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Formatted question description: https://leetcode.ca/all/289.html
289 Game of Life
According to the 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."
Given a board with m by n cells, each cell has an initial state live (1) or dead (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 underpopulation.
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 overpopulation..
Any dead cell with exactly three live neighbors becomes a live cell, as if by reproduction.
Write a function to compute the next state (after one update) of the board given its current state.
The next state is created by applying the above rules simultaneously to every cell in the current state,
where births and deaths occur simultaneously.
Example:
Input:
[
[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]
]
Follow up:
Could you solve it inplace? Remember that the board needs to be updated at the same time:
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 the border of the array.
How would you address these problems?
@tagarray
Algorithm
State 0: Dead cells turned into dead cells
State 1: Live cells are transformed into live cells
State 2: Live cells turned into dead cells
State 3: Dead cells transformed into live cells
Take the remainder of 2 for all states
 Then states 0 and 2 become dead cells,
 State 1 and 3 are living cells
Mission complete.
First scan the original array one by one, for each position, scan eight locations around it
 If you encounter state 1 or 2, the counter will be incremented by 1, and 8 neighbors will be scanned
 If there are less than two living cells or more than three living cells, and the current position is a living cell, mark status 2
 If there are exactly three living cells and currently dead cells, mark status 3 After completing the scan, scan the data again, and the remainder of 2 becomes the result we want
Code
Java

public class Game_of_Life { class Solution { public void gameOfLife(int[][] board) { // Neighbors array to find 8 neighboring cells for a given cell int[] neighbors = {0, 1, 1}; int rows = board.length; int cols = board[0].length; // Iterate through board cell by cell. for (int row = 0; row < rows; row++) { for (int col = 0; col < cols; col++) { // For each cell count the number of live neighbors. int liveNeighbors = 0; for (int i = 0; i < 3; i++) { for (int j = 0; j < 3; j++) { if (!(neighbors[i] == 0 && neighbors[j] == 0)) { int r = (row + neighbors[i]); int c = (col + neighbors[j]); // Check the validity of the neighboring cell. // and whether it was originally a live cell. if ((r < rows && r >= 0) && (c < cols && c >= 0) && (Math.abs(board[r][c]) == 1)) { liveNeighbors += 1; } } } } // Rule 1 to Rule 3 if ((board[row][col] == 1) && (liveNeighbors < 2  liveNeighbors > 3)) { // 1 signifies the cell is now dead but originally was live. board[row][col] = 1; } // Rule 4 if (board[row][col] == 0 && liveNeighbors == 3) { // 2 signifies the cell is now live but was originally dead. board[row][col] = 2; } } } // Get the final representation for the newly updated board. for (int row = 0; row < rows; row++) { for (int col = 0; col < cols; col++) { if (board[row][col] > 0) { board[row][col] = 1; } else { board[row][col] = 0; } } } } } }

// OJ: https://leetcode.com/problems/gameoflife/ // Time: O(N^2) // Space: O(1) class Solution { public: void gameOfLife(vector<vector<int>>& A) { int M = A.size(), N = A[0].size(); auto alive = [&](int x, int y) { return x >= 0 && x < M && y >= 0 && y < N && (A[x][y] & 1); }; for (int i = 0; i < M; ++i) { for (int j = 0; j < N; ++j) { int cnt = 0; for (int dx = 1; dx <= 1; ++dx) { for (int dy = 1; dy <= 1; ++dy) { if (dx == 0 && dy == 0) continue; cnt += alive(i + dx, j + dy); } } if (cnt == 3  (A[i][j] && cnt == 2)) A[i][j] = 2; } } for (int i = 0; i < M; ++i) { for (int j = 0; j < N; ++j) A[i][j] >>= 1; } } };

class Solution: def gameOfLife(self, board: List[List[int]]) > None: """ Do not return anything, modify board inplace instead. """ m, n = len(board), len(board[0]) dirs = [[1, 0], [1, 0], [0, 1], [0, 1], [1, 1], [1, 1], [1, 1], [1, 1]] for i in range(m): for j in range(n): cnt = sum( 1 for a, b in dirs if 0 <= i + a < m and 0 <= j + b < n and board[i + a][j + b] in (1, 2) ) if board[i][j] == 1 and (cnt < 2 or cnt > 3): board[i][j] = 2 elif board[i][j] == 0 and (cnt == 3): board[i][j] = 3 for i in range(m): for j in range(n): board[i][j] %= 2 ############ class Solution(object): def gameOfLife(self, board): """ :type board: List[List[int]] :rtype: void Do not return anything, modify board inplace instead. """ def helper(board, p, q): cnt = 0 for i in range(p  1, p + 2): for j in range(q  1, q + 2): if i == p and j == q: continue if 0 <= i < len(board) and 0 <= j < len(board[0]) and board[i][j] & 1: cnt += 1 if cnt == 3 or (board[p][q] == 1 and cnt == 2): board[p][q] = 2 for i in range(0, len(board)): for j in range(0, len(board[0])): helper(board, i, j) for i in range(0, len(board)): for j in range(0, len(board[0])): board[i][j] >>= 1