LeetCode 솔루션

289. Game of Life

Jack 쪽지보내기 메일보내기 자기소개 아이디로 검색 전체게시물 작성
작성일 2022.04.11 18:15

7,736 조회
6 댓글
0 추천
0 비추천
목록 게시글 쓰기

[ LeetCode 시즌 3 ] 2022년 4월 11일 문제입니다.

Medium 289. Game of Life

According to Wikipedia's article: "The Game of Life, also known simply as Life, is a cellular automaton devised by the British mathematician John Horton Conway in 1970."

The board is made up of an m x n grid of cells, where each cell has an initial state: live (represented by a 1) or dead (represented by a 0). Each cell interacts with its eight neighbors (horizontal, vertical, diagonal) using the following four rules (taken from the above Wikipedia article):

Any live cell with fewer than two live neighbors dies as if caused by under-population.
Any live cell with two or three live neighbors lives on to the next generation.
Any live cell with more than three live neighbors dies, as if by over-population.
Any dead cell with exactly three live neighbors becomes a live cell, as if by reproduction.

The next state is created by applying the above rules simultaneously to every cell in the current state, where births and deaths occur simultaneously. Given the current state of the m x n grid board, return the next state.

Example 1:

Input: board = [[0,1,0],[0,0,1],[1,1,1],[0,0,0]]
Output: [[0,0,0],[1,0,1],[0,1,1],[0,1,0]]

Example 2:

Input: board = [[1,1],[1,0]]
Output: [[1,1],[1,1]]

Constraints:

m == board.length
n == board[i].length
1 <= m, n <= 25
board[i][j] is 0 or 1.

Follow up:

Could you solve it in-place? Remember that the board needs to be updated simultaneously: You cannot update some cells first and then use their updated values to update other cells.
In this question, we represent the board using a 2D array. In principle, the board is infinite, which would cause problems when the active area encroaches upon the border of the array (i.e., live cells reach the border). How would you address these problems?

#leetcode, #문제풀이, #시즌3, #medium, #Opendoor, #Dropbox, #Adobe, #Microsoft, #Amazon, #Snapchat, #Arry, #Matrix, #Simulation

링크

https://leetcode.com/problems/game-of-life/ 2627 회 연결

익명
작성일 2022.04.11 22:42

Runtime: 0 ms, faster than 100.00% of Swift online submissions for Game of Life.
Memory Usage: 14.3 MB, less than 12.12% of Swift online submissions for Game of Life.

class Solution {
    func gameOfLife(_ board: inout [[Int]]) {
        guard board.count >= 1 else { return }
        guard board[0].count <= 25 else { return }
        
        let orgBoard = board
        
        func numOfNeighborsInRow(_ y: Int, _ x: Int, hasCurrentCell: Bool = false) -> Int {
            var neighbors = 0
            
            if x >= 1 {
                neighbors += (orgBoard[y][x-1] == 1 ? 1 : 0)
            }
            
            if !currentCell {
                neighbors += (orgBoard[y][x] == 1 ? 1 : 0)
            }
            
            if x+1 < orgBoard[y].count {
                neighbors += (orgBoard[y][x+1] == 1 ? 1 : 0)
            }
            
            return neighbors
        }
        
        for y in 0..<orgBoard.count {
            for x in 0..<orgBoard[y].count {                
                let isCellLive: Bool = orgBoard[y][x] == 1
                
                var liveNeighbors: Int = 0
                
                if y-1 >= 0 { 
                    liveNeighbors += numOfNeighborsInRow(y-1, x)
                }
                
                if y >= 0 {                   
                    liveNeighbors += numOfNeighborsInRow(y, x, hasCurrentCell: true)
                }
                
                if y+1 < orgBoard.count {                   
                    liveNeighbors += numOfNeighborsInRow(y+1, x)
                }

                if isCellLive {
                    if liveNeighbors < 2 {
                        // live 
                        board[y][x] = 0
                    } else if liveNeighbors == 2 || liveNeighbors == 3 {
                        // next generation
                        board[y][x] = 1
                    } else if liveNeighbors > 3 {
                        // die
                        board[y][x] = 0
                    }             
                } else {
                    if liveNeighbors == 3 {
                        // live
                        board[y][x] = 1
                    }
                }
            }
        }
    }
}

Runtime: 0 ms, faster than 100.00% of Swift online submissions for Game of Life.
Memory Usage: 14.3 MB, less than 12.12% of Swift online submissions for Game of Life.

