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This approach uses Depth-First Search (DFS) to explore each island in grid2, checking if it matches an area in grid1. We iterate through each cell in grid2, initiating a DFS whenever we encounter land (1). During DFS, we verify if the entire island in grid2 can be found in grid1. If during exploration, we encounter a cell that is 1 in grid2 but 0 in grid1, we mark it as not a sub-island.
Time Complexity: O(m * n), where m and n are the dimensions of the grids, as each cell is visited a constant number of times.
Space Complexity: O(1) additional space not counting input and stack space for recursion.
1class Solution {
2 public void dfs(int[][] grid1, int[][] grid2, int i, int j, boolean[] isSubIsland) {
3 if (i < 0 || i >= grid2.length || j < 0 || j >= grid2[0].length || grid2[i][j] == 0) {
4 return;
5 }
6 if (grid1[i][j] == 0) {
7 isSubIsland[0] = false;
8 }
9 grid2[i][j] = 0;
10 dfs(grid1, grid2, i + 1, j, isSubIsland);
11 dfs(grid1, grid2, i - 1, j, isSubIsland);
12 dfs(grid1, grid2, i, j + 1, isSubIsland);
13 dfs(grid1, grid2, i, j - 1, isSubIsland);
14 }
15
16 public int countSubIslands(int[][] grid1, int[][] grid2) {
17 int count = 0;
18 for (int i = 0; i < grid2.length; i++) {
19 for (int j = 0; j < grid2[0].length; j++) {
20 if (grid2[i][j] == 1) {
21 boolean[] isSubIsland = {true};
22 dfs(grid1, grid2, i, j, isSubIsland);
23 if (isSubIsland[0]) {
24 count++;
25 }
26 }
27 }
28 }
29 return count;
30 }
31
32 public static void main(String[] args) {
33 Solution sol = new Solution();
34 int[][] grid1 = {{1,1,1,0,0},{0,1,1,1,1},{0,0,0,0,0},{1,0,0,0,0},{1,1,0,1,1}};
35 int[][] grid2 = {{1,1,1,0,0},{0,0,1,1,1},{0,1,0,0,0},{1,0,1,1,0},{0,1,0,1,0}};
36 System.out.println(sol.countSubIslands(grid1, grid2));
37 }
38}This Java version implements the DFS strategy to find and verify sub-islands within the grid. The solution ensures that matching islands between grid1 and grid2 are identified correctly.
This approach uses the Union-Find data structure to efficiently group connected islands. By initially identifying all islands in both grids, we can ascertain the sub-islands by ensuring that connected components (islands) in grid2 are completely represented within connected components in grid1.
Time Complexity: O(m * n * (α(m*n))), where α is the Inverse Ackermann function, a very slow-growing function.
Space Complexity: O(m * n) to store the parent and rank arrays.
1 private int[] parent;
private int[] rank;
public UnionFind(int size) {
parent = new int[size];
rank = new int[size];
for (int i = 0; i < size; i++) {
parent[i] = i;
rank[i] = 0;
}
}
public int Find(int x) {
if (parent[x] != x) {
parent[x] = Find(parent[x]);
}
return parent[x];
}
public void Union(int x, int y) {
int rootX = Find(x);
int rootY = Find(y);
if (rootX != rootY) {
if (rank[rootX] > rank[rootY]) {
parent[rootY] = rootX;
} else if (rank[rootX] < rank[rootY]) {
parent[rootX] = rootY;
} else {
parent[rootY] = rootX;
rank[rootX]++;
}
}
}
}
public class Solution {
public int CountSubIslands(int[][] grid1, int[][] grid2) {
int m = grid1.Length, n = grid1[0].Length;
UnionFind uf = new UnionFind(m * n);
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid1[i][j] == 1) {
int index = i * n + j;
if (i + 1 < m && grid1[i + 1][j] == 1) uf.Union(index, (i + 1) * n + j);
if (j + 1 < n && grid1[i][j + 1] == 1) uf.Union(index, i * n + j + 1);
}
}
}
bool[] rootVisited = new bool[m * n];
int count = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid2[i][j] == 1) {
int index = i * n + j;
int rootIndex = uf.Find(index);
if (!rootVisited[rootIndex]) {
bool isSubIsland = true;
var directions = new int[][] { new int[] {0, 1}, new int[] {1, 0}, new int[] {0, -1}, new int[] {-1, 0} };
var stack = new Stack<int[]>();
stack.Push(new int[] {i, j});
while (stack.Count > 0) {
var point = stack.Pop();
foreach (var dir in directions) {
int newX = point[0] + dir[0];
int newY = point[1] + dir[1];
if (newX >= 0 && newX < m && newY >= 0 && newY < n && grid2[newX][newY] == 1) {
grid2[newX][newY] = 0;
if (uf.Find(newX * n + newY) != rootIndex) {
isSubIsland = false;
} else {
stack.Push(new int[] {newX, newY});
}
}
}
}
if (isSubIsland) {
rootVisited[rootIndex] = true;
count++;
}
}
}
}
}
return count;
}
public static void Main(string[] args) {
int[][] grid1 = {
new int[] {1, 1, 1, 0, 0},
new int[] {0, 1, 1, 1, 1},
new int[] {0, 0, 0, 0, 0},
new int[] {1, 0, 0, 0, 0},
new int[] {1, 1, 0, 1, 1}
};
int[][] grid2 = {
new int[] {1, 1, 1, 0, 0},
new int[] {0, 0, 1, 1, 1},
new int[] {0, 1, 0, 0, 0},
new int[] {1, 0, 1, 1, 0},
new int[] {0, 1, 0, 1, 0}
};
Solution sol = new Solution();
Console.WriteLine(sol.CountSubIslands(grid1, grid2));
}
}The C# program utilizes a Union-Find class to group islands and test their containment correspondence between the two grids, culminating in determining the number of sub-islands present.