Game of Life
Problem Description
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 of the board is determined by applying the above rules simultaneously to every cell in the current state of the m x n grid board. In this process, births and deaths occur simultaneously.
Given the current state of the board, update the board to reflect its next state.
Note that you do not need to return anything.
Example 1:
Example 2:
Constraints:
m == board.lengthn == board[i].length1 <= m, n <= 25board[i][j]is0or1.
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?
Solve it on LeetCode
Approaches
1. Basic Simulation
Intuition:
The problem involves updating the board according to the rules given, where a cell's next state is conditioned by its current state and the states of its eight neighbors. A straightforward approach is to create a copy of the board to hold the previous state, which can be referenced while we iterate over the original board to apply the game rules.
Code:
- Time Complexity: O(m * n), where m is the number of rows and n is the number of columns. We iterate over each cell and its neighbors.
- Space Complexity: O(m * n), for the copy of the board.
2. In-Place State Tracking
Intuition:
To optimize space usage, we can encode information about both current and next states in the same board. By using integer encoding, we can keep the board's state transitions in a single space. For instance, we can use different integers to represent current and next states:
0->0: 0 (Dead to Dead)1->1: 1 (Live to Live)1->0: 2 (Live to Dead)0->1: 3 (Dead to Live)
Code:
- Time Complexity: O(m * n), where m is the number of rows and n is the number of columns. We iterate over each cell and its neighbors.
- Space Complexity: O(1), as we are modifying the board in-place without using extra space.