From 1a09f2148794cdf51e35bbafefa16e3f7c9eb040 Mon Sep 17 00:00:00 2001 From: Sangeeth Sudheer Date: Wed, 13 Apr 2022 05:37:19 +0530 Subject: [PATCH] add py3 soln for game of life --- 0289_game-of-life/README.md | 38 ++++++++++++++++++ 0289_game-of-life/python3/extra_space.py | 49 ++++++++++++++++++++++++ 0289_game-of-life/python3/solution.py | 0 3 files changed, 87 insertions(+) create mode 100644 0289_game-of-life/README.md create mode 100644 0289_game-of-life/python3/extra_space.py create mode 100644 0289_game-of-life/python3/solution.py diff --git a/0289_game-of-life/README.md b/0289_game-of-life/README.md new file mode 100644 index 0000000..2b612c9 --- /dev/null +++ b/0289_game-of-life/README.md @@ -0,0 +1,38 @@ +According to [Wikipedia's article](https://en.wikipedia.org/wiki/Conway%27s_Game_of_Life): "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](https://en.wikipedia.org/wiki/Moore_neighborhood) (horizontal, vertical, diagonal) using the following four rules (taken from the above Wikipedia article): + +1. Any live cell with fewer than two live neighbors dies as if caused by under-population. +2. Any live cell with two or three live neighbors lives on to the next generation. +3. Any live cell with more than three live neighbors dies, as if by over-population. +4. 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:** + +![](https://assets.leetcode.com/uploads/2020/12/26/grid1.jpg) + + 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:** + +![](https://assets.leetcode.com/uploads/2020/12/26/grid2.jpg) + + 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? \ No newline at end of file diff --git a/0289_game-of-life/python3/extra_space.py b/0289_game-of-life/python3/extra_space.py new file mode 100644 index 0000000..a3a00c9 --- /dev/null +++ b/0289_game-of-life/python3/extra_space.py @@ -0,0 +1,49 @@ +# Time: O(m*n) +# Space: O(m*n) +class Solution: + def gameOfLife(self, board: List[List[int]]) -> None: + """ + Do not return anything, modify board in-place instead. + """ + nrows, ncols = len(board), len(board[0]) + board_copy = [[None] * ncols for _ in range(nrows)] + + for i in range(nrows): + for j in range(ncols): + board_copy[i][j] = next_cell_state(board, i, j) + + # Copy new board values + for i in range(nrows): + for j in range(ncols): + board[i][j] = board_copy[i][j] + +def next_cell_state(board, i, j): + current = board[i][j] + total_alive = sum(get_cell_neighbours(board, i, j)) + + # Rule 4: Exact 3 neighbors will mean dead cell becomes alive + # Live cell remains live + if total_alive == 3: + return 1 + + # Rule 2: Live cell lives onto next generation + # Dead cell remains dead + elif total_alive == 2: + return current + + # Rules 1 and 2: Under-population and over-population + return 0 + +# Return cell at (i, j) or 0 if it's out of bounds +def cell(board, i, j): + nrows, ncols = len(board), len(board[0]) + + return board[i][j] if 0 <= i < nrows and 0 <= j < ncols else 0 + +# Get surrounding 8 neighbours of cell at (i, j) +def get_cell_neighbours(board, i, j): + return ( + cell(board, i - 1, j - 1), cell(board, i - 1, j), cell(board, i - 1, j + 1), + cell(board, i, j - 1), cell(board, i, j + 1), + cell(board, i + 1, j - 1), cell(board, i + 1, j), cell(board, i + 1, j + 1) + ) diff --git a/0289_game-of-life/python3/solution.py b/0289_game-of-life/python3/solution.py new file mode 100644 index 0000000..e69de29