[code=swift]class Solution {
    func gameOfLife(_ board: inout [[Int]]) {
        guard board.count >= 1 else { return }
        guard board[0].count <= 25 else { return }
        
        let orgBoard = board
        
        func numOfNeighborsInRow(_ y: Int, _ x: Int, hasCurrentCell: Bool = false) -> Int {
            var neighbors = 0
            
            if x >= 1 {
                neighbors += (orgBoard[y][x-1] == 1 ? 1 : 0)
            }
            
            if !currentCell {
                neighbors += (orgBoard[y][x] == 1 ? 1 : 0)
            }
            
            if x+1 < orgBoard[y].count {
                neighbors += (orgBoard[y][x+1] == 1 ? 1 : 0)
            }
            
            return neighbors
        }
        
        for y in 0..<orgBoard.count {
            for x in 0..<orgBoard[y].count {                
                let isCellLive: Bool = orgBoard[y][x] == 1
                
                var liveNeighbors: Int = 0
                
                if y-1 >= 0 { 
                    liveNeighbors += numOfNeighborsInRow(y-1, x)
                }
                
                if y >= 0 {                   
                    liveNeighbors += numOfNeighborsInRow(y, x, hasCurrentCell: true)
                }
                
                if y+1 < orgBoard.count {                   
                    liveNeighbors += numOfNeighborsInRow(y+1, x)
                }

if isCellLive {
                    if liveNeighbors < 2 {
                        // live 
                        board[y][x] = 0
                    } else if liveNeighbors == 2 || liveNeighbors == 3 {
                        // next generation
                        board[y][x] = 1
                    } else if liveNeighbors > 3 {
                        // die
                        board[y][x] = 0
                    }             
                } else {
                    if liveNeighbors == 3 {
                        // live
                        board[y][x] = 1
                    }
                }
            }
        }
    }
}[/code]

익명
작성일 2022.04.11 23:29

A C++ constant O(1) space complexity solution
Runtime: 0 ms, faster than 100.00% of C++ online submissions for Game of Life.
Memory Usage: 6.8 MB, less than 84.44% of C++ online submissions for Game of Life.

class Solution {
public:
    void gameOfLife(vector<vector<int>>& board) {
        auto check = [&](int y, int x) {
            return (y < 0 || y == board.size() || x < 0 || x == board[0].size()) ? 0 : board[y][x]&1;
        };
        const pair<int, int> dir[] = {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
        for(int y = 0; y < board.size(); ++y) for(int x = 0; x < board[0].size(); ++x) {
            int alive = 0;
            for(auto [ny,nx] : dir) alive += check(y + ny, x + nx);
            board[y][x] |= board[y][x] ? ((alive & 6) == 2 ? 2 : 0) : (alive == 3 ? 2 : 0);
        }
        for(auto &row : board) for(auto &cell : row) cell >>= 1;
    }
};

익명
작성일 2022.04.12 00:44

C++
Runtime: 0 ms, faster than 100.00% of C++ online submissions for Game of Life.
Memory Usage: 7.1 MB, less than 20.27% of C++ online submissions for Game of Life.

class Solution { 
public:
    void gameOfLife(vector<vector<int>>& board) {
        stack<pair<int, int>> change;
        
        
        for (int i = 0; i < board.size(); ++i) {
            for (int j = 0; j < board[i].size(); ++j) {
                int count = 0;
                
                count += i > 0 && j > 0 ? board[i - 1][j - 1] : 0;
                count += i > 0 ? board[i - 1][j] : 0;
                count += i > 0 && j < board[i - 1].size() - 1 ? board[i - 1][j + 1] : 0;
                count += j > 0 ? board[i][j - 1] : 0;
                count += j < board[i].size() - 1 ? board[i][j + 1] : 0;
                count += i < board.size() - 1 && j > 0 ? board[i + 1][j - 1] : 0;
                count += i < board.size() - 1 ? board[i + 1][j] : 0;
                count += i < board.size() - 1 && j < board[i + 1].size() - 1 ? board[i + 1][j + 1] : 0;
                
                if ((board[i][j] && (count < 2 || count > 3)) ||
                    (!board[i][j] && count == 3)) {
                    change.push(pair<int, int>(i, j));
                }
            }
        }
        while (!change.empty()) {
            pair<int, int> p = change.top(); change.pop();
            board[p.first][p.second] ^= 1;
        }
    }
};

