Skip to content

2343. Count Unguarded Cells In The Grid

Difficulty: Medium

LeetCode Problem View on GitHub

2343. Count Unguarded Cells in the Grid

Medium

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.

 

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 <= rowi, rowj < m
  • 0 <= coli, colj < n
  • All the positions in guards and walls are unique.

Solution

class Solution {
    static class Pair {
        int row, col;
        public Pair(int row, int col) {
            this.row = row;
            this.col = col;
        }
        @Override
        public String toString() {
            return "(" + row + " " + col + ")";
        }
        @Override
        public boolean equals(Object obj) {
            if (this == obj) return true;
            if (obj == null || getClass() != obj.getClass()) return false;
            Pair current = (Pair)(obj);
            return current.row == row && current.col == col;
        }
        @Override
        public int hashCode() {
            return Objects.hash(row, col);
        }
    }
    private HashSet<Pair> bad_cell;
    private HashSet<Pair> wall;
    private HashSet<Pair> guard;
    public int countUnguarded(int n, int m, int[][] guards, int[][] walls) {
        bad_cell = new HashSet<>();
        wall = new HashSet<>();
        guard = new HashSet<>();
        for (int curr[] : guards) guard.add(new Pair(curr[0], curr[1]));
        for (int curr[] : walls) wall.add(new Pair(curr[0], curr[1]));
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (guard.contains(new Pair(i, j))) fill_bad(i, j, n, m);
            }
        }       
        int count = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (!wall.contains(new Pair(i, j)) && !bad_cell.contains(new Pair(i, j)) && !guard.contains(new Pair(i, j))) {
                    count++;
                }
            }
        } 
        return count;
    }

    private void fill_bad(int row, int col, int n , int m) {
        int cr = row, cc = col;
        //up;
        cr--;
        while (cr >= 0) {
            if (!wall.contains(new Pair(cr, cc))) bad_cell.add(new Pair(cr, cc));
            else break;
            if (guard.contains(new Pair(cr, cc))) break;
            cr--;
        }
        //down
        cr = row + 1;
        cc = col;
        while (cr < n) {
            if (!wall.contains(new Pair(cr, cc))) bad_cell.add(new Pair(cr, cc));
            else break;
            if (guard.contains(new Pair(cr, cc))) break;
            cr++;
        }
        //left;
        cr = row;
        cc = col - 1;
        while (cc >= 0) {
            if (!wall.contains(new Pair(cr, cc))) bad_cell.add(new Pair(cr, cc));
            else break;
            if (guard.contains(new Pair(cr, cc))) break;
            cc--;
        }
        //right;
        cr = row;
        cc = col + 1;
        while (cc < m) {
            if (!wall.contains(new Pair(cr, cc))) bad_cell.add(new Pair(cr, cc));
            else break;
            if (guard.contains(new Pair(cr, cc))) break;
            cc++;
        }
    }
}

Complexity Analysis

  • Time Complexity: O(?)
  • Space Complexity: O(?)

Approach

Detailed explanation of the approach will be added here