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Regions Cut By Slashes

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Microsoft

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GraphsArraysStrings

You are given an n x n grid represented as a string array. Each cell in the grid consists of a '/', '\', or a blank space ' '. These characters divide each square into contiguous regions.

Your task is to write a function that takes the grid as input and returns the number of regions.

Example 1:

Input: grid = [" /","/ "]
Output: 2

Explanation: The grid is divided into two regions by the slashes.

Example 2:

Input: grid = [" /","  "]
Output: 1

Explanation: The grid is divided into one region because there is only one slash.

Example 3:

Input: grid = ["/\\","\\/"]
Output: 5

Explanation: Each cell is split into more than one regions.

Constraints:

n == grid.length == grid[i].length
1 <= n <= 30
grid[i][j] is either '/', '\', or ' '.

Could you please provide a solution for this problem with clear and concise code, and also explain the time and space complexity of your approach?

Regions Cut By Slashes

Medium

Microsoft

0 views

Topics:

GraphsArraysStrings

You are given an n x n grid represented as a string array. Each cell in the grid consists of a '/', '\', or a blank space ' '. These characters divide each square into contiguous regions.

Your task is to write a function that takes the grid as input and returns the number of regions.

Example 1:

Input: grid = [" /","/ "]
Output: 2

Explanation: The grid is divided into two regions by the slashes.

Example 2:

Input: grid = [" /","  "]
Output: 1

Explanation: The grid is divided into one region because there is only one slash.

Example 3:

Input: grid = ["/\\","\\/"]
Output: 5

Explanation: Each cell is split into more than one regions.

Constraints:

n == grid.length == grid[i].length
1 <= n <= 30
grid[i][j] is either '/', '\', or ' '.

Could you please provide a solution for this problem with clear and concise code, and also explain the time and space complexity of your approach?

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum size of the grid (i.e., the maximum number of rows/columns)?
Can the input grid be empty or null? If so, what should the return value be?
Will the input grid always be a square grid (number of rows equals number of columns)?
Besides spaces, '/', and ' can the grid contain any other characters?
If the input is invalid (e.g., contains characters other than space, '/', and ', how should I handle it?

Brute Force Solution

Approach

Imagine each slash divides a square into two pieces, and our goal is to count all the connected pieces. The brute force method involves checking every single tiny part of the grid to see which region it belongs to, and then figuring out how many different regions there are.

Here's how the algorithm would work step-by-step:

First, imagine the entire shape is made of very small squares, much smaller than the original squares with slashes.
Start at any small square and mark it as belonging to a particular region.
Look at all the small squares directly next to it (up, down, left, and right).
If a neighboring square is not separated by a slash and hasn't been marked yet, then mark it as belonging to the same region.
Keep repeating this process of finding unmarked neighbors and marking them as belonging to the same region until you can't find any more neighbors that are part of the same region.
Then, find another unmarked small square and mark it as belonging to a new region and repeat the neighbor checking process.
Continue this process until all the small squares are marked as belonging to a region.
Finally, count how many different regions you marked. This is the total number of connected pieces.

Code Implementation

def regions_by_slashes_brute_force(grid):
    grid_size = len(grid)
    extended_grid_size = 3 * grid_size
    extended_grid = [[0] * extended_grid_size for _ in range(extended_grid_size)]

    for row_index in range(grid_size):
        for col_index in range(grid_size):
            if grid[row_index][col_index] == '/':
                for i in range(3):
                    extended_grid[row_index * 3 + i][col_index * 3 + 2 - i] = 1
            elif grid[row_index][col_index] == '
                for i in range(3):
                    extended_grid[row_index * 3 + i][col_index * 3 + i] = 1

    region_count = 0
    visited = [[False] * extended_grid_size for _ in range(extended_grid_size)]

    def depth_first_search(row_index, col_index):
        if row_index < 0 or row_index >= extended_grid_size or \
           col_index < 0 or col_index >= extended_grid_size or \
           visited[row_index][col_index] or extended_grid[row_index][col_index] == 1:
            return

        visited[row_index][col_index] = True
        
        # Explore adjacent cells
        depth_first_search(row_index + 1, col_index)
        depth_first_search(row_index - 1, col_index)
        depth_first_search(row_index, col_index + 1)
        depth_first_search(row_index, col_index - 1)

