LeetCode 959 - Regions Cut By Slashes
Let's dive into a full, detailed technical solution guide for LeetCode 959 - Regions Cut By Slashes following your formatting rules. The problem gives us an n x n grid where each cell contains either a forward slash '/', a backslash '', or a blank space ' '.
Difficulty: 🟡 Medium
Topics: Array, Hash Table, Depth-First Search, Breadth-First Search, Union-Find, Matrix
Solution
Let's dive into a full, detailed technical solution guide for LeetCode 959 - Regions Cut By Slashes following your formatting rules.
Problem Understanding
The problem gives us an n x n grid where each cell contains either a forward slash '/', a backslash '\\', or a blank space ' '. Each slash can be seen as dividing the cell into two triangular regions, and a blank space is just a single undivided region. The goal is to determine how many contiguous regions exist in the entire grid after all the slashes have partitioned the space.
The input is provided as a list of strings, where each string represents a row of the grid. The expected output is a single integer representing the number of distinct regions formed.
Constraints indicate that n is small, up to 30, which allows solutions that are quadratic in n in time and space. Each grid element is guaranteed to be '/', '\\', or ' '; therefore, we do not need to handle invalid characters.
Edge cases that might trip up a naive solution include cells with backslashes (escaped as '\\'), blank spaces (which do not divide a cell), and grids with only one cell. Cells on the borders also need careful handling to avoid indexing errors.
Approaches
A naive or brute-force approach would be to try to interpret each cell as a region and try to combine regions manually using DFS or BFS on the characters themselves. However, the challenge is that slashes do not align to a simple grid; a '/' or '\\' divides a cell diagonally, so connectivity is not along the immediate four neighbors. Trying to implement connectivity directly on the original grid is cumbersome and error-prone.
The key insight for an optimal solution is to subdivide each 1 x 1 cell into four triangles, representing top-left, top-right, bottom-left, and bottom-right parts. Each triangle is connected to others depending on the cell’s character:
- For
'/', top-left connects to bottom-right internally, top-right to bottom-left internally. - For
'\\', top-left connects to top-right, bottom-left connects to bottom-right internally. - For
' ', all four triangles in the cell are connected internally.
After creating these internal connections, we connect triangles to neighboring cells’ triangles where appropriate. Once this is done, the problem reduces to counting connected components in a graph, which can be done via DFS, BFS, or Union-Find.
This subdivision approach allows handling slashes cleanly, and the complexity remains manageable because each n x n cell produces exactly 4 nodes.
| Approach | Time Complexity | Space Complexity | Notes |
|---|---|---|---|
| Brute Force | O(n^2 * k) | O(n^2) | Attempting DFS/BFS directly on characters with diagonal checks is messy and error-prone |
| Optimal (4-cell subdivision + Union-Find) | O(n^2 * α(n^2)) | O(n^2) | Subdivide each cell into 4 triangles, connect them based on slashes and neighbors, count connected components using Union-Find; α is the inverse Ackermann function |
Algorithm Walkthrough
- Subdivide each cell into four triangles: For each
grid[i][j], we represent the cell as 4 nodes labeled 0 (top), 1 (right), 2 (bottom), 3 (left). - Connect triangles within a cell:
If the cell contains ' ', connect all four triangles together.
If the cell contains '/', connect top and left triangles together, and bottom and right together.
If the cell contains '\\', connect top and right triangles together, and bottom and left together.
3. Connect triangles to neighboring cells:
Each triangle may connect to the corresponding triangle in an adjacent cell:
- Top triangle connects to bottom triangle of the cell above
- Right triangle connects to left triangle of the cell to the right
- Bottom triangle connects to top triangle of the cell below
- Left triangle connects to right triangle of the cell to the left
- Use Union-Find (Disjoint Set Union) to track connected components:
Each triangle is a node in Union-Find. Union operations are performed whenever triangles are connected, either within a cell or to neighboring cells. 5. Count unique connected components:
After processing the entire grid, the number of unique connected components in Union-Find gives the number of distinct regions.
Why it works: Each triangle represents a minimal indivisible area. By connecting triangles according to slashes and adjacency, we ensure that all triangles in the same region are connected, and no triangles from different regions are connected. Union-Find efficiently counts the number of connected regions.
Python Solution
from typing import List
class Solution:
def regionsBySlashes(self, grid: List[str]) -> int:
n = len(grid)
parent = [i for i in range(4 * n * n)]
def find(x: int) -> int:
if parent[x] != x:
parent[x] = find(parent[x])
return parent[x]
def union(x: int, y: int) -> None:
parent[find(x)] = find(y)
for i in range(n):
for j in range(n):
root = 4 * (i * n + j)
val = grid[i][j]
# internal connections
if val == '/':
union(root + 0, root + 3)
union(root + 1, root + 2)
elif val == '\\':
union(root + 0, root + 1)
union(root + 2, root + 3)
else:
union(root + 0, root + 1)
union(root + 1, root + 2)
union(root + 2, root + 3)
# connections with neighboring cells
if i + 1 < n: # down
union(root + 2, 4 * ((i + 1) * n + j) + 0)
if j + 1 < n: # right
union(root + 1, 4 * (i * n + j + 1) + 3)
return sum(find(x) == x for x in range(4 * n * n))
Explanation:
We first create a Union-Find array representing all triangles. The internal unions connect triangles according to the cell’s character. Neighboring cell connections ensure that adjacent triangles in different cells are correctly connected. Finally, counting the number of unique parents yields the number of regions.
Go Solution
func regionsBySlashes(grid []string) int {
n := len(grid)
parent := make([]int, 4*n*n)
for i := 0; i < 4*n*n; i++ {
parent[i] = i
}
var find func(int) int
find = func(x int) int {
if parent[x] != x {
parent[x] = find(parent[x])
}
return parent[x]
}
union := func(x, y int) {
parent[find(x)] = find(y)
}
for i := 0; i < n; i++ {
for j := 0; j < n; j++ {
root := 4 * (i*n + j)
val := grid[i][j]
if val == '/' {
union(root+0, root+3)
union(root+1, root+2)
} else if val == '\\' {
union(root+0, root+1)
union(root+2, root+3)
} else {
union(root+0, root+1)
union(root+1, root+2)
union(root+2, root+3)
}
if i+1 < n {
union(root+2, 4*((i+1)*n+j)+0)
}
if j+1 < n {
union(root+1, 4*(i*n+j+1)+3)
}
}
}
count := 0
for i := 0; i < 4*n*n; i++ {
if find(i) == i {
count++
}
}
return count
}
Go-specific notes: Slice initialization differs from Python lists, and we must use make for dynamic arrays. The union-find logic is the same, but Go requires explicit function variables for recursion with closures.
Worked Examples
Example 1: grid = [" /","/ "]
Subdividing each cell into 4 triangles, unioning them internally and externally produces two connected components. DFS or Union-Find correctly counts 2.
Example 2: grid = [" /"," "]
Only one diagonal exists, and the blank cell connects its four triangles internally. Union-Find counts a single connected component. Output: 1.
Example 3: grid = ["/\\","\\/"]
Each cell contributes 4 triangles, and internal plus neighbor unions result in 5 distinct regions.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(n^2 * α(n^2)) | Each cell is processed once; union-find operations are nearly constant time |
| Space | O(n^2) | We store 4 nodes per cell |