A naive or brute-force approach might involve a multi-pass strategy. First, one might traverse grid2 to identify all distinct islands and store the coordinates of every cell belonging to each island in a list of lists. After collecting all islands, the algorithm would iterate through each stored island and check the coordinates against grid1.

Pseudo-code:

text
1. Create a list `islands` to store sets of coordinates.
2. Iterate through grid2:
    If cell is 1 and not visited:
        Perform DFS/BFS to find the full island.
        Store all (r, c) pairs of this island in `islands`.
3. Initialize count = 0.
4. For each island in `islands`:
    is_sub_island = True
    For each (r, c) in island:
        If grid1[r][c] == 0:
            is_sub_island = False
            Break
    If is_sub_island is True:
        count++
5. Return count.

Time Complexity: $O(m \cdot n)$ because we visit nodes a constant number of times. Space Complexity: $O(m \cdot n)$ to store the coordinates of all islands explicitly.

Why it is suboptimal: While the time complexity is technically linear with respect to the grid size, the space complexity is higher than necessary because we explicitly store island coordinates. Furthermore, this approach requires two distinct phases (collection and validation), which adds unnecessary code complexity. In an interview, we want to validate the sub-island property during the traversal to save space and keep the logic concise.

Core Insight for Count Sub Islands

The key insight for LeetCode 1905 is that we can validate the "sub-island" condition simultaneously while traversing the island in grid2. We do not need to store coordinates.

The condition for a sub-island is strict: ALL cells in the grid2 island must be land in grid1. This implies that if we encounter a single cell $(r, c)$ in the current grid2 island where grid1[r][c] is water (0), the entire island is disqualified.

However, a crucial detail often missed is that we cannot simply stop the traversal (short-circuit) when we find an invalid cell. We must continue the DFS to mark the entire island in grid2 as visited. If we stop early, the unvisited parts of the invalid island will be discovered later by the main loop, erroneously treated as a new island, and potentially counted.

Visualizing the Algorithm: Imagine the recursion tree expanding from the first cell of an island in grid2. As the DFS flows into neighbor cells (up, down, left, right), it effectively "colors" the island to mark it as processed. Each node in this recursion tree performs a local check: "Is there land at this position in grid1?" The results of these checks are aggregated. If the local check fails or any child node reports a failure, the root of the recursion tree knows this island is not a sub-island.

Execution Flow

Let's trace the execution for a specific island in grid2:

1. Main Loop: The scanner finds a 1 at grid2[0][0].
2. DFS Start: We call dfs(0, 0).
3. State Check: Inside the DFS:
   • We check if the current cell in grid1 is 1. If grid1[0][0] is 0, we flag this island as invalid (but continue traversing).
   • We mark grid2[0][0] as 0 to mark it as visited.
4. Recursive Expansion: We attempt to move in 4 directions.
   • Move Right to (0, 1): It's a 1. Recurse dfs(0, 1).
   • Move Down to (1, 0): It's a 0. Stop branch.
5. Aggregation:
   • Suppose dfs(0, 1) encounters a cell where grid1 is 0. It returns false.
   • The parent call dfs(0, 0) receives this false.
   • Even if (0, 0) itself was valid, the false from the neighbor propagates up.
6. Completion: The DFS completes, having flipped all 1s in this component to 0. The final result is returned to the main loop. If true, count increases.

Proof of Correctness

The algorithm relies on the property of Connected Components. By definition, a DFS starting at an unvisited node $u$ will visit every node $v$ reachable from $u$ (i.e., the entire island) exactly once, provided we mark nodes as visited.

The invariant we maintain is: $\text{IsSubIsland} = \bigwedge_{\forall (r,c) \in \text{Island}} (\text{grid1}[r][c] == 1)$

Our DFS implements this logical conjunction (AND operation). If any node fails the check (grid1[r][c] == 0), the conjunction becomes false. By visiting every node in the component, we ensure that the check is exhaustive for that island. By marking nodes as visited immediately, we ensure termination and prevent cycles. Thus, the count is strictly the number of components satisfying the invariant.

