Problem Description

You are given an n x n grid where each cell contains one of three characters:

• '/' - a forward slash
• '\' - a backslash (represented as '\\' in the input due to escaping)
• ' ' - a blank space

These characters divide the grid into distinct contiguous regions. When you draw the slashes in each cell, they create boundaries that separate different areas.

For example:

• A '/' in a cell creates a diagonal line from the bottom-left corner to the top-right corner
• A '\' creates a diagonal line from the top-left corner to the bottom-right corner
• A blank space ' ' means no diagonal line in that cell

Your task is to count the total number of separate regions formed by these slashes.

The key insight is that each 1×1 cell can be divided into 4 triangular parts by its diagonals. The solution uses Union-Find to track which triangular parts are connected:

• Each cell is split into 4 triangles (numbered 0-3: top, right, bottom, left)
• Adjacent cells share edges, so their corresponding triangles need to be connected
• The slash characters determine which triangles within a cell are connected
• A '/' connects the top with left triangles, and right with bottom triangles
• A '\' connects the top with right triangles, and left with bottom triangles
• A blank space connects all four triangles within the cell

The algorithm starts with n * n * 4 separate components (4 triangles per cell) and uses union operations to merge connected components. The final count of distinct components gives the number of regions.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The problem involves regions that are connected to each other. Each region can be thought of as a node, and adjacent regions (those that share boundaries) have edges between them. The slashes create boundaries that define how these regions connect.

Is it a tree?

• No: The regions form a general graph structure, not a tree. There can be multiple disconnected components (separate regions), and within each component, there might be cycles depending on how the slashes are arranged.

Is the problem related to directed acyclic graphs?

• No: The problem deals with undirected connectivity between regions. The connections between adjacent areas don't have a direction.

Is the problem related to shortest paths?

• No: We're not trying to find the shortest path between regions. We're counting the total number of distinct regions.

Does the problem involve connectivity?

• Yes: The core of the problem is determining which parts of the grid are connected to form a single region. We need to identify all connected components in the graph.

Disjoint Set Union

• Yes: DSU (Union-Find) is the ideal approach here. We start with many small components (the triangular parts of each cell) and merge them based on the slash patterns and adjacency rules.

Conclusion: The flowchart leads us to using Disjoint Set Union (DSU/Union-Find) for this connectivity problem. While the flowchart suggests DSU rather than DFS directly, both approaches can solve connectivity problems. The solution uses DSU to efficiently track and merge connected components, ultimately counting the number of distinct regions.

Intuition

The key insight is to transform this visual problem into a connectivity problem. When we look at a grid with slashes, we're essentially looking at lines that divide space into separate regions. But how do we represent this computationally?

Imagine each 1×1 cell as being divided into 4 triangular pieces by drawing both possible diagonals (forming an X). Even though we only draw one slash (or none) per cell, thinking of it as 4 triangles gives us a consistent way to model the problem:

• Triangle 0: top
• Triangle 1: right
• Triangle 2: bottom
• Triangle 3: left

Now, when a slash is present:

• A '/' slash runs from bottom-left to top-right, keeping the top-left triangles together and bottom-right triangles together
• A '\' slash runs from top-left to bottom-right, keeping the top-right triangles together and bottom-left triangles together
• A blank space means all 4 triangles in that cell are connected

The brilliant part is realizing that adjacent cells share edges. For example:

• The bottom triangle of cell (i,j) touches the top triangle of cell (i+1,j)
• The right triangle of cell (i,j) touches the left triangle of cell (i,j+1)

This means we need to connect these triangles across cell boundaries.

We start with n * n * 4 separate components (4 triangles for each of the n * n cells). As we process each cell:

1. We connect triangles within the cell based on the slash type
2. We connect triangles to adjacent cells along shared edges

Every union operation reduces the number of components by 1. After processing all cells, the remaining number of components equals the number of distinct regions in the grid.

This approach elegantly converts a geometric visualization problem into a graph connectivity problem that can be solved efficiently with Union-Find.

