Problem Description

In the given problem, you have an 8x8 game board represented by a two-dimensional array board, where each cell can be in one of three states: it can either be free (marked with '.'), hold a white piece ('W'), or hold a black piece ('B'). Your task is to determine if a move, defined as changing a free cell to a particular color (either white or black), is legal.

A move is legal if it turns the selected cell into the endpoint of a "good line". A good line must follow these conditions:

1. It consists of at least three cells, including endpoints.
2. The endpoints of the line must be of one color.
3. The cells between the endpoints must all be of the opposite color and there should be no free cells.

There must be a good line in any of the eight possible directions from the selected cell: horizontal, vertical, or diagonal. The task boils down to checking whether making the move creates at least one good line according to the above conditions.

Intuition

Given the rules for a legal move, the logical approach is to start from the cell at (rMove, cMove) and look in all possible line directions (8 in total) to check if there is a good line with the move being an endpoint.

The solution has the following key steps:

1. Define the eight possible directions to explore from any given cell (up, down, left, right, and the four diagonals).
2. For each direction, keep moving along that line until you either run off the board or encounter a cell that isn't occupied by the opposite color (could be free or the same color as the one you're playing).
3. Count the number of cells of the opposite color you pass in this process.
4. If you find a cell at the end of the sequence that is of the same color as the one you're playing, and there are more than one opposite color cells in between, the move is legal (i.e., a good line is formed).

The provided code creates a list of directions and iterates over them. For each direction, it increments along that path counting the cells until a stopping condition is met. If the final cell checked matches the player's color and there was more than one opposite color cell in between, the function returns true, indicating that the move is legal.

Solution Approach

The implementation of the solution involves systematic exploration in every possible direction from the cell where the move is intended. Here is a breakdown of how the algorithm translates into the solution strategy:

1. Direction Vectors: The solution uses a list of tuples called dirs, where each tuple represents a direction vector in the two-dimensional space of the board. For example, (1, 0) represents moving down, (0, 1) represents moving right, (-1, 0) and (0, -1) represent moving up and left respectively, and the four diagonal directions are represented with combinations of 1 and -1.

2. Iterating Over Directions: The solution iterates over these direction vectors using a for loop. Inside the loop, it sets up two index variables i and j to the rMove and cMove coordinates of the move being queried.

3. Exploring a Direction: For each of the 8 directions, the code enters a while loop which continues as long as the new indices i + a and j + b (representing the next cell in the direction (a, b)) stay within the bounds of the grid (0 to n-1, as the grid size is 8).

4. Counting Opposite Colors: It increments a counter t representing the number of cells traversed. The loop breaks if the next cell is either free or of the same color as the move being played (color), since this means a good line cannot be assured in this direction.

5. Check for Legal Move: After the loop, if the cell that caused the loop to terminate is of the same color as color, and if at least one opposite-colored piece (t > 1) was found in the traversed path, then the move is legal. A true value is immediately returned.

6. Result: If none of the directions leads to a legal move, meaning no good line was formed, the function returns false after exiting the for loop.

The solution effectively uses a pattern common in grid-based problems, iterating over a set of fixed directions to explore adjacent cells. By using the direction vectors with a while loop and boundary checks, it manages to traverse the two-dimensional array efficiently.

The data structures used here are basic; a list for directions and simple variables for indices and counters. The algorithm does not require additional data structures like stacks, queues, or maps.

This iterative approach examines each potential line emanating from the cell (rMove, cMove), handling the board as if it were an infinite plane, clipping off paths that go off-grid or do not meet the conditions for a legal move.

Example Walkthrough

Let's illustrate the solution approach with a small example. Consider the following 8x8 board configuration where 'rMove' is at row 3, and 'cMove' is at column 3 (0-indexed), and we wish to place a black piece ('B'):

. . . . . . . .
. . . . . . . .
. . W B W . . .
. . W . W . . .
. . W B W . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .

We want to determine if placing a black piece at (3,3) is a legal move.

Following the solution approach:

1. Direction Vectors: We have a list of eight possible directions to check - up, down, left, right, and the four diagonals.

2. Iterating Over Directions: We start iterating over these directions. Let's consider checking upwards first (direction vector (-1, 0)).

3. Exploring a Direction: We move up from our desired move position (3,3) to (2,3) and keep going up as long as we're within the bounds and encountering white pieces ('W').

4. Counting Opposite Colors: We find that we indeed encounter white pieces and keep a counter t. Since we find white pieces at positions (2,3) and (1,3), t is now 2.

5. Check for Legal Move: As we move to the next cell upwards (0,3), we find a black piece ('B'). Since t > 1 and the piece found is of the same color as the piece we wish to play, this direction confirms a legal move.

6. Result: We return true before checking the remaining directions because we've found at least one good line (vertical line from (0,3) through (3,3)), and thus the move at (3,3) is legal.

In this example, only one direction was needed to confirm the legality of the move, but in a complete implementation, all eight directions would be checked to evaluate the move fully. This demonstrates how the algorithm systematically checks each direction to determine whether a legal "good line" is formed.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code is determined by how many times we loop over the different directions from the starting move, as well as how far we can go in each direction. We have 8 possible directions to check, and in the worst-case scenario, we could iterate over all n cells in one direction (where n is the size of the board's dimension, which is 8 in this case). Therefore, the worst-case time complexity is O(8n) which simplifies to O(n) because 8 is a constant factor.

In this specific case, since n is fixed at 8, we can also argue that the time complexity is O(1) since the board size doesn't change and there's a maximum number of steps to be taken.

Space Complexity

The space complexity of the code is O(1) since the extra space used does not scale with the size of the input. We have a fixed-size board and the dirs array which consists of 8 directions, and a few variables i, j, and t, which all occupy constant space regardless of input size.

Learn more about how to find time and space complexity quickly using problem constraints.