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-rw-r--r--min_num_days_disconnect_island.cpp56
1 files changed, 34 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
+}