    # Iterate through each cell in the extended grid
    for row_index in range(extended_grid_size):
        for col_index in range(extended_grid_size):
            # If we find an unvisited cell, start DFS
            if extended_grid[row_index][col_index] == 0 and not visited[row_index][col_index]:

                # Need to start a DFS to traverse the region
                depth_first_search(row_index, col_index)

                # Increment the region count after finishing DFS
                region_count += 1

    return region_count

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates over each small square in the grid. If the input grid has dimensions N x N, and we subdivide each square into a finer grid of, say, 3x3 smaller squares, then the total number of small squares becomes (3N) x (3N) = 9N². The algorithm visits each of these small squares in a flood-fill manner. During the flood fill, each small square might be visited multiple times, but in the worst case, we visit each one at most a constant number of times. Therefore, the time complexity is proportional to the total number of small squares, which is O(N²), and N is proportional to the input size, which makes the complexity O(n²).

Space Complexity

O(N^2) – The algorithm utilizes a grid of small squares, where each original square is divided into much smaller squares. Let N be the side length of the original grid. Since the plain English explanation mentions dividing each square into smaller squares, the size of the finer grid is proportional to N * N. The space used is mainly to keep track of which region each small square belongs to, implying an auxiliary data structure such as a matrix or a grid with dimensions proportional to N x N. Therefore, the space complexity is O(N^2).

Optimal Solution

Approach

Imagine each square is divided into four smaller triangles. We then treat the grid as a graph and connect adjacent triangles except where a slash or backslash is present. The number of regions is then the number of connected components in this graph.

Here's how the algorithm would work step-by-step:

Think of each square as being split into four smaller triangles: top, bottom, left, and right.
Imagine drawing invisible lines between these triangles inside and outside the squares.
A slash or backslash acts as a barrier, preventing those invisible lines from being drawn in certain places.
Now, find groups of triangles that are connected by these invisible lines (ignoring the barriers). Each connected group represents a region.
The number of these connected groups is the answer: the number of regions cut by the slashes.

Code Implementation

def regions_by_slashes(grid):
    grid_length = len(grid)
    grid_size = 4 * grid_length * grid_length
    parent = list(range(grid_size))

    def find(triangle_index):
        if parent[triangle_index] != triangle_index:
            parent[triangle_index] = find(parent[triangle_index])
        return parent[triangle_index]

    def union(triangle_index_one, triangle_index_two):
        root_one = find(triangle_index_one)
        root_two = find(triangle_index_two)
        if root_one != root_two:
            parent[root_one] = root_two

    for row_index in range(grid_length):
        for column_index in range(grid_length):
            cell_index = row_index * grid_length + column_index
            base_index = 4 * cell_index

            if grid[row_index][column_index] == '/':
                union(base_index + 0, base_index + 1)
                union(base_index + 2, base_index + 3)
            elif grid[row_index][column_index] == '\
                union(base_index + 0, base_index + 2)
                union(base_index + 1, base_index + 3)
            else:
                union(base_index + 0, base_index + 1)
                union(base_index + 1, base_index + 2)
                union(base_index + 2, base_index + 3)

            # Connect triangles between adjacent cells.
            if column_index + 1 < grid_length:
                union(base_index + 1, 4 * (row_index * grid_length + (column_index + 1)) + 3)

            if row_index + 1 < grid_length:
                union(base_index + 2, 4 * ((row_index + 1) * grid_length + column_index) + 0)

    number_of_regions = 0
    for triangle_index in range(grid_size):
        if parent[triangle_index] == triangle_index:
            number_of_regions += 1

