g0701_0800.s0794_valid_tic_tac_toe_state.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated
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package g0701_0800.s0794_valid_tic_tac_toe_state;

// #Medium #Array #String #2022_03_30_Time_0_ms_(100.00%)_Space_39.6_MB_(91.75%)

/**
 * 794 - Valid Tic-Tac-Toe State\.
 *
 * Medium
 *
 * Given a Tic-Tac-Toe board as a string array `board`, return `true` if and only if it is possible to reach this board position during the course of a valid tic-tac-toe game.
 *
 * The board is a `3 x 3` array that consists of characters `' '`, `'X'`, and `'O'`. The `' '` character represents an empty square.
 *
 * Here are the rules of Tic-Tac-Toe:
 *
 * *   Players take turns placing characters into empty squares `' '`.
 * *   The first player always places `'X'` characters, while the second player always places `'O'` characters.
 * *   `'X'` and `'O'` characters are always placed into empty squares, never filled ones.
 * *   The game ends when there are three 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.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/05/15/tictactoe1-grid.jpg)
 *
 * **Input:** board = ["O "," "," "]
 *
 * **Output:** false
 *
 * **Explanation:** The first player always plays "X". 
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/05/15/tictactoe2-grid.jpg)
 *
 * **Input:** board = ["XOX"," X "," "]
 *
 * **Output:** false
 *
 * **Explanation:** Players take turns making moves. 
 *
 * **Example 3:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/05/15/tictactoe4-grid.jpg)
 *
 * **Input:** board = ["XOX","O O","XOX"]
 *
 * **Output:** true 
 *
 * **Constraints:**
 *
 * *   `board.length == 3`
 * *   `board[i].length == 3`
 * *   `board[i][j]` is either `'X'`, `'O'`, or `' '`.
**/
public class Solution {
    public boolean validTicTacToe(String[] board) {
        // X=1,O=-1,’ ’=0
        int sum = 0;
        int[] winsCol = new int[3];
        int[] winsDiag = new int[2];
        boolean xWin = false;
        boolean oWin = false;
        for (int i = 0; i < 3; i++) {
            String str = board[i];
            int rowSum = 0;
            for (int j = 0; j < 3; j++) {
                // char chr=str.toCharArray()[j];
                int intchr = 0;
                if (str.toCharArray()[j] == 'X') {
                    intchr = 1;
                }
                if (str.toCharArray()[j] == 'O') {
                    intchr = -1;
                }
                rowSum += intchr;
                winsCol[j] += intchr;
                if (i == 2 && winsCol[j] == 3) {
                    xWin = true;
                }
                if (i == 2 && winsCol[j] == -3) {
                    oWin = true;
                }
                if (Math.abs(i - j) != 1) {
                    if (i == j && i == 1) {
                        winsDiag[0] += intchr;
                        winsDiag[1] += intchr;
                    } else if (i == j) {
                        winsDiag[0] += intchr;
                    } else {
                        winsDiag[1] += intchr;
                    }
                }
                if (i == 2 && Math.max(winsDiag[0], winsDiag[1]) == 3) {
                    xWin = true;
                }
                if (i == 2 && Math.min(winsDiag[0], winsDiag[1]) == -3) {
                    oWin = true;
                }
            }
            if (rowSum == 3) {
                xWin = true;
            }
            if (rowSum == -3) {
                oWin = true;
            }
            sum += rowSum;
        }
        if (sum == 0 && !xWin) {
            return true;
        }
        return sum == 1 && !oWin;
    }
}