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// 1568. Minimum Number of Days to Disconnect Island
#include "leetcode.h"
/*
* you are given an 'm * n' binary grid where 1 represents land and 0 represents
* water. an island is a maximal 4 directionally horizontal or vertical
* connected group of 1's. the grid is said to be connected if we have exactly
* one island otherwise it is disconnected. in one day, we are allowed to change
* any single land cell 1 into a water cell 0. return the minimum number of days
* to disconnect the grid.
*/
class Solution {
public:
int minDays(vector<vector<int>> &grid) {
int islands = cnt_islands(grid), n = grid.size(), m = grid[0].size();
if (islands > 1 || !islands)
return 0;
else
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
if (grid[i][j]) {
grid[i][j] = 0;
islands = cnt_islands(grid);
grid[i][j] = 1;
if (islands > 1 || !islands)
return 1;
}
return 2;
}
private:
vector<int> dx = {1, -1, 0, 0}, dy = {0, 0, 1, -1};
void dfs(int x, int y, vector<vector<int>> &grid, vector<vector<int>> &vis) {
int n = grid.size(), m = grid[0].size();
vis[x][y] = 1;
for (int i = 0; i < 4; i++) {
int nx = x + dx[i], ny = y + dy[i];
if (nx >= 0 and ny >= 0 and nx < n and ny < m and !vis[nx][ny] and
grid[nx][ny])
dfs(nx, ny, grid, vis);
}
}
int cnt_islands(vector<vector<int>> &grid) {
int res = 0, n = grid.size(), m = grid[0].size();
vector<vector<int>> vis(n, vector<int>(m, 0));
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
if (!vis[i][j] && grid[i][j]) {
dfs(i, j, grid, vis);
res++;
}
return res;
}
};
int main() {
Solution obj;
vvd(int) g1 = {{0, 1, 1, 0}, {0, 1, 1, 0}, {0, 0, 0, 0}}, g2 = {{1, 1}};
printf("%d\n", obj.minDays(g1)); // expect: 2
printf("%d\n", obj.minDays(g2)); // expect: 2
}
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