    # Count connected components to determine number of regions
    return number_of_regions

Big(O) Analysis

Time Complexity

O(n^2) – The core of the algorithm involves iterating through each cell of the n x n grid and performing a Union-Find operation on the four triangles within each cell, and potentially connecting them to neighboring cells. The Union-Find operations (find and union) typically take near-constant time, especially with path compression and union by rank optimizations, essentially making them O(1) on average. However, in the worst-case without optimizations, they could be considered O(α(n)), where α(n) is the inverse Ackermann function, which grows extremely slowly and is practically constant for all practical input sizes. Thus, iterating over all the n^2 cells and performing Union-Find operations results in a time complexity of O(n^2 * α(4 * n^2)) which can be effectively simplified to O(n^2) as α(n) is nearly constant. Therefore, the dominant factor is the grid traversal.

Space Complexity

O(N^2) – The plain English explanation implies using a data structure (implicitly a union-find or similar) to track connected components of triangles. Since each square is divided into four triangles, and we have an N x N grid of squares, we have 4 * N^2 triangles. The union-find data structure will store information about each of these triangles, using at most a space proportional to the number of triangles. Therefore the auxiliary space used is proportional to 4 * N^2, which simplifies to O(N^2).

Edge Cases

Case	How to Handle
Null or empty grid	Return 1, as an empty grid technically has one region
1x1 grid	Handle the cases of '/', ' or ' ' correctly based on how they split the single cell
Large grid (n approaching constraints)	Ensure the solution's space and time complexity (likely O(n^2)) remain within acceptable limits for the maximum grid size
Grid filled entirely with spaces	The grid will have only one region and needs to be handled efficiently without excessive processing
Grid filled entirely with forward slashes	This represents a maximal division and should be handled without stack overflow if using recursion
Grid filled entirely with backslashes	Similar to the forward slash case, ensure efficient processing of maximal division
Grid with mixed slashes causing cycle formation in DFS/UnionFind	Algorithms like DFS or UnionFind need to correctly handle cycles to avoid infinite loops or incorrect region counts.
Integer overflow in any calculations based on grid size (if applicable)	Use appropriate data types (e.g., long) to prevent potential overflow issues when calculating indices or sizes if related to coordinate manipulation

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum size of the grid (i.e., the maximum number of rows/columns)?
Can the input grid be empty or null? If so, what should the return value be?
Will the input grid always be a square grid (number of rows equals number of columns)?
Besides spaces, '/', and ' can the grid contain any other characters?
If the input is invalid (e.g., contains characters other than space, '/', and ', how should I handle it?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

First, imagine the entire shape is made of very small squares, much smaller than the original squares with slashes.
Start at any small square and mark it as belonging to a particular region.
Look at all the small squares directly next to it (up, down, left, and right).
If a neighboring square is not separated by a slash and hasn't been marked yet, then mark it as belonging to the same region.
Keep repeating this process of finding unmarked neighbors and marking them as belonging to the same region until you can't find any more neighbors that are part of the same region.
Then, find another unmarked small square and mark it as belonging to a new region and repeat the neighbor checking process.
Continue this process until all the small squares are marked as belonging to a region.
Finally, count how many different regions you marked. This is the total number of connected pieces.

Code Implementation

def regions_by_slashes_brute_force(grid):
    grid_size = len(grid)
    extended_grid_size = 3 * grid_size
    extended_grid = [[0] * extended_grid_size for _ in range(extended_grid_size)]

    for row_index in range(grid_size):
        for col_index in range(grid_size):
            if grid[row_index][col_index] == '/':
                for i in range(3):
                    extended_grid[row_index * 3 + i][col_index * 3 + 2 - i] = 1
            elif grid[row_index][col_index] == '
                for i in range(3):
                    extended_grid[row_index * 3 + i][col_index * 3 + i] = 1

    region_count = 0
    visited = [[False] * extended_grid_size for _ in range(extended_grid_size)]

    def depth_first_search(row_index, col_index):
        if row_index < 0 or row_index >= extended_grid_size or \
           col_index < 0 or col_index >= extended_grid_size or \
           visited[row_index][col_index] or extended_grid[row_index][col_index] == 1:
            return

        visited[row_index][col_index] = True
        
        # Explore adjacent cells
        depth_first_search(row_index + 1, col_index)
        depth_first_search(row_index - 1, col_index)
        depth_first_search(row_index, col_index + 1)
        depth_first_search(row_index, col_index - 1)

