How do you solve Count Unguarded Cells in the Grid on LeetCode?

Count Unguarded Cells in the Grid (LeetCode #2257) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Guarded cells can be efficiently tracked using a set instead of a full grid.

What is the brute force solution for Count Unguarded Cells in the Grid?

The brute force approach for Count Unguarded Cells in the Grid is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves creating a grid and marking all cells that can be seen by guards. This is straightforward but inefficient for larger grids. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Unguarded Cells in the Grid?

Count Unguarded Cells in the Grid uses the following concepts: Array, Matrix, Simulation. The recommended patterns to study are: Hash Set for efficient membership checking., Grid traversal techniques (like BFS/DFS) can be useful in similar problems..

What are the interview tips for Count Unguarded Cells in the Grid?

Always clarify the problem and constraints before jumping into coding. Think about edge cases, such as all cells being guarded or walls blocking all views. Discuss your thought process with the interviewer to demonstrate your understanding.

What are common mistakes when solving Count Unguarded Cells in the Grid?

Failing to account for walls or other guards when marking cells as guarded. Not using a set for efficient lookups, leading to unnecessary complexity.

#2257

Count Unguarded Cells in the Grid

Medium

ArrayMatrixSimulationHash Set for efficient membership checking.Grid traversal techniques (like BFS/DFS) can be useful in similar problems.

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses a set to track guarded cells directly, avoiding the need for a full grid representation. This is more efficient in terms of both time and space.

⚙️

Algorithm

3 steps

1Step 1: Create a set to store the positions of guarded cells.
2Step 2: Add all guard and wall positions to the set to avoid counting them later.
3Step 3: For each guard, iterate in all four directions, adding cells to the set until hitting a wall or another guard.

solution.py19 lines

1def countUnguarded(m, n, guards, walls):
2    guarded = set()
3    for r, c in guards:
4        guarded.add((r, c))
5    for r, c in walls:
6        guarded.add((r, c))
7    
8    directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
9    for r, c in guards:
10        for dr, dc in directions:
11            nr, nc = r, c
12            while 0 <= nr < m and 0 <= nc < n:
13                nr += dr
14                nc += dc
15                if (nr, nc) in guarded:
16                    break
17                guarded.add((nr, nc))
18    
19    return m * n - len(guarded)

ℹ

Complexity note: The time complexity is O(n) because we only traverse the grid a limited number of times based on the number of guards and walls. The space complexity is O(n) due to the set storing guarded cell positions.

1Guarded cells can be efficiently tracked using a set instead of a full grid.
2Iterating in all four directions from each guard allows us to mark visible cells until blocked.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.