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-rw-r--r--min_num_days_disconnect_island.cpp56
-rw-r--r--min_num_days_disconnect_island.py62
2 files changed, 96 insertions, 22 deletions
diff --git a/min_num_days_disconnect_island.cpp b/min_num_days_disconnect_island.cpp
index f1598c8..992600a 100644
--- a/min_num_days_disconnect_island.cpp
+++ b/min_num_days_disconnect_island.cpp
@@ -1,25 +1,31 @@
+// 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 = islandCount(grid);
- int n = grid.size(), m = grid[0].size();
- if (islands > 1 || islands == 0)
+ 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++) {
+ else
+ for (int i = 0; i < n; i++)
+ for (int j = 0; j < m; j++)
if (grid[i][j]) {
grid[i][j] = 0;
- islands = islandCount(grid);
+ islands = cnt_islands(grid);
grid[i][j] = 1;
- if (islands > 1 || islands == 1)
+ if (islands > 1 || !islands)
return 1;
}
- }
- }
- }
return 2;
}
@@ -28,23 +34,29 @@ private:
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 a = 0; a < 4; a++) {
- int nx = x + dx[a], ny = y + dy[a];
- if (nx >= 0 && ny >= 0 && nx < n && ny < m && !vis[nx][ny] &&
+ 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, nx, grid, vis);
+ dfs(nx, ny, grid, vis);
}
}
- int islandCount(vector<vector<int>> &grid) {
- int count = 0, n = grid.size(), m = grid[0].size();
+ 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++) {
+ 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 count;
+ 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
+}
diff --git a/min_num_days_disconnect_island.py b/min_num_days_disconnect_island.py
new file mode 100644
index 0000000..14dfd44
--- /dev/null
+++ b/min_num_days_disconnect_island.py
@@ -0,0 +1,62 @@
+# 1568. Minimum Number of Days to Disconnect Island
+
+"""
+you are given an 'm * n' binary grid where 1 repansents land and 0 repansents
+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(object):
+ def minDays(self, grid):
+ """
+ :type grid: List[List[int]]
+ :rtype: int
+ """
+ cnt = 0
+ for i in range(len(grid)):
+ for j in range(len(grid[0])):
+ if grid[i][j]:
+ cnt += 1
+ root = (i, j)
+ if cnt <= 1:
+ return cnt
+ vis, low, time, ans = {root}, {}, {}, []
+
+ def articulation_points(curr, parent):
+ low[curr] = time[curr] = len(vis)
+ children = 0
+ i, j = curr
+ for x, y in [(i + 1, j), (i - 1, j), (i, j + 1), (i, j - 1)]:
+ if (x, y) == parent:
+ continue
+ if 0 <= x < len(grid) and 0 <= y < len(grid[0]) and grid[x][y]:
+ if (x, y) not in vis:
+ vis.add((x, y))
+ articulation_points((x, y), curr)
+ low[curr] = min(low[curr], low[(x, y)])
+ children += 1
+ if low[(x, y)] >= time[(curr)] and parent != (-1, -1):
+ ans.append([i, j])
+ else:
+ low[curr] = min(low[curr], time[(x, y)])
+
+ if parent == (-1, -1) and children > 1:
+ ans.append([x, y])
+
+ articulation_points(root, (-1, -1))
+ if len(vis) != cnt:
+ return 0
+ elif ans:
+ return 1
+ else:
+ return 2
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.minDays(grid=[[0, 1, 1, 0], [0, 1, 1, 0], [0, 0, 0, 0]]))
+ print(obj.minDays(grid=[[1, 1]]))