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Count Unguarded Cells in the Grid LeetCode Solution
Problem – Count Unguarded Cells in the Grid LeetCode Solution
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 <= 1052 <= m * n <= 1051 <= guards.length, walls.length <= 5 * 1042 <= guards.length + walls.length <= m * nguards[i].length == walls[j].length == 20 <= rowi, rowj < m0 <= coli, colj < n- All the positions in
guardsandwallsare unique.
Count Unguarded Cells in the Grid LeetCode Solution in C++
class Solution {
public:
int countUnguarded(int m, int n, vector<vector<int>>& g, vector<vector<int>>& w) {
// m is no. of rows, n is no. of columns, g is guards vector and w is walls vector
vector<vector<int>> v(m, vector<int> (n,0));
int k = w.size();
for(int i=0;i<k;i++){
int x = w[i][0], y = w[i][1];
v[x][y] = -2;
}
k = g.size();
for(int i=0;i<k;i++){
int x = g[i][0], y = g[i][1];
v[x][y] = 2;
}
for(int j=0;j<k;j++){
int x = g[j][0], y = g[j][1];
for(int i=x-1;i>=0;i--){ // up
if(v[i][y]==-2 || v[i][y]==2) break;
v[i][y] = 1;
}
for(int i=x+1;i<m;i++){ // down
if(v[i][y]==-2 || v[i][y]==2) break;
v[i][y] = 1;
}
for(int i=y-1;i>=0;i--){ // left
if(v[x][i]==-2 || v[x][i]==2) break;
v[x][i] = 1;
}
for(int i=y+1;i<n;i++){ // right
if(v[x][i]==-2 || v[x][i]==2) break;
v[x][i] = 1;
}
}
int ans = 0;
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
if(!v[i][j]) ans++;
}
}
return ans;
}
};
Count Unguarded Cells in the Grid LeetCode Solution in Python
class Solution:
def countUnguarded(self, m: int, n: int, guards: List[List[int]], walls: List[List[int]]) -> int:
dp = [[0] * n for _ in range(m)]
for x, y in guards+walls:
dp[x][y] = 1
directions = [(0, 1), (1, 0), (-1, 0), (0, -1)]
for x, y in guards:
for dx, dy in directions:
curr_x = x
curr_y = y
while 0 <= curr_x+dx < m and 0 <= curr_y+dy < n and dp[curr_x+dx][curr_y+dy] != 1:
curr_x += dx
curr_y += dy
dp[curr_x][curr_y] = 2
return sum(1 for i in range(m) for j in range(n) if dp[i][j] == 0)
Count Unguarded Cells in the Grid LeetCode Solution in Java
class Solution {
public int countUnguarded(int m, int n, int[][] guards, int[][] walls) {
int[][] res = new int[m][n]; //We create the matrix with the proper dimension
int result = 0, cnti, cntj;
for(int[] i : walls){ //We insert all the walls
res[i[0]][i[1]] = 2;
}
for(int[] i : guards){ //We insert all the guards
res[i[0]][i[1]] = 1;
}
for(int i = 0; i < res.length;i++){
for(int j = 0; j < res[i].length;j++){
if(res[i][j] == 1){ //If we found a guard...
cnti = i; //Position of the guard
cntj = j; //Position of the guard
while(cnti-1 != -1 && res[cnti-1][cntj] != 2 && res[cnti-1][cntj] != 1){ //If we can go up in the matrix...
res[cnti-1][cntj] = 3;
cnti--;
}
cnti = i; //We reset the value to the initial one
while(cnti+1 != m && res[cnti+1][cntj] != 2 && res[cnti+1][cntj] != 1){ //If we can go down in the matrix...
res[cnti+1][cntj] = 3;
cnti++;
}
cnti = i; //We reset the value to the initial one
while(cntj-1 != -1 && res[cnti][cntj-1] != 2 && res[cnti][cntj-1] != 1){ //If we can go to left in the matrix...
res[cnti][cntj-1] = 3;
cntj--;
}
cntj = j; //We reset the value to the initial one
while(cntj+1 != n && res[cnti][cntj+1] != 2 && res[cnti][cntj+1] != 1){ //If we can go to rigth in the matrix...
res[cnti][cntj+1] = 3;
cntj++;
}
}
}
}
for(int[] i : res){ //Once we have marked the correct squares in the matrix...
for(int j : i){
if(j == 0){ //If we find a '0', we add one to the counter as it is an unguarded cell
result++;
}
}
}
return result; //We return the number of unguarded cells
}
}
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