    # Iterate through each cell in the extended grid
    for row_index in range(extended_grid_size):
        for col_index in range(extended_grid_size):
            # If we find an unvisited cell, start DFS
            if extended_grid[row_index][col_index] == 0 and not visited[row_index][col_index]:

                # Need to start a DFS to traverse the region
                depth_first_search(row_index, col_index)

                # Increment the region count after finishing DFS
                region_count += 1

    return region_count

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Think of each square as being split into four smaller triangles: top, bottom, left, and right.
Imagine drawing invisible lines between these triangles inside and outside the squares.
A slash or backslash acts as a barrier, preventing those invisible lines from being drawn in certain places.
Now, find groups of triangles that are connected by these invisible lines (ignoring the barriers). Each connected group represents a region.
The number of these connected groups is the answer: the number of regions cut by the slashes.

Code Implementation

def regions_by_slashes(grid):
    grid_length = len(grid)
    grid_size = 4 * grid_length * grid_length
    parent = list(range(grid_size))

    def find(triangle_index):
        if parent[triangle_index] != triangle_index:
            parent[triangle_index] = find(parent[triangle_index])
        return parent[triangle_index]

    def union(triangle_index_one, triangle_index_two):
        root_one = find(triangle_index_one)
        root_two = find(triangle_index_two)
        if root_one != root_two:
            parent[root_one] = root_two

    for row_index in range(grid_length):
        for column_index in range(grid_length):
            cell_index = row_index * grid_length + column_index
            base_index = 4 * cell_index

            if grid[row_index][column_index] == '/':
                union(base_index + 0, base_index + 1)
                union(base_index + 2, base_index + 3)
            elif grid[row_index][column_index] == '\
                union(base_index + 0, base_index + 2)
                union(base_index + 1, base_index + 3)
            else:
                union(base_index + 0, base_index + 1)
                union(base_index + 1, base_index + 2)
                union(base_index + 2, base_index + 3)

            # Connect triangles between adjacent cells.
            if column_index + 1 < grid_length:
                union(base_index + 1, 4 * (row_index * grid_length + (column_index + 1)) + 3)

            if row_index + 1 < grid_length:
                union(base_index + 2, 4 * ((row_index + 1) * grid_length + column_index) + 0)

    number_of_regions = 0
    for triangle_index in range(grid_size):
        if parent[triangle_index] == triangle_index:
            number_of_regions += 1

    # Count connected components to determine number of regions
    return number_of_regions

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty grid	Return 1, as an empty grid technically has one region
1x1 grid	Handle the cases of '/', ' or ' ' correctly based on how they split the single cell
Large grid (n approaching constraints)	Ensure the solution's space and time complexity (likely O(n^2)) remain within acceptable limits for the maximum grid size
Grid filled entirely with spaces	The grid will have only one region and needs to be handled efficiently without excessive processing
Grid filled entirely with forward slashes	This represents a maximal division and should be handled without stack overflow if using recursion
Grid filled entirely with backslashes	Similar to the forward slash case, ensure efficient processing of maximal division
Grid with mixed slashes causing cycle formation in DFS/UnionFind	Algorithms like DFS or UnionFind need to correctly handle cycles to avoid infinite loops or incorrect region counts.
Integer overflow in any calculations based on grid size (if applicable)	Use appropriate data types (e.g., long) to prevent potential overflow issues when calculating indices or sizes if related to coordinate manipulation