Introduction

In this blog post, we will explore how to solve the LeetCode problem "1034" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an m x n integer matrix grid, and three integers row, col, and color. Each value in the grid represents the color of the grid square at that location. Two squares are called adjacent if they are next to each other in any of the 4 directions. Two squares belong to the same connected component if they have the same color and they are adjacent. The border of a connected component is all the squares in the connected component that are either adjacent to (at least) a square not in the component, or on the boundary of the grid (the first or last row or column). You should color the border of the connected component that contains the square grid[row][col] with color. Return the final grid. Example 1: Input: grid = [[1,1],[1,2]], row = 0, col = 0, color = 3 Output: [[3,3],[3,2]] Example 2: Input: grid = [[1,2,2],[2,3,2]], row = 0, col = 1, color = 3 Output: [[1,3,3],[2,3,3]] Example 3: Input: grid = [[1,1,1],[1,1,1],[1,1,1]], row = 1, col = 1, color = 2 Output: [[2,2,2],[2,1,2],[2,2,2]] Constraints: m == grid.length n == grid[i].length 1 <= m, n <= 50 1 <= grid[i][j], color <= 1000 0 <= row < m 0 <= col < n

Explanation

Here's the breakdown of the solution:

• Identify Connected Component: Use Depth-First Search (DFS) to traverse and identify all cells belonging to the connected component containing grid[row][col].
• Identify Border: During the DFS, check for border cells. A cell is a border cell if it's on the grid boundary or adjacent to a cell with a different color.

• Color Border: After the DFS, update the color of the identified border cells.

• Runtime Complexity: O(m * n), where 'm' is the number of rows and 'n' is the number of columns.

• Storage Complexity: O(m * n) due to the call stack in the worst-case DFS scenario and the visited set.

Code

    def colorBorder(grid, row, col, color):
    m, n = len(grid), len(grid[0])
    initial_color = grid[row][col]
    visited = set()
    border_cells = []

    def dfs(i, j):
        if (i, j) in visited:
            return

        if i < 0 or i >= m or j < 0 or j >= n or grid[i][j] != initial_color:
            return

        visited.add((i, j))

        is_border = False
        if i == 0 or i == m - 1 or j == 0 or j == n - 1:
            is_border = True
        else:
            neighbors = [(i + 1, j), (i - 1, j), (i, j + 1), (i, j - 1)]
            for ni, nj in neighbors:
                if grid[ni][nj] != initial_color:
                    is_border = True
                    break

        if is_border:
            border_cells.append((i, j))

        dfs(i + 1, j)
        dfs(i - 1, j)
        dfs(i, j + 1)
        dfs(i, j - 1)

    dfs(row, col)

    for i, j in border_cells:
        grid[i][j] = color

    return grid