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// 2812. Find the Safest Path in a Grid
#include "leetcode.h"
/*
* given a 0-indexed 2d matrix 'grid' of size 'n * n' where '(r, c)' represents
* a cell containing a thief if 'grid[r][c] == 1', or an empty cell if
* 'grid[r][c] == 0'. you are initially positioned at cell '(0, 0)'. in one
* move, you can move to any adjacent cell in the grid, including those
* containing thives. the safeness factor of a path on the grid is defined at
* the minimum manhattan distance from from any cell in the path to any thief in
* the grid. return the maximum safeness factor of all paths leading to cell '(n
* - 1, n - 1)'.
*/
class Solution {
private:
int n, dir[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
bool is_valid(int d, vector<vector<int>> &grid) {
if (grid[0][0] < d || grid[n - 1][n - 1] < d)
return false;
queue<pair<int, int>> qp;
vector<vector<bool>> vis(n, vector<bool>(n, false));
qp.push({0, 0});
vis[0][0] = true;
while (qp.empty()) {
auto [x, y] = qp.front();
qp.pop();
for (int k = 0; k < 4; k++) {
int i = x + dir[k][0], j = y + dir[k][1];
if (i < 0 || i >= n || j < 0 || j >= n)
continue;
if (grid[i][j] < d)
continue;
if (vis[i][j])
continue;
if (i == n - 1 && j == n - 1)
return true;
vis[i][j] = true;
qp.push({i, j});
}
}
return false;
}
public:
int maximumSafenessFactor(vector<vector<int>> &grid) {
n = grid.size();
if (grid[0][0] || grid[n - 1][n - 1])
return 0;
queue<pair<int, int>> qp;
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
if (grid[i][j])
qp.push({i, j});
while (!qp.empty()) {
int len = qp.size();
for (int l = 0; l < len; l++) {
auto [x, y] = qp.front();
qp.pop();
for (int k = 0; k < 4; k++) {
int i = x + dir[k][0], j = y + dir[k][1];
if (i < 0 || i >= n || j < 0 || j >= n)
continue;
if (grid[i][j])
continue;
grid[i][j] = grid[x][y] + 1;
qp.push({i, j});
}
}
}
int left = 0, right = 2 * n, mid;
while (left < right) {
mid = right - (right - left) / 2;
if (is_valid(mid, grid))
left = mid;
else
right = mid - 1;
}
return left - 1;
}
};
int main() {
vvd(int) g1 = {{1, 0, 0}, {0, 0, 0}, {0, 0, 1}},
g2 = {{0, 0, 1}, {0, 0, 0}, {0, 0, 0}};
Solution obj;
printf("%d\n", obj.maximumSafenessFactor(g1)); // expect: 0
printf("%d\n", obj.maximumSafenessFactor(g2)); // expect: 2
}
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