코딩테스트 / 혼자서 하는 틱택토

https://school.programmers.co.kr/learn/courses/30/lessons/160585

이런저런 케이스를 다 생각해서 풀어야 하는 문제... 

1. 선공인 O와 후공인 X의 개수가 1 이상 나면 안 된다. 
2. O와 X가 동시에 이겨서는 안 된다. 
3. O가 이겼을 때 다음 X는 없다. 그러므로 O가 X보다 1개 많아야.. 
4. X가 이겼을 때 다음 O는 없다. 그러므로 O와 X의 개수가 같아야.. 

코틀린

class Solution {
    private fun countMarker(board: Array<String>): Pair<Int, Int> {
        var (oCounter, xCounter) = arrayOf(0, 0)
        for (line in board)
            for (each in line)
                if (each == 'O')
                    oCounter++
                else if (each == 'X')
                    xCounter++
        return Pair(oCounter, xCounter)
    }

    private fun countWin(board: Array<String>): Pair<Int, Int> {
        var (oWin, xWin) = arrayOf(0, 0)
        for (i in 0..2) {
            var (oRow, xRow, oCol, xCol) = arrayOf(0, 0, 0, 0)
            for (j in 0..2) {
                when (board[i][j]) {
                    'O' -> oRow++
                    'X' -> xRow++
                }
                when (board[j][i]) {
                    'O' -> oCol++
                    'X' -> xCol++
                }
            }
            oWin += (if (oRow == 3) 1 else 0) + if (oCol == 3) 1 else 0
            xWin += (if (xRow == 3) 1 else 0) + if (xCol == 3) 1 else 0
        }
        var (oCnt1, xCnt1, oCnt2, xCnt2) = arrayOf(0, 0, 0, 0)
        for (i in 0..2) {
            oCnt1 += if (board[i][i] == 'O') 1 else 0
            xCnt1 += if (board[i][i] == 'X') 1 else 0
            oCnt2 += if (board[i][2 - i] == 'O') 1 else 0
            xCnt2 += if (board[i][2 - i] == 'X') 1 else 0
        }
        oWin += (if (oCnt1 == 3) 1 else 0) + if (oCnt2 == 3) 1 else 0
        xWin += (if (xCnt1 == 3) 1 else 0) + if (xCnt2 == 3) 1 else 0
        return Pair(oWin, xWin)
    }

    fun solution(board: Array<String>): Int {
        val (oCounter, xCounter) = countMarker(board)
        if (oCounter - xCounter !in 0..1)
            return 0
        val (oWin, xWin) = countWin(board)
        if ((oWin > 0 && xWin > 0) || (oWin > 0 && oCounter != xCounter + 1) || (xWin > 0 && oCounter != xCounter))
            return 0
        return 1
    }
}

class Solution {
    private fun countWin(board: Array<String>, marker: Char): Int {
        var win = board.map { row -> row.count { it == marker } }.count { it == 3 }
        win += (0..2).map { i -> ((0..2).map { j -> board[j][i] }) }.map { col -> col.count { it == marker } }.count { it == 3 }
        win += if ((0..2).count { board[it][it] == marker } == 3) 1 else 0
        win += if ((0..2).count { board[it][2 - it] == marker } == 3) 1 else 0
        return win
    }

    fun solution(board: Array<String>): Int {
        val countMarker = { array: Array<String>, marker: Char -> array.sumOf { row -> row.count { it == marker } } }
        val (oCounter, xCounter) = Pair(countMarker(board, 'O'), countMarker(board, 'X'))
        val (oWin, xWin) = Pair(countWin(board, 'O'), countWin(board, 'X'))
        return if ((oCounter - xCounter !in 0..1) || (oWin > 0 && xWin > 0) || (oWin > 0 && oCounter != xCounter + 1) || (xWin > 0 && oCounter != xCounter)) 0 else 1
    }
}

 

파이썬

리스트 컴프리헨션에서 이중포문을 쓰는 건 좋지 않다. (하지만 썼다.)

