Problem Description

The problem presents a two-dimensional grid which consists of both land (0s) and water (1s). An island is defined as a group of land cells (or 0s) that are connected horizontally or vertically (4-directionally). A closed island is an island that is completely surrounded by water (1s), meaning it does not touch the edge of the grid at any point.

The goal is to count how many closed islands are present in the grid. For example, a 0 that is at the edge of the grid could be part of an island, but since it is touching the boundary, it cannot be considered a closed island.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The grid in the problem is akin to a graph where each cell functions as a node, and connections exist between vertically and horizontally adjacent cells.

Is it a tree?

• No: The grid is not inherently hierarchical and can contain multiple isolated groups of cells (islands), which do not form a single spanning tree.

Is the problem related to directed acyclic graphs (DAGs)?

• No: The issue revolves around finding closed structures, which is unrelated to the properties of directed acyclic graphs.

Is the problem related to shortest paths?

• No: The challenge involves identifying regions (closed islands), not determining the shortest path between points.

Does the problem involve connectivity?

• Yes: The core task is to discern and count distinct, fully enclosed land areas, which is a connectivity issue.

Does the problem have small constraints?

• Yes: Given typical grid-based problems like this, the constraints tend to be manageable, allowing for exhaustive searches.

Conclusion: The flowchart implies using DFS or backtracking for this problem due to the need for exploring all possible paths in relatively small constraints, making Depth-First Search (DFS) perfect for traversing each potential closed island thoroughly.

Intuition

To solve this problem, we can perform a depth-first search (DFS) on each cell that is land (0). When we perform a DFS, we recursively visit each connected land cell and mark them as water (1) to avoid revisiting them. While performing the DFS, we also check if we ever touch the boundary of the grid (i.e., the row index i is 0 or m-1 or the column index j is 0 or n-1). If the DFS starts from a cell that never touches the boundary, it means we are on a closed island, and thus we should count it.

The function dfs(i, j) is designed to perform this task, where i and j are the row and column indices, respectively. This function marks the current cell as visited by setting grid[i][j] to 1. It then checks in all four directions and calls itself (the dfs function) on any connected land cell it finds. If any recursive call to dfs encounters the boundary, it propagates back a failure to recognize the island as closed by returning 0 (when bitwise & operator is applied). If an island is completely surrounded by water and doesn't touch the boundary at any point, dfs returns 1.

The pairwise(dirs) function combined with the surrounding for loop is a slight mistake in the code, possibly due to a misunderstanding of Python's itertools.pairwise, which doesn’t apply here. Instead, it should loop over pairs of directions (like (dirs[k], dirs[k + 1]) for k in range 4) to check all four neighboring cells.

The final count of closed islands is obtained by initializing a sum with a generator expression that iterates over every cell in the grid and calls the dfs function if a land cell (0) is found. The result of dfs will only be 1 for cells that are part of a closed island, and thus the sum represents the total number of closed islands in the grid.

Note: The code provided seems to have a mistake with the pairwise(dirs) portion, which does not align with standard Python functions. The intention here might be to iterate over the directions in pairs to traverse up, right, down, and left in the grid.

Learn more about Depth-First Search, Breadth-First Search and Union Find patterns.

Solution Approach

The solution applies the depth-first search (DFS) algorithm to explore the grid and identify islands that are closed. Here's a step-by-step breakdown of the implementation:

1. Initialization: A DFS function is defined, which will be used to explore the grid. Additionally, the size of the grid is noted in variables m (for rows) and n (for columns). The dirs array (which should contain pairs of directions to move up, right, down, and left) is used to facilitate the exploration of neighboring cells.

2. Depth-First Search (DFS): The core of the solution is in the dfs function, which takes a cell (i, j) and:

   • Marks it as visited by setting grid[i][j] = 1.
   • Checks if the cell is within the boundary but not on it, meaning 0 < i < m - 1 and 0 < j < n - 1.
   • Iterates in all four directions based on the dirs array, and if an adjoining cell is unvisited land (grid[x][y] == 0), it recurs with these new coordinates.
   • If the recursive exploration touches the grid boundary, it will return 0, indicating this is not a closed island. If the exploration completes without touching the boundary, it returns 1, indicating a closed island.

3. Counting Closed Islands: The main function now initializes a sum with a generator expression that loops over each cell in the grid. It calls dfs on cells that are land (0) and adds up the results. Each 1 returned by dfs denotes a closed island.

