Leetcode 959: Regions Cut By Slashes
You are given an n x n grid where each cell contains one of the following characters: ‘/’, ‘', or ’ ‘. These characters divide the grid into different regions. Your task is to determine how many distinct regions the grid is divided into.
Problem
Approach
Steps
Complexity
Input: The input consists of a list of strings representing the n x n grid, where each string corresponds to a row in the grid. Each character in the grid can be '/', '\', or ' '.
Example: Input: grid = [" /", "\ "]
Constraints:
• 1 <= n <= 30
• grid[i][j] is either '/', '\', or ' '.
Output: Return the total number of distinct regions formed by the characters in the grid.
Example: Output: 2
Constraints:
• The function should return a non-negative integer representing the number of regions.
Goal: The goal is to count the number of distinct regions formed by '/' and '\' characters in the grid, with the characters dividing the grid into different regions.
Steps:
• 1. Convert the grid into a larger grid where each cell in the original grid becomes a 3x3 subgrid.
• 2. Mark regions in the larger grid based on the '/' and '\' characters.
• 3. Use depth-first search (DFS) to count the number of distinct regions by traversing the grid and marking cells that are part of the same region.
Goal: The input grid must adhere to the specified constraints: the grid is of size n x n, where 1 <= n <= 30, and each character in the grid is either '/', '\', or ' '.
Steps:
• The grid will contain between 1 and 30 rows and columns.
• Each character in the grid will be either '/', '\', or ' '.
Assumptions:
• The grid is a valid n x n matrix with characters '/', '\', and ' '.
• Input: Input: grid = [" /", "\ "]
• Explanation: In this case, the grid is divided into two distinct regions, one formed by the '/' and the other formed by the '\'.
• Input: Input: grid = [" /", " "]
• Explanation: Here, the grid is divided into one region formed by the '/' character, and the second row is empty space.
• Input: Input: grid = ["/\\", "\/ "]
• Explanation: In this case, the grid is divided into five distinct regions due to the positioning of the '\' and '/' characters.
Approach: The approach involves transforming the grid into a 3x3 matrix for each original grid cell, then counting the number of connected regions using depth-first search (DFS).
Observations:
• Each cell in the grid can contribute to a new region or be part of an existing region depending on its connections with neighboring cells.
• By expanding each cell into a 3x3 subgrid, we can simplify the problem of region counting by treating connected components as parts of the same region.
Steps:
• 1. Create a larger grid where each cell in the original grid corresponds to a 3x3 block.
• 2. Based on whether the character is '/', '\', or ' ', mark the relevant cells in the larger grid as part of the region.
• 3. Use depth-first search (DFS) to explore the grid, marking all cells that belong to the same region.
• 4. Count the number of distinct regions by counting the number of DFS explorations that start from unvisited cells.
Empty Inputs:
• The grid will never be empty as per the constraints.
Large Inputs:
• For larger grids, the solution must efficiently handle grids with dimensions up to 30x30.
Special Values:
• Consider grids with no '/' or '\' characters, where all cells are empty space.
Constraints:
• The grid will always be a square of size n x n with 1 <= n <= 30.
int regionsBySlashes(vector<string>& grid) {
int n = grid.size();
vector<vector<int>> g(n * 3, vector<int>(n * 3, 0));
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
if(grid[i][j] == '/') {
g[i * 3 + 2][j * 3] = g[i * 3 + 1][j * 3 + 1] = g[i * 3][j * 3 + 2] = 1;
} else if(grid[i][j] == '\\') {
g[i * 3][j * 3] = g[i * 3 + 1][j * 3 + 1] = g[i * 3 + 2][j * 3 + 2] = 1;
}
}
}
int reg = 0;
for(int i = 0; i < n * 3; i++) {
for(int j = 0; j < n * 3; j++) {
reg += dfs(g, i, j) ? 1 : 0;
}
}
return reg;
}
int dfs(vector<vector<int>> &g, int i, int j) {
if (min(i, j) < 0 || max(i, j) >= g.size() || g[i][j] != 0) {
return 0;
}
g[i][j] = 1;
return 1 + dfs(g, i + 1, j) + dfs(g, i - 1, j) + dfs(g, i, j + 1) + dfs(g, i, j - 1);
}
1 : Function Declaration
int regionsBySlashes(vector<string>& grid) {
Declares the function that computes the number of regions formed by slashes in a given grid.
2 : Variable Initialization
int n = grid.size();
Initializes the variable `n` with the size of the grid.
3 : Matrix Initialization
vector<vector<int>> g(n * 3, vector<int>(n * 3, 0));
Expands the grid by a factor of 3 to better represent slashes and backslashes.
4 : Outer Loop
for(int i = 0; i < n; i++) {
Outer loop to iterate over rows of the grid.
5 : Inner Loop
for(int j = 0; j < n; j++) {
Inner loop to iterate over columns of the grid.
6 : Slash Condition
if(grid[i][j] == '/') {
Checks if the current cell contains a forward slash and marks corresponding positions in the expanded grid.
7 : Mark Grid for Slash
g[i * 3 + 2][j * 3] = g[i * 3 + 1][j * 3 + 1] = g[i * 3][j * 3 + 2] = 1;
Marks the positions of the forward slash in the expanded grid.
8 : Backslash Condition
} else if(grid[i][j] == '\\') {
Checks if the current cell contains a backslash and marks corresponding positions in the expanded grid.
9 : Mark Grid for Backslash
g[i * 3][j * 3] = g[i * 3 + 1][j * 3 + 1] = g[i * 3 + 2][j * 3 + 2] = 1;
Marks the positions of the backslash in the expanded grid.
10 : Region Counter
int reg = 0;
Initializes a counter for the number of distinct regions.
11 : Outer Loop for DFS
for(int i = 0; i < n * 3; i++) {
Outer loop to iterate over rows of the expanded grid for DFS.
12 : Inner Loop for DFS
for(int j = 0; j < n * 3; j++) {
Inner loop to iterate over columns of the expanded grid for DFS.
13 : Region Identification
reg += dfs(g, i, j) ? 1 : 0;
Performs DFS to identify and count a new region if it exists.
14 : Return Statement
return reg;
Returns the total number of regions identified in the grid.
15 : DFS Function Declaration
int dfs(vector<vector<int>> &g, int i, int j) {
Declares the depth-first search (DFS) helper function.
16 : Base Condition
if (min(i, j) < 0 || max(i, j) >= g.size() || g[i][j] != 0) {
Checks if the current cell is out of bounds or already visited.
17 : Base Condition Return
return 0; }
Returns 0 if the base condition is met.
18 : Mark Cell as Visited
g[i][j] = 1;
Marks the current cell as visited in the grid.
19 : Recursive DFS Calls
return 1 + dfs(g, i + 1, j) + dfs(g, i - 1, j) + dfs(g, i, j + 1) + dfs(g, i, j - 1);
Recursively explores all neighboring cells to calculate the region size.
Best Case: O(n^2)
Average Case: O(n^2)
Worst Case: O(n^2)
Description: In all cases, the time complexity is O(n^2) because we need to explore every cell in the grid at least once.
Best Case: O(1)
Worst Case: O(n^2)
Description: The space complexity is O(n^2) due to the storage of the larger grid for processing and the space used by the depth-first search stack.
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