Solution Explanation

This problem asks us to color the border of a connected component in a grid. A connected component is a group of cells with the same color that are adjacent (horizontally or vertically). The border consists of cells that are either adjacent to a cell of a different color or are on the edge of the grid.

The most efficient approach is using Depth-First Search (DFS). We perform a DFS to traverse the connected component starting from the given cell (row, col). During the traversal, we mark visited cells using a vis array. Crucially, we identify border cells while exploring. A cell is a border cell if:

1. It's adjacent to a cell outside the connected component (different color).
2. It's on the boundary of the grid (first or last row/column).

We mark these border cells with the new color during the DFS.

Algorithm

1. Initialization: Create a vis matrix (boolean array) of the same dimensions as the grid to track visited cells. Initialize all cells to false.

2. DFS Traversal: Implement a recursive DFS function:

   • It takes the row index i, column index j, and the original color c of the connected component as input.
   • Mark the current cell (i, j) as visited (vis[i][j] = True).
   • Explore the four adjacent cells:
     • If an adjacent cell is within the grid bounds:
       • If the adjacent cell is not visited:
         • If the adjacent cell has the same color c, recursively call DFS on it.
         • If the adjacent cell has a different color, it's a border cell; change the current cell's color to color.
     • If an adjacent cell is outside the grid bounds, it's a border cell; change the current cell's color to color.

3. Start DFS: Call the DFS function starting from the given (row, col).

4. Return: Return the modified grid.

Time and Space Complexity Analysis

• Time Complexity: O(M*N), where M and N are the dimensions of the grid. In the worst case, we visit every cell in the grid.

• Space Complexity: O(MN) in the worst case. This is due to the vis matrix used to track visited cells. The recursion depth of the DFS can also reach O(MN) in the worst-case scenario (a fully connected component). However, in practice, the space used by the recursive call stack is often smaller than the vis matrix.

Code Implementation (Python)

from itertools import pairwise
 
class Solution:
    def colorBorder(self, grid: List[List[int]], row: int, col: int, color: int) -> List[List[int]]:
        m, n = len(grid), len(grid[0])
        vis = [[False] * n for _ in range(m)]
        def dfs(i: int, j: int, c: int) -> None:
            vis[i][j] = True
            for a, b in pairwise((-1, 0, 1, 0, -1)):
                x, y = i + a, j + b
                if 0 <= x < m and 0 <= y < n:
                    if not vis[x][y]:
                        if grid[x][y] == c:
                            dfs(x, y, c)
                        else:
                            grid[i][j] = color
                else:
                    grid[i][j] = color
        dfs(row, col, grid[row][col])
        return grid

The code in other languages (Java, C++, Go, TypeScript) follows the same algorithmic approach, differing only in syntax and data structure implementations. The pairwise function in Python efficiently iterates through the four adjacent cells. Other languages may use arrays or loops to achieve the same.