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Java-based LeetCode algorithm problem solutions, regularly updated
package g1201_1300.s1254_number_of_closed_islands;
// #Medium #Array #Depth_First_Search #Breadth_First_Search #Matrix #Union_Find
// #Graph_Theory_I_Day_2_Matrix_Related_Problems
// #2022_03_12_Time_3_ms_(55.59%)_Space_46.4_MB_(60.84%)
/**
* 1254 - Number of Closed Islands\.
*
* Medium
*
* Given a 2D `grid` consists of `0s` (land) and `1s` (water). An _island_ is a maximal 4-directionally connected group of `0s` and a _closed island_ is an island **totally** (all left, top, right, bottom) surrounded by `1s.`
*
* Return the number of _closed islands_.
*
* **Example 1:**
*
* ![](https://assets.leetcode.com/uploads/2019/10/31/sample_3_1610.png)
*
* **Input:** grid = \[\[1,1,1,1,1,1,1,0],[1,0,0,0,0,1,1,0],[1,0,1,0,1,1,1,0],[1,0,0,0,0,1,0,1],[1,1,1,1,1,1,1,0]]
*
* **Output:** 2
*
* **Explanation:** Islands in gray are closed because they are completely surrounded by water (group of 1s).
*
* **Example 2:**
*
* ![](https://assets.leetcode.com/uploads/2019/10/31/sample_4_1610.png)
*
* **Input:** grid = \[\[0,0,1,0,0],[0,1,0,1,0],[0,1,1,1,0]]
*
* **Output:** 1
*
* **Example 3:**
*
* **Input:** grid = \[\[1,1,1,1,1,1,1],
* [1,0,0,0,0,0,1],
* [1,0,1,1,1,0,1],
* [1,0,1,0,1,0,1],
* [1,0,1,1,1,0,1],
* [1,0,0,0,0,0,1],
* [1,1,1,1,1,1,1]]
*
* **Output:** 2
*
* **Constraints:**
*
* * `1 <= grid.length, grid[0].length <= 100`
* * `0 <= grid[i][j] <=1`
**/
public class Solution {
private int rows;
private int cols;
private boolean isLand;
public int closedIsland(int[][] grid) {
rows = grid.length;
cols = grid[0].length;
int result = 0;
for (int r = 0; r < rows; r++) {
for (int c = 0; c < cols; c++) {
if (grid[r][c] == 0) {
isLand = true;
dfs(grid, r, c);
if (isLand) {
result++;
}
}
}
}
return result;
}
private void dfs(int[][] grid, int r, int c) {
if (r == 0 || c == 0 || r == rows - 1 || c == cols - 1) {
isLand = false;
}
grid[r][c] = 'k';
if (r > 0 && grid[r - 1][c] == 0) {
dfs(grid, r - 1, c);
}
if (c > 0 && grid[r][c - 1] == 0) {
dfs(grid, r, c - 1);
}
if (r < rows - 1 && grid[r + 1][c] == 0) {
dfs(grid, r + 1, c);
}
if (c < cols - 1 && grid[r][c + 1] == 0) {
dfs(grid, r, c + 1);
}
}
}