Learn more about Depth-First Search, Breadth-First Search and Union Find patterns.

Solution Approach

The solution implements a Union-Find (Disjoint Set Union) data structure to track and merge connected triangular components.

Data Structure Setup:

• Create a parent array p of size n * n * 4 where each element initially points to itself
• Each cell (i, j) has 4 triangles indexed as: 4 * (i * n + j) + triangle_number
• Initialize size to n * n * 4 (total number of components)

Union-Find Operations:

• find(x): Returns the root parent of component x with path compression for optimization
• union(a, b): Merges two components if they have different roots, decreasing size by 1

Processing Each Cell: For each cell at position (i, j) with index k = i * n + j:

1. Connect to Adjacent Cells:

   • If not at bottom edge (i < n - 1): Connect bottom triangle of current cell (4*k + 2) with top triangle of cell below ((k + n) * 4)
   • If not at right edge (j < n - 1): Connect right triangle of current cell (4*k + 1) with left triangle of cell to the right ((k + 1) * 4 + 3)

2. Handle Internal Connections Based on Character:

   • '/': Connect triangles 0↔3 (top↔left) and 1↔2 (right↔bottom)
   • '\': Connect triangles 0↔1 (top↔right) and 2↔3 (bottom↔left)
   • ' ': Connect all four triangles (0↔1, 1↔2, 2↔3)

Algorithm Flow:

1. Initialize n*n*4 components
2. For each cell (i,j):
   a. Connect to adjacent cells (bottom and right neighbors)
   b. Based on slash type, connect internal triangles
   c. Each union reduces component count by 1
3. Return final component count (size)

The time complexity is O(n² * α(n)) where α is the inverse Ackermann function (practically constant), and space complexity is O(n²) for the parent array.

Example Walkthrough

Let's walk through a small 2×2 grid example:

grid = [" /",
        "/ "]

Initial Setup:

• We have 2×2 = 4 cells, each divided into 4 triangles
• Total components = 2 * 2 * 4 = 16 components
• Each triangle gets an ID from 0 to 15

Cell Numbering and Triangle IDs:

Cell (0,0): triangles 0,1,2,3    Cell (0,1): triangles 4,5,6,7
Cell (1,0): triangles 8,9,10,11  Cell (1,1): triangles 12,13,14,15

Visual representation of triangles in each cell:

   0        (each cell divided like this)
  / \
 3   1
  \ /
   2

Step 1: Process Cell (0,0) - contains ' ' (blank)

• Connect to adjacent cells:
  • Bottom triangle 2 connects to top triangle 8 of cell (1,0) below
  • Right triangle 1 connects to left triangle 7 of cell (0,1) to the right
• Internal connections for blank space:
  • Connect 0↔1, 1↔2, 2↔3 (all four triangles connected)
• Components after: 16 - 5 unions = 11

Step 2: Process Cell (0,1) - contains '/'

• Connect to adjacent cells:
  • Bottom triangle 6 connects to top triangle 12 of cell (1,1) below
  • No right neighbor (edge of grid)
• Internal connections for '/':
  • Connect 4↔7 (top↔left) and 5↔6 (right↔bottom)
• Components after: 11 - 3 unions = 8

Step 3: Process Cell (1,0) - contains '/'

• Connect to adjacent cells:
  • No bottom neighbor (edge of grid)
  • Right triangle 9 connects to left triangle 15 of cell (1,1) to the right
• Internal connections for '/':
  • Connect 8↔11 (top↔left) and 9↔10 (right↔bottom)
• Components after: 8 - 3 unions = 5

Step 4: Process Cell (1,1) - contains ' ' (blank)

• Connect to adjacent cells:
  • No bottom or right neighbors (corner of grid)
• Internal connections for blank space:
  • Connect 12↔13, 13↔14, 14↔15 (all four triangles connected)
• Components after: 5 - 3 unions = 2

Final Result: 2 regions

The algorithm correctly identifies that this grid forms 2 separate regions - one large connected region and one triangular region in the corner.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n² × α(n²))

The algorithm processes each cell in the n × n grid exactly once. For each cell, it performs a constant number of union operations (at most 4 unions per cell). Each union operation involves two find operations, and each find operation has an amortized time complexity of α(m) where α is the inverse Ackermann function and m is the total number of elements. Since we have 4n² total elements (4 triangles per cell), the find operation takes O(α(4n²)) = O(α(n²)) amortized time. Therefore, the overall time complexity is O(n² × α(n²)), which is nearly linear in practice since α(n) grows extremely slowly.

