1034. Coloring A Border

Problem Description

You are given a matrix of integers where each integer represents a color. You need to identify a connected component in the matrix based on a starting cell given by its row and column indexes. A connected component entails all adjacent cells (up, down, left, or right) with the same color. The border of this connected component includes all cells in the component adjacent to cells with different colors or cells lying on the grid's boundary.

Your task is to recolor only the cells on the border of this connected component with a new color. The index of the starting cell within the connected component and the new color are given. The updated grid should be returned, showing the new border color while leaving the interior of the connected component and the rest of the grid unchanged.

Intuition

The solution to this problem uses Depth-First Search (DFS), an algorithm well-suited for exploring and marking the structure of a connected component within a grid. We begin at a cell (grid[row][col]) and look for all adjacent cells that share the same color, which signifies that they are part of the same connected component.

The intuition behind using DFS is to traverse through all the cells in the connected component starting from the specified cell (grid[row][col]). As we do this, we need to determine which cells are on the border of the component. A border cell is defined as a cell that:

1. Is on the edge of the grid, or
2. Is adjacent to a cell that has a different color than the one we started with.

The DFS function checks all four directions around a cell. For every cell that we visit:

• If the cell is within the grid and the adjacent cell is not visited yet, we check:
  • If the adjacent cell has the same color, we continue DFS on that cell since it's part of the same component.
  • If the adjacent cell has a different color, we know that we are on the border, and thus we change the color of the current cell.
• If the cell is on the edge of the grid, it's a border cell, so we change the color of the current cell.

We iterate in this manner until all the cells in the connected component have been visited and the appropriate border cells have been recolored. An auxiliary vis matrix is used to keep track of visited cells to avoid processing a cell more than once, which helps in preventing an infinite loop.

Finally, we return the modified grid with the newly colored borders, while the interior of the connected component and all other parts of the grid remain unchanged.

Learn more about Depth-First Search and Breadth-First Search patterns.

Solution Approach

The solution utilizes a Depth-First Search (DFS) approach to explore the grid and identify border cells of the connected component. The following are the key aspects of the implementation:

1. DFS Method: The solution defines a nested dfs function which is responsible for the recursive search. It accepts the current cell's indices i and j, as well as the original color c of the connected component.

2. Tracking Visited Cells: A two-dimensional list vis is created with the same dimensions as the input grid grid, initialized with False values. This list is used to keep track of cells that have already been visited to prevent infinite recursion and repeated processing.

3. Checking Neighbors: In each call to the dfs function, it iterates over the four adjacent cells (up, down, left, right) using a for-loop and the pairwise helper to generate pairs of directions.

4. Edge Conditions: The algorithm checks if a neighboring cell is within the boundaries of the grid. If it is outside, it means the current cell is on the border of the grid and thus should be recolored.

5. Coloring Border Cells:

   • If the adjacent cell is within the grid and hasn't been visited (vis[x][y] is False), there are two conditions:
     • If the adjacent cell has the same color as the original (grid[x][y] == c), the dfs function is called recursively for that cell to continue the component traversal.
     • If the adjacent cell has a different color, the current cell is on the border of the connected component and its color is changed to color.

6. Calling DFS: After initializing the vis grid, the dfs function is invoked starting from the cell grid[row][col], using its color as the reference for the connected component.

7. Returning the Result: Once the DFS traversal is complete and all border cells are recolored, the modified grid is returned.

The implementation creatively uses recursion to deeply explore the connected component and the auxiliary vis matrix to manage state across recursive calls. The determination of border cells is carried out in a granular fashion by checking each neighbor, which effectively allows the coloring to occur only where it is required.

Example Walkthrough

Let's consider a small 4x4 matrix and walk through the solution approach with a given starting cell and new color:

1Grid:
21 1 2 3
31 1 2 4
45 1 3 4
56 5 5 4
6
7Starting Cell: (1, 1) (Zero-indexed, meaning the cell in row 1, column 1)
8Original Color of Starting Cell: 1
9New Border Color: 9

First, we initialize an auxiliary 2D list vis of the same size as the grid with False values to keep track of whether a cell has been visited. Next, we define a dfs function to explore the grid starting from the given cell.

1. We call the dfs function on the starting cell (1, 1) with its color 1. Since it matches the original color, we continue with DFS. The vis grid will be updated with True at this cell to denote it's been visited.

2. The dfs function now checks adjacent cells (1, 0), (1, 2), (0, 1), and (2, 1).

3. Whenever visiting an adjacent cell, the border conditions are checked to ensure we are within the grid. Cells at (1, 0) and (1, 2) have the same color as the original and are within the grid; they're not border cells, and the dfs is called on them.

4. Cells (0, 1) and (2, 1) are also part of the connected component, so DFS continues. Meanwhile, vis is continuously updated.

5. By the time we check (0, 1), we find that the cell (0, 0) is of a different color. This means (0, 1) is a border cell because it's adjacent to a differently colored cell.

6. The cells on the actual edge of the matrix are by default border cells, hence their colors are changed to the new border color 9.

7. As we continue this process recursively, all border cells are identified and changed. For this example, the border consists of cells (1, 0), (0, 1), (1, 2), and (2, 1).

8. After all recursion steps complete, the grid is updated as follows:

11 9 2 3
29 1 9 4
35 9 3 4
46 5 5 4

9. The dfs call terminates, and we return the modified grid with the borders recolored to 9.

The steps in the dfs ensure that all cells that are in the connected component of the original color are explored, but only the actual border cells are recolored as per the conditions. This is effective at preserving the original structure of the component while updating the border as specified.

Solution Implementation

Time and Space Complexity

The provided Python code performs a depth-first search (DFS) to color the borders of a connected component in a grid based on the given conditions. The complexity analysis is as follows:

Time Complexity:

The time complexity of this DFS algorithm is mainly determined by the number of cells in the grid that are visited. In the worst case, the algorithm will visit every cell in the grid once. There are m * n cells, where m is the number of rows and n is the number of columns in the grid. Therefore, the DFS will have a time complexity of O(m * n).

Space Complexity:

The space complexity includes the memory taken by the vis matrix to keep track of visited cells, as well as the recursion call stack for DFS. The vis matrix has the same dimensions as the input grid, so it takes O(m * n) space.

The recursion depth can go as deep as the maximum size of the connected component in the grid. In the worst case, this could be the entire grid, which would mean that the call stack can potentially grow up to m * n calls deep. Hence, the space complexity due to the recursion stack could also be O(m * n).

Therefore, the overall space complexity is O(m * n) as well.

Note: The provided code contains a pairwise function which is not defined within the code snippet or as part of the standard Python library up to the knowledge cutoff date. Assuming it's intended to generate pairs of coordinates for traversing adjacent cells, its behavior seems to be that of incorrectly iterating over the grid coordinates without actually yielding pairs.

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