Leetcode #1034: Coloring A Border
In this guide, we solve Leetcode #1034 Coloring A Border in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Depth-First Search, Breadth-First Search, Array, Matrix
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
Input: grid = [[1,1],[1,2]], row = 0, col = 0, color = 3
Output: [[3,3],[3,2]]
Python Solution
class Solution:
def colorBorder(
self, grid: List[List[int]], row: int, col: int, color: int
) -> List[List[int]]:
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
m, n = len(grid), len(grid[0])
vis = [[False] * n for _ in range(m)]
dfs(row, col, grid[row][col])
return grid
Complexity
The time complexity is O(V+E). The space complexity is O(V).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.