def count_win(board):
    win_o = win_x = 0
    for i in range(3):
        row_o = row_x = col_o = col_x = 0
        for j in range(3):
            if board[i][j] == 'O':
                row_o += 1
            elif board[i][j] == 'X':
                row_x += 1
            if board[j][i] == 'O':
                col_o += 1
            elif board[j][i] == 'X':
                col_x += 1
        win_o += (1 if row_o == 3 else 0) + (1 if col_o == 3 else 0)
        win_x += (1 if row_x == 3 else 0) + (1 if col_x == 3 else 0)
    cnt_o_1 = cnt_x_1 = cnt_o_2 = cnt_x_2 = 0
    for i in range(3):
        cnt_o_1 += 1 if board[i][i] == 'O' else 0
        cnt_x_1 += 1 if board[i][i] == 'X' else 0
        cnt_o_2 += 1 if board[i][2 - i] == 'O' else 0
        cnt_x_2 += 1 if board[i][2 - i] == 'X' else 0
    win_o += (1 if cnt_o_1 == 3 else 0) + (1 if cnt_o_2 == 3 else 0)
    win_x += (1 if cnt_x_1 == 3 else 0) + (1 if cnt_x_2 == 3 else 0)
    return win_o, win_x


def solution(board):
    count_o, count_x = tuple(map(sum, zip(*((each == 'O', each == 'X') for line in board for each in line))))
    if count_o - count_x not in (0, 1):
        return 0
    win_o, win_x = count_win(board)
    if (win_o > 0 and win_x > 0) or (win_o > 0 and count_o != count_x + 1) or (win_x > 0 and count_o != count_x):
        return 0
    return 1

 

(O(n)을 고려하여) 한번에 'O'와 'X'를 처리하니 코드가 지저분하다. O(n)을 포기했다.
(과연 이래도 좋은가 애매하다. 읽기 좋은 코드가 선호되는 추세이긴 하다.)

def count_win(board, marker):
    win = cnt1 = cnt2 = 0
    for i in range(3):
        row = col = 0
        for j in range(3):
            if board[i][j] == marker:
                row += 1
            if board[j][i] == marker:
                col += 1
        win += (1 if row == 3 else 0) + (1 if col == 3 else 0)
    for i in range(3):
        cnt1 += 1 if board[i][i] == marker else 0
        cnt2 += 1 if board[i][2 - i] == marker else 0
    win += (1 if cnt1 == 3 else 0) + (1 if cnt2 == 3 else 0)
    return win


def solution(board):
    count_o, count_x = tuple(map(sum, zip(*((each == 'O', each == 'X') for line in board for each in line))))
    win_o, win_x = count_win(board, 'O'), count_win(board, 'X')
    if count_o - count_x not in (0, 1) or (win_o > 0 and win_x > 0) \
            or (win_o > 0 and count_o != count_x + 1) or (win_x > 0 and count_o != count_x):
        return 0
    return 1

 

깔끔하게.. 

def count_win(board, marker):
    win = sum(all(each == marker for each in row) for row in board)
    win += sum(all(each == marker for each in col) for col in zip(*board))
    win += all(board[i][i] == marker for i in range(3))
    win += all(board[i][2 - i] == marker for i in range(3))
    return win


def solution(board):
    count_o, count_x = tuple(map(sum, zip(*((each == 'O', each == 'X') for line in board for each in line))))
    win_o, win_x = count_win(board, 'O'), count_win(board, 'X')
    if count_o - count_x not in (0, 1) or (win_o > 0 and win_x > 0) \
            or (win_o > 0 and count_o != count_x + 1) or (win_x > 0 and count_o != count_x):
        return 0
    return 1

*board에 주의하자. 

한 줄 더 줄여보자.. 

def count_win(board, marker):
    win = sum(all(each == marker for each in row) for row in board)
    win += sum(all(each == marker for each in col) for col in zip(*board))
    win += sum(map(all, zip(*((board[i][i] == marker, board[i][2 - i] == marker) for i in range(3)))))
    return win


def solution(board):
    count_o, count_x = tuple(map(sum, zip(*((each == 'O', each == 'X') for line in board for each in line))))
    win_o, win_x = count_win(board, 'O'), count_win(board, 'X')
    if count_o - count_x not in (0, 1) or (win_o > 0 and win_x > 0) \
            or (win_o > 0 and count_o != count_x + 1) or (win_x > 0 and count_o != count_x):
        return 0
    return 1

그만하자.