LeetCode 1568: Minimum Number of Days to Disconnect Island
LeetCode 1568 Solution Explanation
Explanation:
To solve this problem, we can simulate disconnecting the island by converting each land cell to water one by one and checking if the grid becomes disconnected. We can use a depth-first search (DFS) to check if the grid is connected after each conversion. We need to consider two special cases - when the grid has only one island and when the grid is already disconnected.
- Count the number of islands in the initial grid.
- If there is only one island, return 0 because the grid is already disconnected.
- Iterate through each land cell in the grid.
- For each land cell, convert it to water and perform a DFS to check if the grid becomes disconnected.
- If the grid becomes disconnected, return the number of days it took to disconnect.
- If no disconnection occurs after trying all land cells, return 1 (at least 1 day is needed to disconnect).
Time complexity: O(m * n * m * n), where m is the number of rows and n is the number of columns in the grid. Space complexity: O(m * n) for the recursive DFS stack.
:
LeetCode 1568 Solutions in Java, C++, Python
class Solution {
public int minDays(int[][] grid) {
int islands = countIslands(grid);
if (islands != 1) {
return 0;
}
int minDays = 2;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] == 1) {
grid[i][j] = 0;
if (countIslands(grid) != 1) {
return 1;
}
grid[i][j] = 1;
}
}
}
return minDays;
}
private int countIslands(int[][] grid) {
// Implement DFS to count the number of islands
}
}Interactive Code Editor for LeetCode 1568
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