C++
Runtime: 0 ms, faster than 100.00% of C++ online submissions for Game of Life.
Memory Usage: 7.1 MB, less than 20.27% of C++ online submissions for Game of Life.
[code=C++]class Solution { 
public:
    void gameOfLife(vector<vector<int>>& board) {
        stack<pair<int, int>> change;
        
        
        for (int i = 0; i < board.size(); ++i) {
            for (int j = 0; j < board[i].size(); ++j) {
                int count = 0;
                
                count += i > 0 && j > 0 ? board[i - 1][j - 1] : 0;
                count += i > 0 ? board[i - 1][j] : 0;
                count += i > 0 && j < board[i - 1].size() - 1 ? board[i - 1][j + 1] : 0;
                count += j > 0 ? board[i][j - 1] : 0;
                count += j < board[i].size() - 1 ? board[i][j + 1] : 0;
                count += i < board.size() - 1 && j > 0 ? board[i + 1][j - 1] : 0;
                count += i < board.size() - 1 ? board[i + 1][j] : 0;
                count += i < board.size() - 1 && j < board[i + 1].size() - 1 ? board[i + 1][j + 1] : 0;
                
                if ((board[i][j] && (count < 2 || count > 3)) ||
                    (!board[i][j] && count == 3)) {
                    change.push(pair<int, int>(i, j));
                }
            }
        }
        while (!change.empty()) {
            pair<int, int> p = change.top(); change.pop();
            board[p.first][p.second] ^= 1;
        }
    }
};[/code]

익명
작성일 2022.04.13 12:54

class Solution {
public:
    void gameOfLife(vector<vector<int>>& board) {
        int m=board.size();
        int n=board[0].size();
        int sum=0;
        
        vector<vector<int>> Tmpboard(m+2,vector<int>(n+2,0));
        
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                Tmpboard[i+1][j+1] = board[i][j];  
            };
        };
        
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                sum=Tmpboard[i][j]+Tmpboard[i][j+1]+Tmpboard[i][j+2]+Tmpboard[i+1][j]+Tmpboard[i+1][j+2]+Tmpboard[i+2][j]+Tmpboard[i+2][j+1]+Tmpboard[i+2][j+2];
                if(sum!=2){
                    if(sum==3){
                        board[i][j]=1;
                    }else if( sum>3 || sum<2 ){
                        board[i][j]=0;
                    };
                };
            };
        };
        
    }
};

익명
작성일 2022.04.14 21:36

익명
작성일 2022.04.14 21:52

class Solution:
    def gameOfLife(self, board: List[List[int]]) -> None:
        cp = copy.deepcopy(board)
        dir = [
            [1, 0],
            [1, 1],
            [0, 1],
            [-1, 1],
            [-1, 0],
            [-1, -1],
            [0, -1],
            [1, -1]
        ]
        ylen = len(board)
        xlen = len(board[0])
        for y in range(ylen):
            for x in range(xlen):
                cnt = 0
                for i in range(len(dir)):
                    ny = y + dir[i][0]
                    nx = x + dir[i][1]
                    if ny < 0 or ny >= ylen or nx < 0 or nx >= xlen: 
                        continue
                    if cp[ny][nx] == 1:
                        cnt += 1 
                if cp[y][x] == 1:
                    if cnt < 2 or cnt > 3:
                        board[y][x] = 0
                if cp[y][x] == 0:
                    if cnt == 3:
                        board[y][x] = 1
        return board

[code=python]class Solution:
    def gameOfLife(self, board: List[List[int]]) -> None:
        cp = copy.deepcopy(board)
        dir = [
            [1, 0],
            [1, 1],
            [0, 1],
            [-1, 1],
            [-1, 0],
            [-1, -1],
            [0, -1],
            [1, -1]
        ]
        ylen = len(board)
        xlen = len(board[0])
        for y in range(ylen):
            for x in range(xlen):
                cnt = 0
                for i in range(len(dir)):
                    ny = y + dir[i][0]
                    nx = x + dir[i][1]
                    if ny < 0 or ny >= ylen or nx < 0 or nx >= xlen: 
                        continue
                    if cp[ny][nx] == 1:
                        cnt += 1 
                if cp[y][x] == 1:
                    if cnt < 2 or cnt > 3:
                        board[y][x] = 0
                if cp[y][x] == 0:
                    if cnt == 3:
                        board[y][x] = 1
        return board[/code]