 
            (board[2][0] == player && board[1][1] == player && board[0][2] == player);
    }
}
```

{% endtab %}

{% tab title="Solution 2" %}

```java
/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    private final int n = 3;
    
    public String tictactoe(int[][] moves) {
        if(moves.length > 4) {
            int[] rows = new int[n];
            int[] cols = new int[n];
            int diagonal = 0, antiDiagonal = 0;
            
            int player = 1;
            
            for(int[] move: moves) {
                int x = move[0], y = move[1];
                
                if(x == y) {
                    diagonal += player;
                }
                if(x+y == n-1) {
                    antiDiagonal += player;
                }
                
                rows[x] += player;
                cols[y] += player;
                
                if(Math.abs(rows[x]) == n || Math.abs(cols[y]) == n ||
                   Math.abs(diagonal) == n || Math.abs(antiDiagonal) == n) {
                    return (player == 1)? "A": "B";
                }
                
                player *= -1;
            }
        }
        
        return moves.length == n*n? "Draw": "Pending";
    }
}
```

{% endtab %}
{% endtabs %}

### **Follow up**

*