Reference Implementation

C++ Solution for LeetCode 1905

cpp
class Solution {
public:
    int m, n;
    
    // DFS returns true if the connected component is a valid sub-island
    bool dfs(vector<vector<int>>& grid1, vector<vector<int>>& grid2, int r, int c) {
        // Boundary checks and water check
        if (r < 0 || r >= m || c < 0 || c >= n || grid2[r][c] == 0) {
            return true;
        }
        
        // Mark as visited in grid2 to avoid cycles and recounting
        grid2[r][c] = 0;
        
        // Check validity for the current cell
        bool isSub = (grid1[r][c] == 1);
        
        // Recurse in all 4 directions.
        // IMPORTANT: We must NOT use short-circuit operators (&&).
        // We need to traverse the entire island to mark all cells as visited.
        bool up = dfs(grid1, grid2, r - 1, c);
        bool down = dfs(grid1, grid2, r + 1, c);
        bool left = dfs(grid1, grid2, r, c - 1);
        bool right = dfs(grid1, grid2, r, c + 1);
        
        // The island is valid only if the current cell is valid AND all parts are valid
        return isSub && up && down && left && right;
    }

    int countSubIslands(vector<vector<int>>& grid1, vector<vector<int>>& grid2) {
        m = grid1.size();
        n = grid1[0].size();
        int count = 0;
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                // If we find an unvisited land cell in grid2
                if (grid2[i][j] == 1) {
                    if (dfs(grid1, grid2, i, j)) {
                        count++;
                    }
                }
            }
        }
        return count;
    }
};

Java Solution for LeetCode 1905

java
class Solution {
    private int m;
    private int n;

    public int countSubIslands(int[][] grid1, int[][] grid2) {
        m = grid1.length;
        n = grid1[0].length;
        int count = 0;

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                // Start DFS if we find unvisited land in grid2
                if (grid2[i][j] == 1) {
                    // If the entire island is valid, increment count
                    if (dfs(grid1, grid2, i, j)) {
                        count++;
                    }
                }
            }
        }
        return count;
    }

    private boolean dfs(int[][] grid1, int[][] grid2, int r, int c) {
        // Base case: check boundaries or if cell is water/visited
        if (r < 0 || r >= m || c < 0 || c >= n || grid2[r][c] == 0) {
            return true;
        }

        // Mark current cell as visited
        grid2[r][c] = 0;

        // Determine if current cell is valid (must be land in grid1)
        boolean isCurrentValid = (grid1[r][c] == 1);

        // Visit all neighbors.
        // We use bitwise AND (&) or separate boolean variables to ensure 
        // DFS executes for ALL directions. Logical AND (&&) would short-circuit.
        boolean up = dfs(grid1, grid2, r - 1, c);
        boolean down = dfs(grid1, grid2, r + 1, c);
        boolean left = dfs(grid1, grid2, r, c - 1);
        boolean right = dfs(grid1, grid2, r, c + 1);

        return isCurrentValid && up && down && left && right;
    }
}

Python Solution for LeetCode 1905

python
class Solution:
    def countSubIslands(self, grid1: List[List[int]], grid2: List[List[int]]) -> int:
        m, n = len(grid1), len(grid1[0])
        
        def dfs(r, c):
            # Base case: out of bounds or water/visited
            if r < 0 or r >= m or c < 0 or c >= n or grid2[r][c] == 0:
                return True
            
            # Mark as visited
            grid2[r][c] = 0
            
            # Check if this cell is valid (must be land in grid1)
            is_valid = (grid1[r][c] == 1)
            
            # Recurse all 4 directions
            # We must execute all DFS calls to ensure the whole island is marked visited.
            res1 = dfs(r + 1, c)
            res2 = dfs(r - 1, c)
            res3 = dfs(r, c + 1)
            res4 = dfs(r, c - 1)
            
            # Return True only if current cell and all connected parts are valid
            return is_valid and res1 and res2 and res3 and res4

        count = 0
        for i in range(m):
            for j in range(n):
                if grid2[i][j] == 1:
                    if dfs(i, j):
                        count += 1
                        
        return count