Have you ever wondered how to ensure maximum coverage with minimal resources, like setting up surveillance cameras in a building or guarding the boundaries of a field? 🏢🔍 While it might seem like a logistics puzzle, problems like these often have practical solutions grounded in computational thinking. One such fascinating challenge is LeetCode Problem #2257: Count Unguarded Cells in the Grid. 🧩👀

Imagine a grid 📏 representing a room, with guards 👮 stationed to monitor the area and walls 🧱 creating obstacles. Each guard can “see” in all four cardinal directions—north, south, east, and west—until their line of sight is blocked. Your task? Figure out how many cells remain unguarded and unblocked after accounting for all the guards’ visibility.

This problem combines elements of grid simulation, spatial traversal, and efficient line-of-sight calculations. While it might sound complex at first, breaking it down step by step makes it surprisingly approachable! ✨

In this blog post, we’ll:

• Unpack the problem’s requirements 🔍,
• Simulate guard visibility step by step 👁️‍🗨️,
• Solve the problem with a clean Python implementation 🐍,
• Analyze the solution’s performance to ensure efficiency 🏎️.

By the end, you’ll have a crystal-clear understanding of how to tackle grid-based problems using computational strategies. Ready to dive in? Let’s get started! 🚀

Problem Statement 🚀

In this exciting 🧩 medium-level challenge, you’re tasked with calculating how many unoccupied cells 🟦 are neither guarded nor blocked by walls in an m x n grid. Here’s the setup:

• Guards 👮 and walls 🧱 occupy specific grid cells.
• Each guard can see (and guard) all cells in the four cardinal directions—north, south, east, and west—until their view is blocked by a wall or another guard.
• Your mission is to count all unguarded cells 🟦 while considering guards’ views and obstacles.

✨ Key Intuition

The solution to this problem involves simulating the guards’ lines of sight in all four directions. Here’s how we approach it:

1. Mark the grid: Walls 🧱 and guards 👮 occupy some cells, so we first mark these in our grid.
2. Simulate guards’ visibility: For each guard, simulate the cells they can “see” in the four directions until blocked by a wall or another guard.
3. Count the leftovers: Any cell still unmarked after processing all guards is unguarded 🟦.

This approach uses basic simulation and grid traversal, which keeps it simple and efficient. Let’s break it down! 🧑‍💻

🖍️ Solution Approach

Step-by-Step Breakdown

1. Initialize the Grid 🗂️: Create a grid g of size m x n with all cells initialized to 0 🟦, representing unguarded and empty cells.

2. Mark Guards and Walls 🚧: Loop through the lists of guards 👮 and walls 🧱 to mark their positions with 2 in the grid.

3. Simulate Guard Visibility 👀:
   • For each guard, simulate visibility in all four directions using simple movement vectors.
   • Stop marking cells as guarded when you encounter a wall 🧱, another guard 👮, or reach the boundary of the grid.
4. Count Unguarded Cells 🔢: Finally, iterate through the grid and count all cells still marked as 0 🟦.

🌟 Example Walkthrough

Let’s take an example grid of size 4 x 4 with the following inputs:

Input:

• guards = [[1, 1], [2, 3]] 👮
• walls = [[0, 2], [2, 1]] 🧱

🟦 Initial Grid:

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

📋 Step 1: Mark Guards and Walls

• Place guards 👮 at (1, 1) and (2, 3).
• Place walls 🧱 at (0, 2) and (2, 1).

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

👀 Step 2: Simulate Guard Visibility

Let’s process each guard one by one:

1. Guard at (1, 1):
   • North (upwards): Guards (0, 1).
   • East (rightwards): Guards (1, 2) and stops at (1, 3).
   • South (downwards): Stops at (2, 1) due to a wall.
   • West (leftwards): Guards (1, 0).

[ ][1][W][ ]
[1][G][1][ ]
[ ][W][ ][G]
[ ][ ][ ][ ]

1. Guard at (2, 3):
   • North: Guards (1, 3) and stops at (0, 3).
   • East: Stops immediately at the grid boundary.
   • South: Guards (3, 3).
   • West: Guards (2, 2) and stops at (2, 1) (wall).

[ ][1][W][ ]
[1][G][1][1]
[ ][W][1][G]
[ ][ ][ ][1]

🔢 Step 3: Count Unguarded Cells

After processing all guards, the grid looks like this:

[ ][1][W][ ]
[1][G][1][1]
[ ][W][1][G]
[ ][ ][ ][1]

Count the number of 0s 🟦 (unguarded and unoccupied cells):

• Answer: 5 unguarded cells.

🐍 Python Implementation

Here’s the Python code for the solution:

from typing import List

class Solution:
    def countUnguarded(self, m: int, n: int, guards: List[List[int]], walls: List[List[int]]) -> int:
        # Initialize the grid
        grid = [[0] * n for _ in range(m)]
        
        # Mark guards and walls
        for r, c in guards:
            grid[r][c] = 2
        for r, c in walls:
            grid[r][c] = 2
        
        # Define movement directions
        directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]  # Up, Down, Left, Right
        
        # Simulate guards' visibility
        for r, c in guards:
            for dr, dc in directions:
                nr, nc = r, c
                while 0 <= nr + dr < m and 0 <= nc + dc < n and grid[nr + dr][nc + dc] == 0:
                    nr, nc = nr + dr, nc + dc
                    grid[nr][nc] = 1
        
        # Count unguarded cells
        return sum(cell == 0 for row in grid for cell in row)

📈 Complexity Analysis

Time Complexity:

1. Grid Initialization: \(O(m \times n)\)
2. Marking Guards/Walls: \(O(g + w)\), where \(g\) is the number of guards and \(w\) is the number of walls.
3. Simulating Visibility: \(O(g \times (m + n))\) for guards’ line-of-sight simulation.

Total: \(O(m \times n + g \times (m + n))\)

Space Complexity:

• \(O(m \times n)\) for the grid representation.

🏁 Conclusion

In this problem, we’ve explored how to simulate guards’ lines of sight effectively using a grid-based approach. By marking walls and guards, simulating visibility, and counting leftover cells, we arrive at a clean and efficient solution. 🚀