Formatted question description: https://leetcode.ca/all/1254.html
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:
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:
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 <= 1000 <= grid[i][j] <=1
Related Topics: Depth-first Search
Solution 1. DFS
// OJ: https://leetcode.com/problems/number-of-closed-islands/
// Time: O(MN)
// Space: O(1)
class Solution {
int dirs[4][2] = { {0,1},{0,-1},{-1,0},{1,0} }, M, N;
bool dfs(vector<vector<int>> &G, int x, int y, int id) {
G[x][y] = id;
bool ans = true;
for (auto &[dx, dy] : dirs) {
int a = x + dx, b = y + dy;
bool oob = a < 0 || b < 0 || a >= M || b >= N;
ans = ans && !oob;
if (oob || G[a][b] != 0) continue;
ans = dfs(G, a, b, id) && ans;
}
return ans;
}
public:
int closedIsland(vector<vector<int>>& G) {
int id = 2, ans = 0;
M = G.size(), N = G[0].size();
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (G[i][j] != 0) continue;
if (dfs(G, i, j, id++)) ++ans;
}
}
return ans;
}
};
Java
-
class Solution { final int WATER = -1; final int WHITE = 0; final int GRAY = 1; final int BLACK = 2; public int closedIsland(int[][] grid) { if (grid == null || grid.length == 0 || grid[0].length == 0) return 0; int rows = grid.length, columns = grid[0].length; int insideRows = rows - 1, insideColumns = columns - 1; int[][] colors = new int[rows][columns]; for (int i = 0; i < rows; i++) { for (int j = 0; j < columns; j++) colors[i][j] = grid[i][j] == 1 ? WATER : WHITE; } for (int i = 0; i < rows; i++) { if (colors[i][0] == WHITE) breadthFirstSearch(grid, colors, i, 0); if (colors[i][columns - 1] == WHITE) breadthFirstSearch(grid, colors, i, columns - 1); } for (int i = 1; i < insideColumns; i++) { if (colors[0][i] == WHITE) breadthFirstSearch(grid, colors, 0, i); if (colors[rows - 1][i] == WHITE) breadthFirstSearch(grid, colors, rows - 1, i); } int islandCount = 0; for (int i = 1; i < insideRows; i++) { for (int j = 1; j < insideColumns; j++) { if (colors[i][j] == WHITE) { breadthFirstSearch(grid, colors, i, j); islandCount++; } } } return islandCount; } public void breadthFirstSearch(int[][] grid, int[][] colors, int row, int column) { int rows = grid.length, columns = grid[0].length; int[][] directions = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; Queue<int[]> queue = new LinkedList<int[]>(); colors[row][column] = GRAY; queue.offer(new int[]{row, column}); while (!queue.isEmpty()) { int[] cell = queue.poll(); int curRow = cell[0], curColumn = cell[1]; for (int[] direction : directions) { int newRow = curRow + direction[0], newColumn = curColumn + direction[1]; if (newRow >= 0 && newRow < rows && newColumn >= 0 && newColumn < columns && colors[newRow][newColumn] == WHITE) { colors[newRow][newColumn] = GRAY; queue.offer(new int[]{newRow, newColumn}); } } colors[curRow][curColumn] = BLACK; } } } -
// OJ: https://leetcode.com/problems/number-of-closed-islands/ // Time: O(MN) // Space: O(1) class Solution { int dirs[4][2] = { {0,1},{0,-1},{-1,0},{1,0} }, M, N; bool dfs(vector<vector<int>> &G, int x, int y, int id) { G[x][y] = id; bool ans = true; for (auto &[dx, dy] : dirs) { int a = x + dx, b = y + dy; bool oob = a < 0 || b < 0 || a >= M || b >= N; ans = ans && !oob; if (oob || G[a][b] != 0) continue; ans = dfs(G, a, b, id) && ans; } return ans; } public: int closedIsland(vector<vector<int>>& G) { int id = 2, ans = 0; M = G.size(), N = G[0].size(); for (int i = 0; i < M; ++i) { for (int j = 0; j < N; ++j) { if (G[i][j] != 0) continue; if (dfs(G, i, j, id++)) ++ans; } } return ans; } }; -
# 1254. Number of Closed Islands # https://leetcode.com/problems/number-of-closed-islands/ class Solution: def closedIsland(self, grid: List[List[int]]) -> int: rows, cols = len(grid), len(grid[0]) def dfs(i,j): if 0 <= i < rows and 0 <= j < cols and grid[i][j] != 1: grid[i][j] = 1 dfs(i, j-1) dfs(i, j+1) dfs(i-1, j) dfs(i+1, j) res = 0 for i in range(rows): for j in range(cols): if i == 0 or j == 0 or i == rows-1 or j == cols-1: dfs(i,j) for i in range(rows): for j in range(cols): if grid[i][j] == 0: dfs(i,j) res += 1 return res