Problem Description

In this problem, we are given a m x n matrix called board, composed of characters 'X' and 'O'. The goal is to identify all the regions formed by 'O' that are completely surrounded by 'X' in all four directions (up, down, left, and right) without any breaks. Once these regions are identified, we need to flip all the 'O's in those regions to 'X's. It's important to note that an 'O' on the boundary of the board isn't considered surrounded, and hence, the regions connected to an 'O' on the board boundary are safe and should not be flipped.

Flowchart Walkthrough

Let's use the algorithm flowchart to navigate through the problem of Surrounded Regions in Leetcode 130, using the Flowchart. Here is the step-by-step analysis to determine the appropriate algorithm:

Is it a graph?

• Yes: The board in the problem can be visualized as a graph where each cell is a node and edges connect adjacent cells.

Is it a tree?

• No: The board isn't inherently hierarchical or branched out like a tree. It potentially has loops and is more grid-structured.

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

• No: The problem does not involve processing or organizing data in a manner characteristic of DAGs. It's about identifying enclosed regions.

Is the problem related to shortest paths?

• No: We are not trying to find the shortest path between points; the goal is to identify and modify surrounded regions.

Does the problem involve connectivity?

• Yes: The core of the problem is to determine which 'O' cells are connected to the boundary and hence not fully surrounded.

Since the follow-up questions for connectivity involving weighted graphs and disjoint set union are not applicable (the grid/graph is unweighted and there’s no union-find specific operations required), we stop at connectivity.

Conclusion: The flowchart pinpoints the use of Depth-First Search (DFS) as it efficiently allows us to explore each 'O' cell connection to edges from any starting point on the grid. DFS is optimal for marking connected regions starting from boundary 'O's to avoid conversion to 'X'.

Intuition

The solution is based on the idea that an 'O' region would not be captured if it's connected directly or indirectly to an 'O' on the boundary of the board since it cannot be surrounded entirely by 'X's. To implement this, we apply depth-first search (DFS) to mark all 'O's that are connected to the boundary 'O's with a temporary marker (in this case, '.'). This multi-step approach ensures we only flip those 'O's into 'X's that are truly surrounded:

1. Identify Boundary-Connected 'O's: Iterate over the border rows and columns. For any 'O' found on the boundary, perform a DFS search to mark not only this 'O' but also any other 'O' connected to this one, directly or indirectly. We replace these 'O's temporarily with '.' to indicate they're safe from flipping.

2. Capture Surrounded Regions: After the DFS marking step, we go through the entire board to flip the remaining 'O's (those that are not marked with '.') into 'X's as they are surrounded by 'X's.

3. Restore Boundary-Connected Regions: Finally, we make another pass to convert all temporary markers '.' back into 'O's to restore the initial state for all the 'O's that were connected to the board boundary.

By taking this approach, we ensure to flip only those 'O's that are truly surrounded by 'X's, while preserving the boundary-connected 'O's.

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

Solution Approach

The solution implements a depth-first search (DFS) algorithm to modify the matrix in place. Here's a walk-through of the code and the algorithms/data structures/patterns used in the solution:

1. Depth-First Search (DFS) Algorithm: The dfs function is a classic implementation of the DFS algorithm. When it encounters an 'O', it changes that 'O' to a temporary marker '.' to indicate that it has been visited and should not be flipped later. Then, it checks adjacent cells (in all four directions) and recursively calls itself to mark connected 'O's.

   def dfs(i, j):
       board[i][j] = '.'
       for a, b in [[0, -1], [0, 1], [1, 0], [-1, 0]]:
           x, y = i + a, j + b
           if 0 <= x < m and 0 <= y < n and board[x][y] == 'O':
               dfs(x, y)

2. Boundary Cells Check: We iterate over the matrix looking for 'O's on the boundary edges. For every 'O' found, the dfs function is called to start the marking process.

