LeetCode #2257 — MEDIUM

Count Unguarded Cells in the Grid

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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.

Return the number of unoccupied cells that are not guarded.

Example 1:

Input: m = 4, n = 6, guards = [[0,0],[1,1],[2,3]], walls = [[0,1],[2,2],[1,4]]
Output: 7
Explanation: The guarded and unguarded cells are shown in red and green respectively in the above diagram.
There are a total of 7 unguarded cells, so we return 7.

Example 2:

Input: m = 3, n = 3, guards = [[1,1]], walls = [[0,1],[1,0],[2,1],[1,2]]
Output: 4
Explanation: The unguarded cells are shown in green in the above diagram.
There are a total of 4 unguarded cells, so we return 4.

Constraints:

1 <= m, n <= 10⁵
2 <= m * n <= 10⁵
1 <= guards.length, walls.length <= 5 * 10⁴
2 <= guards.length + walls.length <= m * n
guards[i].length == walls[j].length == 2
0 <= row_i, row_j < m
0 <= col_i, col_j < n
All the positions in guards and walls are unique.

Step 01

Brute Force Baseline

Problem summary: 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] = [rowi, coli] and walls[j] = [rowj, colj] represent the positions of the ith guard and jth 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. Return the number of unoccupied cells that are not guarded.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

4
6
[[0,0],[1,1],[2,3]]
[[0,1],[2,2],[1,4]]

Example 2

3
3
[[1,1]]
[[0,1],[1,0],[2,1],[1,2]]

Core Insight

What unlocks the optimal approach

Create a 2D array to represent the grid. Can you mark the tiles that can be seen by a guard?
Iterate over the guards, and for each of the 4 directions, advance the current tile and mark the tile. When should you stop advancing?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2257: Count Unguarded Cells in the Grid
class Solution {
    public int countUnguarded(int m, int n, int[][] guards, int[][] walls) {
        int[][] g = new int[m][n];
        for (var e : guards) {
            g[e[0]][e[1]] = 2;
        }
        for (var e : walls) {
            g[e[0]][e[1]] = 2;
        }
        int[] dirs = {-1, 0, 1, 0, -1};
        for (var e : guards) {
            for (int k = 0; k < 4; ++k) {
                int x = e[0], y = e[1];
                int a = dirs[k], b = dirs[k + 1];
                while (x + a >= 0 && x + a < m && y + b >= 0 && y + b < n && g[x + a][y + b] < 2) {
                    x += a;
                    y += b;
                    g[x][y] = 1;
                }
            }
        }
        int ans = 0;
        for (var row : g) {
            for (int v : row) {
                if (v == 0) {
                    ++ans;
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2257: Count Unguarded Cells in the Grid
func countUnguarded(m int, n int, guards [][]int, walls [][]int) (ans int) {
	g := make([][]int, m)
	for i := range g {
		g[i] = make([]int, n)
	}
	for _, e := range guards {
		g[e[0]][e[1]] = 2
	}
	for _, e := range walls {
		g[e[0]][e[1]] = 2
	}
	dirs := [5]int{-1, 0, 1, 0, -1}
	for _, e := range guards {
		for k := 0; k < 4; k++ {
			x, y := e[0], e[1]
			a, b := dirs[k], dirs[k+1]
			for x+a >= 0 && x+a < m && y+b >= 0 && y+b < n && g[x+a][y+b] < 2 {
				x, y = x+a, y+b
				g[x][y] = 1
			}
		}
	}
	for _, row := range g {
		for _, v := range row {
			if v == 0 {
				ans++
			}
		}
	}
	return
}

# Accepted solution for LeetCode #2257: Count Unguarded Cells in the Grid
class Solution:
    def countUnguarded(
        self, m: int, n: int, guards: List[List[int]], walls: List[List[int]]
    ) -> int:
        g = [[0] * n for _ in range(m)]
        for i, j in guards:
            g[i][j] = 2
        for i, j in walls:
            g[i][j] = 2
        dirs = (-1, 0, 1, 0, -1)
        for i, j in guards:
            for a, b in pairwise(dirs):
                x, y = i, j
                while 0 <= x + a < m and 0 <= y + b < n and g[x + a][y + b] < 2:
                    x, y = x + a, y + b
                    g[x][y] = 1
        return sum(v == 0 for row in g for v in row)

// Accepted solution for LeetCode #2257: Count Unguarded Cells in the Grid
impl Solution {
    pub fn count_unguarded(m: i32, n: i32, guards: Vec<Vec<i32>>, walls: Vec<Vec<i32>>) -> i32 {
        let m = m as usize;
        let n = n as usize;
        let mut g = vec![vec![0; n]; m];
        for e in &guards {
            g[e[0] as usize][e[1] as usize] = 2;
        }
        for e in &walls {
            g[e[0] as usize][e[1] as usize] = 2;
        }
        let dirs = [-1, 0, 1, 0, -1];
        for e in &guards {
            let (x0, y0) = (e[0] as i32, e[1] as i32);
            for k in 0..4 {
                let (mut x, mut y) = (x0, y0);
                let (a, b) = (dirs[k], dirs[k + 1]);
                while x + a >= 0
                    && x + a < m as i32
                    && y + b >= 0
                    && y + b < n as i32
                    && g[(x + a) as usize][(y + b) as usize] < 2
                {
                    x += a;
                    y += b;
                    g[x as usize][y as usize] = 1;
                }
            }
        }
        let mut ans = 0;
        for row in g {
            for v in row {
                if v == 0 {
                    ans += 1;
                }
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #2257: Count Unguarded Cells in the Grid
function countUnguarded(m: number, n: number, guards: number[][], walls: number[][]): number {
    const g: number[][] = Array.from({ length: m }, () => Array.from({ length: n }, () => 0));
    for (const [i, j] of guards) {
        g[i][j] = 2;
    }
    for (const [i, j] of walls) {
        g[i][j] = 2;
    }
    const dirs: number[] = [-1, 0, 1, 0, -1];
    for (const [i, j] of guards) {
        for (let k = 0; k < 4; ++k) {
            let [x, y] = [i, j];
            let [a, b] = [dirs[k], dirs[k + 1]];
            while (x + a >= 0 && x + a < m && y + b >= 0 && y + b < n && g[x + a][y + b] < 2) {
                x += a;
                y += b;
                g[x][y] = 1;
            }
        }
    }
    let ans = 0;
    for (const row of g) {
        for (const v of row) {
            ans += v === 0 ? 1 : 0;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n)

Space

O(m × n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.