How do you solve Coloring A Border on LeetCode?

Coloring A Border (LeetCode #1034) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n) time complexity and O(m * n) space complexity. Key insight: Understanding connected components is crucial.

What is the brute force solution for Coloring A Border?

The brute force approach for Coloring A Border is Brute Force, with O(m * n) time complexity and O(m * n) space complexity. The brute force approach involves scanning the entire grid to identify connected components and their borders. We then color the identified border squares. This method is straightforward but can be inefficient for larger grids. The optimal Optimal Solution approach improves this to O(m * n).

What algorithm or data structure is used to solve Coloring A Border?

Coloring A Border uses the following concepts: Array, Depth-First Search, Breadth-First Search, Matrix. The recommended patterns to study are: Depth-First Search, Graph Traversal.

What are the interview tips for Coloring A Border?

Clarify the problem requirements and constraints. Discuss your thought process while coding. Test your solution with edge cases.

What are common mistakes when solving Coloring A Border?

Not marking cells as visited, leading to infinite loops. Confusing border conditions with component conditions.

#1034

Coloring A Border

Medium

ArrayDepth-First SearchBreadth-First SearchMatrixDepth-First SearchGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(m * n)	O(m * n)
Space	O(m * n)	O(m * n)

💡

Intuition

Time O(m * n)Space O(m * n)

The optimal solution uses a similar DFS approach but optimizes the border detection by marking cells during the DFS traversal. This reduces the need for a separate pass to identify borders, making it more efficient.

⚙️

Algorithm

3 steps

1Step 1: Perform a DFS to explore the connected component starting from grid[row][col].
2Step 2: During the DFS, check if a cell is a border cell and mark it accordingly.
3Step 3: After the DFS completes, color all marked border cells with the new color.

solution.py33 lines

1# Full working Python code
2from collections import deque
3
4def colorBorder(grid, row, col, color):
5    m, n = len(grid), len(grid[0])
6    original_color = grid[row][col]
7    visited = [[False] * n for _ in range(m)]
8    border = set()
9
10    def is_border(r, c):
11        if r < 0 or r >= m or c < 0 or c >= n:
12            return True
13        return grid[r][c] != original_color
14
15    def dfs(r, c):
16        if visited[r][c]:
17            return
18        visited[r][c] = True
19        if is_border(r, c):
20            border.add((r, c))
21        for dr, dc in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
22            nr, nc = r + dr, c + dc
23            if 0 <= nr < m and 0 <= nc < n and grid[nr][nc] == original_color:
24                dfs(nr, nc)
25            elif is_border(nr, nc):
26                border.add((r, c))
27
28    dfs(row, col)
29
30    for r, c in border:
31        grid[r][c] = color
32
33    return grid

ℹ

Complexity note: The time complexity remains O(m * n) due to the need to traverse the entire grid. The space complexity is also O(m * n) due to the visited array and border set.

1Understanding connected components is crucial.
2Identifying border cells efficiently can reduce unnecessary computations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.