LeetCode #289 — MEDIUM

Game of Life

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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:

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?

Step 01

Brute Force Baseline

Problem summary: 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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[0,1,0],[0,0,1],[1,1,1],[0,0,0]]

Example 2

[[1,1],[1,0]]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #289: Game of Life
class Solution {
    public void gameOfLife(int[][] board) {
        int m = board.length, n = board[0].length;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int live = -board[i][j];
                for (int x = i - 1; x <= i + 1; ++x) {
                    for (int y = j - 1; y <= j + 1; ++y) {
                        if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] > 0) {
                            ++live;
                        }
                    }
                }
                if (board[i][j] == 1 && (live < 2 || live > 3)) {
                    board[i][j] = 2;
                }
                if (board[i][j] == 0 && live == 3) {
                    board[i][j] = -1;
                }
            }
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (board[i][j] == 2) {
                    board[i][j] = 0;
                } else if (board[i][j] == -1) {
                    board[i][j] = 1;
                }
            }
        }
    }
}

// Accepted solution for LeetCode #289: Game of Life
func gameOfLife(board [][]int) {
	m, n := len(board), len(board[0])
	for i := 0; i < m; i++ {
		for j, v := range board[i] {
			live := -v
			for x := i - 1; x <= i+1; x++ {
				for y := j - 1; y <= j+1; y++ {
					if x >= 0 && x < m && y >= 0 && y < n && board[x][y] > 0 {
						live++
					}
				}
			}
			if v == 1 && (live < 2 || live > 3) {
				board[i][j] = 2
			}
			if v == 0 && live == 3 {
				board[i][j] = -1
			}
		}
	}
	for i := 0; i < m; i++ {
		for j, v := range board[i] {
			if v == 2 {
				board[i][j] = 0
			}
			if v == -1 {
				board[i][j] = 1
			}
		}
	}
}

# Accepted solution for LeetCode #289: Game of Life
class Solution:
    def gameOfLife(self, board: List[List[int]]) -> None:
        m, n = len(board), len(board[0])
        for i in range(m):
            for j in range(n):
                live = -board[i][j]
                for x in range(i - 1, i + 2):
                    for y in range(j - 1, j + 2):
                        if 0 <= x < m and 0 <= y < n and board[x][y] > 0:
                            live += 1
                if board[i][j] and (live < 2 or live > 3):
                    board[i][j] = 2
                if board[i][j] == 0 and live == 3:
                    board[i][j] = -1
        for i in range(m):
            for j in range(n):
                if board[i][j] == 2:
                    board[i][j] = 0
                elif board[i][j] == -1:
                    board[i][j] = 1

// Accepted solution for LeetCode #289: Game of Life
const DIR: [(i32, i32); 8] = [
    (-1, 0),
    (1, 0),
    (0, -1),
    (0, 1),
    (-1, -1),
    (-1, 1),
    (1, -1),
    (1, 1),
];

impl Solution {
    #[allow(dead_code)]
    pub fn game_of_life(board: &mut Vec<Vec<i32>>) {
        let n = board.len();
        let m = board[0].len();
        let mut weight_vec: Vec<Vec<i32>> = vec![vec![0; m]; n];

        // Initialize the weight vector
        for i in 0..n {
            for j in 0..m {
                if board[i][j] == 0 {
                    continue;
                }
                for (dx, dy) in DIR {
                    let x = (i as i32) + dx;
                    let y = (j as i32) + dy;
                    if Self::check_bounds(x, y, n as i32, m as i32) {
                        weight_vec[x as usize][y as usize] += 1;
                    }
                }
            }
        }

        // Update the board
        for i in 0..n {
            for j in 0..m {
                if weight_vec[i][j] < 2 {
                    board[i][j] = 0;
                } else if weight_vec[i][j] <= 3 {
                    if board[i][j] == 0 && weight_vec[i][j] == 3 {
                        board[i][j] = 1;
                    }
                } else {
                    board[i][j] = 0;
                }
            }
        }
    }

    #[allow(dead_code)]
    fn check_bounds(i: i32, j: i32, n: i32, m: i32) -> bool {
        i >= 0 && i < n && j >= 0 && j < m
    }
}

// Accepted solution for LeetCode #289: Game of Life
/**
 Do not return anything, modify board in-place instead.
 */
function gameOfLife(board: number[][]): void {
    const m = board.length;
    const n = board[0].length;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            let live = -board[i][j];
            for (let x = i - 1; x <= i + 1; ++x) {
                for (let y = j - 1; y <= j + 1; ++y) {
                    if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] > 0) {
                        ++live;
                    }
                }
            }
            if (board[i][j] === 1 && (live < 2 || live > 3)) {
                board[i][j] = 2;
            }
            if (board[i][j] === 0 && live === 3) {
                board[i][j] = -1;
            }
        }
    }
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (board[i][j] === 2) {
                board[i][j] = 0;
            }
            if (board[i][j] === -1) {
                board[i][j] = 1;
            }
        }
    }
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.