An n x n grid is composed of 1 x 1 squares where each 1 x 1 square consists of a '/', '\', or blank space ' '. These characters divide the square into contiguous regions.
Given the grid grid represented as a string array, return the number of regions.
Note that backslash characters are escaped, so a '\' is represented as '\\'.
Example 1:
Input: grid = [" /","/ "] Output: 2
Example 2:
Input: grid = [" /"," "] Output: 1
Example 3:
Input: grid = ["/\\","\\/"] Output: 5 Explanation: Recall that because \ characters are escaped, "\\/" refers to \/, and "/\\" refers to /\.
Constraints:
n == grid.length == grid[i].length1 <= n <= 30grid[i][j] is either '/', '\', or ' '.In this approach, each 1x1 square is divided into four smaller triangles. The / and \ slashes correspond to boundaries between these triangles. We can represent these triangles using a graph and use DFS to count the number of connected components in this graph. Each connection between triangles denotes a passage within the same region.
The implementation involves creating a larger 3n x 3n grid where each grid cell is expanded into a 3x3 grid based on present slashes. We then perform a DFS for each unvisited triangle marking it visited, effectively counting the disconnected regions in the grid.
Java
Time Complexity: O((n * 3) * (n * 3)), where n is the grid size as we traverse each 3x3 cell.
Space Complexity: O((n * 3) * (n * 3)) due to the graph representation.
This approach uses the Union-Find data structure to manage and merge connected components. Each cell in the grid is broken into four triangles and represented as nodes in the DSU. Connectivity relationships are established and merged as per slashes present in the cell, and the number of disconnected regions is determined.
The C++ solution employs the Union-Find data structure to manage the four sub-triangles per cell. It efficiently merges them based on the presence of slashes to identify distinct regions. The number of unique sets determines the count of regions.
JavaScript
Time Complexity: O(N^2 * α(N^2)), with α being a very slow-growing function (inverse Ackermann function), coming from the Union-Find operations.
Space Complexity: O(N^2) due to storing parent and rank structures for each cell division.
| Approach | Complexity |
|---|---|
| Approach 1: Represent Each Grid Cell by a Subdivision into Smaller Triangles | Time Complexity: O((n * 3) * (n * 3)), where n is the grid size as we traverse each 3x3 cell. |
| Approach 2: Union-Find to Track Regions | Time Complexity: O(N^2 * α(N^2)), with α being a very slow-growing function (inverse Ackermann function), coming from the Union-Find operations. |
Regions Cut By Slashes - Leetcode 959 - Python • NeetCodeIO • 18,516 views views
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