LeetCode in Kotlin

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:

Input: board = [“O “,” “,” “]

Output: false

Explanation: The first player always plays “X”.

Example 2:

Input: board = [“XOX”,” X “,” “]

Output: false

Explanation: Players take turns making moves.

Example 3:

Input: board = [“XOX”,”O O”,”XOX”]

Output: true

Constraints:

• board.length == 3
• board[i].length == 3
• board[i][j] is either 'X', 'O', or ' '.

Solution

import kotlin.math.abs

class Solution {
    fun validTicTacToe(board: Array<String>): Boolean {
        // X=1,O=-1,’ ’=0
        var sum = 0
        val winsCol = IntArray(3)
        val winsDig = IntArray(2)
        var xWin = false
        var oWin = false
        for (i in 0..2) {
            val str = board[i]
            var rowSum = 0
            for (j in 0..2) {
                // char chr=str.toCharArray()[j];
                var 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 (abs(i - j) != 1) {
                    if (i == j && i == 1) {
                        winsDig[0] += intchr
                        winsDig[1] += intchr
                    } else if (i == j) {
                        winsDig[0] += intchr
                    } else {
                        winsDig[1] += intchr
                    }
                }
                if (i == 2 && winsDig[0].coerceAtLeast(winsDig[1]) == 3) {
                    xWin = true
                }
                if (i == 2 && winsDig[0].coerceAtMost(winsDig[1]) == -3) {
                    oWin = true
                }
            }
            if (rowSum == 3) {
                xWin = true
            }
            if (rowSum == -3) {
                oWin = true
            }
            sum += rowSum
        }
        return if (sum == 0 && !xWin) {
            true
        } else sum == 1 && !oWin
    }
}

This site is open source. Improve this page.