Here, it is worth noting that the "Reference Solution Approach" mentioned Union find. While the given solution does not utilize this pattern, it's relevant to acknowledge that Union find is another approach that can solve this problem. Union find, or disjoint-set data structure, would allow us to group land cells into disjoint sets and then count sets that do not touch the boundary of the grid, which effectively would identify closed islands. However, in the provided solution, DFS is used instead, which is an equally valid method for solving this problem.

The effectiveness of the DFS approach lies in its ability to mark and visit each cell only once, ensuring that the solution is efficient, with the time complexity being O(m*n), where m is the number of rows and n is the number of columns in the grid. The DFS algorithm is a natural fit for problems related to navigating and exploring grids, particularly in searching for connected components, such as islands in this context.

Example Walkthrough

Let's take a small grid as an example to illustrate how the solution approach works. Consider the following grid:

1 1 1 1 0 1
1 0 0 1 0 1
1 1 1 1 1 1
1 1 1 0 0 1
1 0 0 1 1 1
1 1 1 1 1 1

In this grid, 1 represents water and 0 represents land. We want to find how many closed islands (areas of 0s that do not touch the grid's boundary) there are.

1. We start by defining a dfs function and initializing the grid size, m and n, respectively. In this case, m = 6 (rows) and n = 6 (columns).

2. We iterate over each cell in the grid. When we come across a land cell (0), we start the DFS process from it by calling our dfs function.

3. We call the dfs function on the land cell in the second row, second column (grid[1][1]). The cell is marked as visited by setting it to 1. DFS checks the neighboring cells in four directions (up, down, left, right) based on a pair of directions.

4. The DFS explores all connected land cells. For this example, it would change the following cells to 1:

1 1 1 1 0 1
1 1 1 1 0 1
1 1 1 1 1 1
1 1 1 0 0 1
1 0 0 1 1 1
1 1 1 1 1 1

5. The DFS does not reach the boundary for this group of land cells, so it counts as one closed island.

6. The next unvisited land cell initiates another DFS call which is on the fourth row, fourth column (grid[3][3]). The process is similar; it explores connected land cells:

1 1 1 1 0 1
1 1 1 1 0 1
1 1 1 1 1 1
1 1 1 1 1 1
1 0 0 1 1 1
1 1 1 1 1 1

7. However, this time, the DFS touches the boundary at cell (grid[4][1]), so this island is not closed.

8. The final unvisited land cell group starts at grid[4][1] and is similarly adjacent to the boundary.

At the end of this process, we see that only the first group of land cells (grid[1][1]) constitutes a closed island. Thus, the final count is 1.

The DFS algorithm efficiently identified areas that constitute closed islands, ensuring no cell is visited more than once, resulting in a time complexity of O(m*n). While the pairwise(dirs) part in the example content is confusing, the logic of the DFS here applies the correct four-directional checks by iterating over each direction pair to look at neighboring cells.

Solution Implementation

Time and Space Complexity

The given Python code is a solution for finding the number of "closed islands" in a 2D grid, where 0 represents land and 1 represents water. A "closed island" is one that is surrounded by water and doesn't touch the edge of the grid. The algorithm uses Depth-First Search (DFS) to explore the grid.

Time Complexity

The time complexity of the code is O(m * n), where m is the number of rows and n is the number of columns in the grid. This is because the algorithm potentially visits each cell in the grid once. The DFS is started only if the current cell is a 0 (land), and then it marks every visited land cell as 1 (water) to prevent revisiting. In the worst case, all cells in the grid will be traversed once.

Space Complexity

The space complexity of the DFS is O(m * n) in the worst case. This might happen when the depth of the recursion stack grows proportionally to the number of cells in the worst-case scenario, such as when the grid forms a large area of contiguous land. The space is used by the recursion call stack during the depth-first search.

However, if we consider that modern Python implementations use a technique called "tail-recursion elimination" in some cases, the space complexity can be effectively less than O(m * n) for some grids because not all calls will result in an additional stack frame. But in the analysis of DFS, it is conventional to consider the worst-case scenario regarding recursion.

The additional space used by the dirs variable and a few integer variables is negligible compared to the recursive stack space, thus not affecting the overall space complexity.

Learn more about how to find time and space complexity quickly using problem constraints.