What is the time complexity of Count Unguarded Cells in the Grid?

The optimal solution has a time complexity of O(m*n).

What is the space complexity of Count Unguarded Cells in the Grid?

The space complexity is O(m*n).

What data structures are used to solve Count Unguarded Cells in the Grid?

This problem uses Array, Matrix, Simulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Count Unguarded Cells in the Grid?

Insight 1: The core idea is to simulate the guards' line of sight by iterating through the grid in four directions (up, down, left, right) from each cell. Insight 2: Using a 2D array (grid) to represent the state of each cell (unguarded, guarded, wall, guard) simplifies the simulation and counting process.

Which companies ask LeetCode 2257. Count Unguarded Cells in the Grid in interviews?

This problem is frequently asked at Poshmark.

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Count Unguarded Cells in the Grid - LeetCode 2257 Solution

Count Unguarded Cells in the Grid - Complete Solution Guide

Count Unguarded Cells in the Grid is LeetCode problem 2257, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given two integers m and n representing a 0-indexed m x n grid. You are also given two 2D integer arrays guards and walls where guards[i] = [row i , col i ] and walls[j] = [row j , col j ] represent the positions of the i th guard and j th wall respectively. A guard can see every cell in the four cardinal directions (north, east, south, or west) starting from their position unless obstructed by a wall or another guard. A cell is guarded if there is at least one guard that can see it. Ret

Detailed Explanation

The problem asks us to determine the number of unguarded cells in a rectangular grid, given the grid's dimensions (m x n), the positions of guards, and the positions of walls. Guards can 'see' in all four cardinal directions until their line of sight is blocked by a wall or another guard. A cell is guarded if any guard can see it. The goal is to count the cells that are neither walls, guards, nor guarded cells.

Solution Approach

The solution uses a simulation approach. First, it initializes a 2D array (grid) to represent the grid's state, marking walls and guards. Then, for each row and column, it scans in both directions (left to right, right to left for rows; top to bottom, bottom to top for columns) simulating the guards' line of sight. If a guard is 'seeing' (not blocked by a wall or another guard) and encounters an unguarded cell, that cell is marked as guarded. Finally, the solution iterates through the grid and counts the remaining unguarded cells.

Step-by-Step Algorithm

Step 1: Initialize a 2D array `grid` of size m x n, initially filled with 0, representing empty and unguarded cells.
Step 2: Iterate through the `walls` array and update the `grid` with 2 at each wall's coordinates.
Step 3: Iterate through the `guards` array and update the `grid` with 3 at each guard's coordinates.
Step 4: Perform row-wise scans. Iterate through each row of the `grid`:
Step 4.1: Scan from left to right. Maintain a boolean variable `sees` to indicate if a guard is currently 'seeing'. Update `sees` to true if a guard is encountered and reset to false if a wall or guard is encountered. Mark unguarded cells as guarded (1) if `sees` is true.
Step 4.2: Scan from right to left. Repeat the process as in Step 4.1.
Step 5: Perform column-wise scans. Iterate through each column of the `grid`:
Step 5.1: Scan from top to bottom. Maintain a boolean variable `sees` to indicate if a guard is currently 'seeing'. Update `sees` to true if a guard is encountered and reset to false if a wall or guard is encountered. Mark unguarded cells as guarded (1) if `sees` is true.
Step 5.2: Scan from bottom to top. Repeat the process as in Step 5.1.
Step 6: Iterate through the `grid` and count the cells that are still marked as 0 (unguarded).
Step 7: Return the count of unguarded cells.

Key Insights

Insight 1: The core idea is to simulate the guards' line of sight by iterating through the grid in four directions (up, down, left, right) from each cell.
Insight 2: Using a 2D array (grid) to represent the state of each cell (unguarded, guarded, wall, guard) simplifies the simulation and counting process.
Insight 3: Recognizing that a guard's line of sight is stopped by a wall or another guard is crucial for correctly marking guarded cells.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

This problem involves: Array, Matrix, Simulation.

Companies

Asked at: Poshmark.

Count Unguarded Cells in the Grid | LeetCode 2257

Medium

Array Matrix Simulation

Time: O(m*n)Space: O(m*n)

Solution Code

Available in:

class Solution {
    public int countUnguarded(int m, int n, int[][] guards, int[][] walls) {
        // State representation in the grid:
        // 0: empty and unguarded
        // 1: guarded
        // 2: wall
        // 3: guard
        
        int[][] grid = new int[m][n];
        
        for (int[] wall : walls) {
            int r = wall[0];
            int c = wall[1];
            grid[r][c] = 2;
        }
        for (int[] guard : guards) {
            int r = guard[0];
            int c = guard[1];
            grid[r][c] = 3;
        }
        
        // Mark guarded cells by scanning in four directions for each row and column.
        // A cell is guarded if there is an active line of sight from a guard.
        // A line of sight is started by a guard and stopped by a wall or another guard.
        
