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authorJack Sangdahl <[email protected]>2025-03-10 00:36:13 -0600
committerJack Sangdahl <[email protected]>2025-03-10 00:36:13 -0600
commit7c60dee83b436e8e8a667bf9f191beced8de4fc3 (patch)
treeeb7a24ee2067f92245c57bb118e85a1baf15df50 /min_num_days_disconnect_island.cpp
parent5d7cdc215e4ba1ec4cbb24c084cccb5e5e641468 (diff)
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chore: move cpp and rust code to directory
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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 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
-}