diff options
Diffstat (limited to 'min_num_days_disconnect_island.cpp')
-rw-r--r-- | min_num_days_disconnect_island.cpp | 56 |
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 +} |