Space Complexity: O(n²)

The algorithm uses a parent array p of size 4n² to implement the Union-Find data structure, where each cell is divided into 4 triangular regions. This requires O(4n²) = O(n²) space. The recursion stack for the find operation with path compression can go up to O(log(n²)) = O(log n) in the worst case before path compression optimizes it. Since O(n²) > O(log n), the overall space complexity is O(n²).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Triangle Numbering/Mapping

One of the most common mistakes is incorrectly mapping the triangle positions within each cell. The triangles are numbered 0-3 representing top, right, bottom, and left respectively. Getting this mapping wrong will lead to incorrect connections between adjacent cells.

Pitfall Example:

# Wrong: Confusing which triangle connects to which neighbor
if row_idx < grid_size - 1:
    # Incorrectly connecting top triangle with bottom neighbor
    union(4 * cell_base_idx + 0, (cell_base_idx + grid_size) * 4 + 2)

Solution: Always visualize the cell division clearly:

    0 (top)
   /|\
  / | \
3   |   1
(left) (right)
  \ | /
   \|/
    2 (bottom)

• Bottom triangle (2) of current cell connects to top triangle (0) of cell below
• Right triangle (1) of current cell connects to left triangle (3) of right neighbor

2. Forgetting Path Compression in Find Operation

Without path compression, the find operation can degrade to O(n) time complexity in worst case, making the overall solution inefficient.

Pitfall Example:

def find(node: int) -> int:
    # Missing path compression - inefficient!
    while parent[node] != node:
        node = parent[node]
    return node

Solution: Always implement path compression:

def find(node: int) -> int:
    if parent[node] != node:
        parent[node] = find(parent[node])  # Path compression
    return parent[node]

3. Incorrect Slash Character Connections

Misunderstanding which triangles should be connected for each slash type is a critical error.

Pitfall Example:

if slash_char == '/':
    # Wrong connections for '/' 
    union(4 * cell_base_idx + 0, 4 * cell_base_idx + 2)  # Top with bottom (incorrect!)
    union(4 * cell_base_idx + 1, 4 * cell_base_idx + 3)  # Right with left (incorrect!)

Solution: Remember the visual representation:

• / divides the cell diagonally from bottom-left to top-right:
  • Connects top (0) with left (3)
  • Connects right (1) with bottom (2)
• \ divides the cell diagonally from top-left to bottom-right:
  • Connects top (0) with right (1)
  • Connects left (3) with bottom (2)

4. Off-by-One Errors in Boundary Checks

Forgetting to check boundaries before connecting to adjacent cells will cause index out of bounds errors.

Pitfall Example:

# Missing boundary check - will fail at grid edges
union(4 * cell_base_idx + 2, (cell_base_idx + grid_size) * 4 + 0)
union(4 * cell_base_idx + 1, (cell_base_idx + 1) * 4 + 3)

Solution: Always check boundaries before connecting to neighbors:

if row_idx < grid_size - 1:  # Not at bottom edge
    union(4 * cell_base_idx + 2, (cell_base_idx + grid_size) * 4 + 0)
if col_idx < grid_size - 1:  # Not at right edge
    union(4 * cell_base_idx + 1, (cell_base_idx + 1) * 4 + 3)

5. Incorrect Initial Component Count

Starting with the wrong number of components will give an incorrect final answer.

Pitfall Example:

# Wrong: Only counting cells, not triangular regions
region_count = grid_size * grid_size  # Should be multiplied by 4!

Solution: Each cell is divided into 4 triangular regions:

region_count = grid_size * grid_size * 4