        // Row-wise scans
        for (int r = 0; r < m; r++) {
            // Left to right
            boolean sees = false;
            for (int c = 0; c < n; c++) {
                if (grid[r][c] == 2 || grid[r][c] == 3) {
                    sees = (grid[r][c] == 3);
                } else if (sees && grid[r][c] == 0) {
                    grid[r][c] = 1;
                }
            }
            
            // Right to left
            sees = false;
            for (int c = n - 1; c >= 0; c--) {
                if (grid[r][c] == 2 || grid[r][c] == 3) {
                    sees = (grid[r][c] == 3);
                } else if (sees && grid[r][c] == 0) {
                    grid[r][c] = 1;
                }
            }
        }
        
        // Column-wise scans
        for (int c = 0; c < n; c++) {
            // Top to bottom
            boolean sees = false;
            for (int r = 0; r < m; r++) {
                if (grid[r][c] == 2 || grid[r][c] == 3) {
                    sees = (grid[r][c] == 3);
                } else if (sees && grid[r][c] == 0) {
                    grid[r][c] = 1;
                }
            }
            
            // Bottom to top
            sees = false;
            for (int r = m - 1; r >= 0; r--) {
                if (grid[r][c] == 2 || grid[r][c] == 3) {
                    sees = (grid[r][c] == 3);
                } else if (sees && grid[r][c] == 0) {
                    grid[r][c] = 1;
                }
            }
        }
        
        // Count the remaining empty (state 0) cells
        int unguarded_count = 0;
        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                if (grid[r][c] == 0) {
                    unguarded_count++;
                }
            }
        }
        
        return unguarded_count;
    }
}

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Problem Info

Description

Problem 2257: Count Unguarded Cells in the Grid - Detailed explanation and solution approaches.

Topics

Array Matrix Simulation

Companies

Poshmark

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2257

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Initialize a 2D array `grid` of size m x n, initially filled with 0, representing empty and unguarded cells.
2
Step 2: Iterate through the `walls` array and update the `grid` with 2 at each wall's coordinates.
3
Step 3: Iterate through the `guards` array and update the `grid` with 3 at each guard's coordinates.
4
Step 4: Perform row-wise scans. Iterate through each row of the `grid`:
5
Step 4.1: Scan from left to right. Maintain a boolean variable `sees` to indicate if a guard is currently 'seeing'. Update `sees` to true if a guard is encountered and reset to false if a wall or guard is encountered. Mark unguarded cells as guarded (1) if `sees` is true.
6
Step 4.2: Scan from right to left. Repeat the process as in Step 4.1.
7
Step 5: Perform column-wise scans. Iterate through each column of the `grid`:
8
Step 5.1: Scan from top to bottom. Maintain a boolean variable `sees` to indicate if a guard is currently 'seeing'. Update `sees` to true if a guard is encountered and reset to false if a wall or guard is encountered. Mark unguarded cells as guarded (1) if `sees` is true.
9
Step 5.2: Scan from bottom to top. Repeat the process as in Step 5.1.
10
Step 6: Iterate through the `grid` and count the cells that are still marked as 0 (unguarded).
11
Step 7: Return the count of unguarded cells.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n)

💾

Space Complexity

O(m*n)

📝

Why?

The algorithm iterates through the entire grid multiple times (once for initialization, once for placing walls and guards, and four times for simulating the line of sight - row-wise left to right, row-wise right to left, column-wise top to bottom, column-wise bottom to top and finally for counting). Each iteration takes O(m*n) time. Thus the overall Time Complexity is O(m*n). The algorithm uses a 2D array `grid` of size m x n to store the state of each cell. This takes O(m*n) space. Thus, the overall Space Complexity is O(m*n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using Directional Vectors

Instead of looping multiple times, one can use directional vectors (e.g., (0, 1), (0, -1), (1, 0), (-1, 0)) to simulate the guards' vision in all four directions from their starting positions. This might seem slightly more concise but offers little to no performance benefits.

Time: O(m*n)Space: O(m*n)

Trade-offs: This method can be slightly more readable for some but might be harder to grasp initially. The time and space complexity remains the same because, at worst, each cell needs to be visited multiple times.

✨

Key Insights

1Insight 1: The core idea is to simulate the guards' line of sight by iterating through the grid in four directions (up, down, left, right) from each cell.
2Insight 2: Using a 2D array (grid) to represent the state of each cell (unguarded, guarded, wall, guard) simplifies the simulation and counting process.
3Insight 3: Recognizing that a guard's line of sight is stopped by a wall or another guard is crucial for correctly marking guarded cells.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: Empty grid (no guards or walls). The solution correctly identifies all cells as unguarded and returns m * n.
Edge case 2: Grid filled entirely with walls and/or guards. The solution correctly identifies that there are no unguarded cells and returns 0.
Edge case 3: A guard is positioned at the very edge of the grid. The code correctly handles the boundary condition, as the scanning loops don't cause out-of-bounds accesses.
Edge case 4: Multiple guards and walls are adjacent. The scan properly stops the guard line of sight based on the first obstacle encountered in that direction.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to reset the 'sees' flag when a wall or guard is encountered, resulting in cells being incorrectly marked as guarded.

✗Common mistake 2: Not scanning in all four directions, causing some cells to be missed and incorrectly counted as unguarded.

✗Common mistake 3: Not handling the edge cases, such as an empty grid or a grid fully occupied by guards/walls.

✗Common mistake 4: Incorrectly indexing the grid, leading to array out-of-bounds errors.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Surveillance systems where the objective is to identify blind spots (unguarded areas) given the locations and ranges of surveillance cameras (guards) and obstacles (walls).

🏦Real-world scenario 2: Security planning for large venues or events, where the goal is to optimize the placement of security personnel (guards) to minimize unguarded areas.

🎮Real-world scenario 3: Wireless signal coverage optimization. Guards are analogous to WiFi access points, and walls are physical obstructions. Unguarded cells represent areas with poor or no wireless signal.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 6, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Count Unguarded Cells in the Grid"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether matrix techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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