   for i in range(m):
       for j in range(n):
           if board[i][j] == 'O' and (
               i == 0 or i == m - 1 or j == 0 or j == n - 1
           ):
               dfs(i, j)

3. Flipping Surrounded 'O's Inside the Matrix: After marking all the boundary-connected 'O's with '.', we iterate over the matrix again. This time, we flip the unmarked 'O's to 'X's because they're surrounded by 'X's. The marked '.' cells retain information about boundary-connected 'O's and are not flipped.

   for i in range(m):
       for j in range(n):
           if board[i][j] == 'O':
               board[i][j] = 'X'

4. Restoring the Boundary-Connected 'O's: In the last iteration over the matrix, we revert our temporary markers '.' back to 'O's, restoring their original state as they were not supposed to be flipped.

   for i in range(m):
       for j in range(n):
           if board[i][j] == '.':
               board[i][j] = 'O'

In this implementation, no additional data structures are needed as we modify the board in place. The pattern used here mainly revolves around the DFS algorithm, which is a powerful tool for searching or traversing through an adjacency graph, or in this case, a matrix, especially when we are dealing with connected components.

The solution has linear complexity with respect to the number of cells in the matrix since each cell is visited at most twice. Once during the DFS marking stage and once during the flipping stage. This ensures an efficient solution to the problem.

Example Walkthrough

Let's consider a small 3x4 board as an example:

Board:
O X X X
X O O X
X X X X

Step 1: Identify Boundary-Connected 'O's

We find that the 'O' in the top left corner is on the boundary. Applying the DFS algorithm starting from this 'O', we change it to a '.' to mark it as visited and safe:

Board:
. X X X
X O O X
X X X X

Since there are no other 'O's directly connected to the boundary 'O's, our board remains the same after the first step.

Step 2: Capture Surrounded Regions

Now, we scan through the rest of the board. The 'O's in the middle are surrounded by 'X's and there is no DFS path from any boundary 'O' to them. So we flip these 'O's to 'X's:

Board:
. X X X
X X X X
X X X X

Step 3: Restore Boundary-Connected Regions

Finally, we need to revert the '.' back to 'O', since it was marked only for the purpose of identification and not flipping:

Board:
O X X X
X X X X
X X X X

Our final board shows the 'O's in the middle flipped to 'X's, while the 'O' on the boundary remains intact. This concludes our walkthrough of how the depth-first search algorithm can be used to solve the problem of flipping all 'O's surrounded by 'X's on a board, while leaving the boundary-connected 'O's untouched.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of this algorithm is O(M*N), where M is the number of rows in the board and N is the number of columns. This is because the algorithm must visit every cell in the grid to perform the depth-first search (DFS) and the subsequent replacement of 'O's with 'X's.

DFS is invoked on cells that are on the border of the grid and are marked with 'O'. In the worst case, all border cells could be 'O's and the DFS would travel into every reachable 'O' from the border which can potentially cover the whole grid. Hence, the complexity considering the entire DFS calls collectively is linear with the number of cells, O(M*N).

After the DFS, the algorithm goes through all cells to replace remaining 'O's with 'X's and convert all temporary '.' markers back to 'O's. This is a straightforward double for-loop over the cells, which again results in a complexity of O(M*N).

Putting this together, despite having nested loops, the DFS does not visit cells more than once. Overall, the time complexity is O(M*N) as each cell is entered only once by the DFS.

Space Complexity

The space complexity of the algorithm is also O(M*N) in the worst-case scenario, due to the recursive nature of DFS. If the grid is filled with 'O's, the call stack for the DFS can grow as large as the number of cells in the grid before backtracking. This would be the case in a grid where one 'O' is connected to all other 'O's, thus the entire grid would be traversed without backtracking.

It is important to note that the space complexity assumes that the depth of the recursion stack is considered part of the space complexity. If the space used by the recursion stack is not considered, then the space complexity would be O(1) since the algorithm modifies the input grid in place and does not utilize additional space proportional to the input size.

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