1034. Coloring A Border
Given a 2-dimensional grid
of integers, each value in the grid represents the color of the grid square at that location.
Two squares belong to the same connected component if and only if they have the same color and are next to each other in any of the 4 directions.
The border of a connected component is all the squares in the connected component that are either 4-directionally adjacent to a square not in the component, or on the boundary of the grid (the first or last row or column).
Given a square at location (r0, c0)
in the grid and a color
, color the border of the connected component of that square with the given color
, and return the final grid
.
Example 1:
Input: grid = [[1,1],[1,2]], r0 = 0, c0 = 0, color = 3
Output: [[3, 3], [3, 2]]
Example 2:
Input: grid = [[1,2,2],[2,3,2]], r0 = 0, c0 = 1, color = 3
Output: [[1, 3, 3], [2, 3, 3]]
Example 3:
Input: grid = [[1,1,1],[1,1,1],[1,1,1]], r0 = 1, c0 = 1, color = 2
Output: [[2, 2, 2], [2, 1, 2], [2, 2, 2]]
Note:
1 <= grid.length <= 50
1 <= grid[0].length <= 50
1 <= grid[i][j] <= 1000
0 <= r0 < grid.length
0 <= c0 < grid[0].length
1 <= color <= 1000
# @lc code=start
using LeetCode
function color_border(grid::Matrix{Int}, r0::Int, c0::Int, color::Int)
visited = fill(false, size(grid))
cis = CartesianIndices(grid)
dirs = CartesianIndex.(((-1, 0), (1, 0), (0, -1), (0, 1)))
cs = CartesianIndex{2}[]
function dfs(grid::Matrix{Int}, pos)
visited[pos] = true
cnt = 0
for d in dirs
I = pos + d
if I ∈ cis && grid[I] == grid[pos]
cnt += 1
!visited[I] && dfs(grid, I)
end
end
(cnt != 4) && push!(cs, pos)
end
dfs(grid, CartesianIndex(r0, c0))
for I in cs
grid[I] = color
end
grid
end
# @lc code=end
color_border (generic function with 1 method)
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