Welcome to Subscribe On Youtube
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;
}
};
-
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; } } } ############ class Solution { private int m; private int n; private int[][] grid; public int closedIsland(int[][] grid) { m = grid.length; n = grid[0].length; this.grid = grid; int ans = 0; for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { if (grid[i][j] == 0) { ans += dfs(i, j); } } } return ans; } private int dfs(int i, int j) { int res = i > 0 && i < m - 1 && j > 0 && j < n - 1 ? 1 : 0; grid[i][j] = 1; int[] dirs = {-1, 0, 1, 0, -1}; for (int k = 0; k < 4; ++k) { int x = i + dirs[k], y = j + dirs[k + 1]; if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 0) { res &= dfs(x, y); } } return res; } } -
// 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; } }; -
class Solution: def closedIsland(self, grid: List[List[int]]) -> int: m, n = len(grid), len(grid[0]) p = list(range(m * n)) def find(x): if p[x] != x: p[x] = find(p[x]) return p[x] for i in range(m): for j in range(n): if grid[i][j] == 1: continue idx = i * n + j if i < m - 1 and grid[i + 1][j] == 0: p[find(idx)] = find((i + 1) * n + j) if j < n - 1 and grid[i][j + 1] == 0: p[find(idx)] = find(i * n + j + 1) s = [0] * (m * n) for i in range(m): for j in range(n): if grid[i][j] == 0: s[find(i * n + j)] = 1 for i in range(m): for j in range(n): root = find(i * n + j) if not s[root]: continue if i == 0 or i == m - 1 or j == 0 or j == n - 1: s[root] = 0 return sum(s) ############ # 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 -
func closedIsland(grid [][]int) (ans int) { m, n := len(grid), len(grid[0]) dirs := [5]int{-1, 0, 1, 0, -1} var dfs func(i, j int) int dfs = func(i, j int) int { res := 1 if i == 0 || i == m-1 || j == 0 || j == n-1 { res = 0 } grid[i][j] = 1 for k := 0; k < 4; k++ { x, y := i+dirs[k], j+dirs[k+1] if x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 0 { res &= dfs(x, y) } } return res } for i, row := range grid { for j, v := range row { if v == 0 { ans += dfs(i, j) } } } return }