Problem Description

This LeetCode problem requires us to calculate the number of unoccupied cells in a 0-indexed m x n grid that are not guarded by any guards and not blocked by walls. We are given two lists, guards and walls, where each element is a pair of coordinates representing the location of a guard or a wall, respectively. A guard can see and thus guard all cells in the four cardinal directions (north, east, south, or west) unless there is a wall or another guard obstructing the view. Cells are considered guarded if at least one guard can see them. Our goal is to count how many cells are left unguarded, considering that walls and guards themselves occupy some of the cells.

Intuition

The solution to this problem is based on simulating the guards' line of sight and marking cells they can "see" as guarded. By creating a grid, we can simulate the guards' lines of sight in each of the four cardinal directions. We can iterate through each direction from the guard's position until we either reach the grid's boundary, encounter another guard, or a wall, which blocks the guard's line of sight. The key things to keep in mind are:

• Guards and walls occupy some cells, so these should be marked differently to indicate they are not empty and cannot be guarded by other guards.
• We need to check the visibility in all directions from each guard until we are obstructed or out of bounds.
• After marking all the guarded cells, anything left as unmarked is unguarded, and we just need to count those cells.

By iterating over the guards' positions and simulating their view lines while bypassing walls and other guards, we can then easily count the number of unobserved cells to get our answer.

Solution Approach

The implementation of the solution follows a step-by-step approach which relies on simulation and proper labeling of the cells within the grid.

Here's a walkthrough of the algorithm:

• Initialization: We start by creating a grid g sized m x n, which will serve as a representation of our problem space. The value of each cell in g initially is 0, which stands for unguarded and unoccupied by either a guard or a wall.

• Marking Guards and Walls: We loop through the guards and walls arrays, marking the corresponding cells in our grid g as 2. This 2 is a special label indicating cells that are occupied by either a guard or a wall and cannot be further guarded.

• Guarding Cells: Next, we iterate over all the guard positions. Since guards can see in the four cardinal directions, we simulate this visibility. To do this, we use the dirs tuple which defines the directional steps for north, east, south, and west. For each direction:

  • We start from the guard's position and move cell by cell in the current direction.
  • The movement continues until we reach a cell that is either out of bounds, marked as 2 (a wall or another guard), or has already been guarded.
  • As we move through cells, we change their status to 1 to signify that they are guarded.

  This simulation uses the pairwise iterator from the dirs tuple to get the direction vectors, which define our movement across the grid from the guard's current position.

• Counting Unguarded Cells: Finally, we go through the entire grid g and count all cells still marked as 0, which denotes they are unoccupied and unguarded.

Therefore, the algorithm employs a simulation method and utilizes a matrix to keep track of the state of each cell accurately. This is an efficient way to solve the problem as it does not involve any complex data structures or algorithms, just straightforward iterative logic and condition checking.

Example Walkthrough

Let's walk through a simple example to illustrate the solution approach. Suppose we have a 3x3 grid represented by an m x n matrix, where m = n = 3. Imagine our grid looks something like this:

[ ][ ][ ]
[ ][ ][ ]
[ ][ ][ ]

Now, let's say we have one guard and one wall, with the guards list as [(1, 1)] and the walls list as [(0, 2)]. The guard is located in the cell at row 1 column 1 (use 0-indexing), and the wall is located at row 0 column 2. The grid now looks like this, with 'G' representing the guard and 'W' the wall:

[ ][ ][W]
[ ][G][ ]
[ ][ ][ ]

Following the solution approach:

• Initialization: We create an initial grid g of size 3x3 with all cells initialized to 0 to represent unguarded cells.

[0][0][0]
[0][0][0]
[0][0][0]

• Marking Guards and Walls: We mark the guard's and wall's positions on the grid with 2. The grid now looks like this:

[0][0][2]
[0][2][0]
[0][0][0]

• Guarding Cells: Since the guard can look in all four directions, we will simulate this for our single guard at (1, 1):

  • North Direction (upwards): We check the cell (0, 1). Since it's a 0, we mark it as 1. The cell (0, 2) won't be checked because there's a wall at (0, 2).
  • East Direction (to the right): The guard checks the cell (1, 2). It's a 0, so we mark it as 1.
  • South Direction (downwards): The guard checks the cell (2, 1). It's a 0, so we mark it as 1.
  • West Direction (to the left): The guard checks the cell (1, 0). It's a 0, so we mark it as 1.

After the guard has scanned all four directions, the grid now looks like:

[0][1][2]
[1][2][1]
[0][1][0]

• Counting Unguarded Cells: We count all 0s in the grid to find the unguarded cells. There are three 0s, which means there are three unguarded cells.

This simple example demonstrates how the solution approach accurately simulates guards’ visibility to find unguarded cells in the grid.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code is O(m * n + g * (m + n)), where m and n correspond to the number of rows and columns of the grid, and g is the number of guards. The reasoning behind this is as follows:

• The initial setup of the grid g with size m * n takes O(m * n) time.
• The loops for placing guards and walls each run at most O(g + w), where w is the number of walls, however, these are negligible compared to other terms when g and w are much smaller than m * n.
• The nested loops iterate through each guard, and for each guard, scan across the grid in four directions up to m or n times (whichever direction taken), hence 4*(m + n) for each guard, aggregating to g * (m + n) for all guards.

Space Complexity

The space complexity of the algorithm is O(m * n). This is because only a single m * n grid is used as extra space to store the state of each cell (whether it is free, guarded, or a wall).

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