How do you solve Regions Cut By Slashes on LeetCode?

Regions Cut By Slashes (LeetCode #959) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n² * α(n)) time complexity and O(n) space complexity. Key insight: Understanding how slashes divide the grid is crucial.

What is the brute force solution for Regions Cut By Slashes?

The brute force approach for Regions Cut By Slashes is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each cell in the grid to see if it belongs to a new region. We can use a depth-first search (DFS) to explore all connected cells starting from any unvisited cell. The optimal Optimal Solution approach improves this to O(n² * α(n)).

What are the interview tips for Regions Cut By Slashes?

Explain your thought process clearly. Discuss the trade-offs of different approaches. Practice implementing union-find as it’s a common pattern.

What are common mistakes when solving Regions Cut By Slashes?

Not handling the escaped backslashes correctly. Overlooking edge cases with grid boundaries.

#959

Regions Cut By Slashes

Medium

ArrayHash TableDepth-First SearchBreadth-First SearchUnion-FindMatrixUnion-FindGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n² * α(n))
Space	O(1)	O(n)

💡

Intuition

Time O(n² * α(n))Space O(n)

The optimal approach uses a union-find data structure to efficiently group connected regions. By treating each cell as a node, we can quickly unite regions based on the slashes and spaces.

⚙️

Algorithm

3 steps

1Step 1: Initialize a union-find structure to manage the regions.
2Step 2: Traverse each cell in the grid and determine connections based on the characters ('/', '\', ' ').
3Step 3: For each cell, union the appropriate neighboring cells based on the presence of slashes.

solution.py35 lines

1class UnionFind:
2    def __init__(self, n):
3        self.parent = list(range(n))
4        self.rank = [1] * n
5    def find(self, x):
6        if self.parent[x] != x:
7            self.parent[x] = self.find(self.parent[x])
8        return self.parent[x]
9    def union(self, x, y):
10        rootX = self.find(x)
11        rootY = self.find(y)
12        if rootX != rootY:
13            if self.rank[rootX] > self.rank[rootY]:
14                self.parent[rootY] = rootX
15            elif self.rank[rootX] < self.rank[rootY]:
16                self.parent[rootX] = rootY
17            else:
18                self.parent[rootY] = rootX
19                self.rank[rootX] += 1
20
21def regions_by_slashes(grid):
22    n = len(grid)
23    uf = UnionFind(n * n)
24    for i in range(n):
25        for j in range(n):
26            if grid[i][j] == ' ':
27                if i > 0: uf.union(i * n + j, (i - 1) * n + j)
28                if j > 0: uf.union(i * n + j, i * n + (j - 1))
29            elif grid[i][j] == '/':
30                if i > 0: uf.union(i * n + j, (i - 1) * n + (j + 1))
31                if j < n - 1: uf.union(i * n + j, i * n + (j + 1))
32            elif grid[i][j] == '\':
33                if i > 0: uf.union(i * n + j, (i - 1) * n + j)
34                if j > 0: uf.union(i * n + j, i * n + (j - 1))
35    return len(set(uf.find(x) for x in range(n * n)))

ℹ

Complexity note: The time complexity is O(n² * α(n)), where α is the inverse Ackermann function, which grows very slowly, making it nearly constant for practical input sizes. The space complexity is O(n) for the union-find structure.

1Understanding how slashes divide the grid is crucial.
2Union-Find can efficiently manage connected components.

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