Problem Statement
In this problem, you're presented with a binary grid where the value 1 indicates land and the value 0 indicates water. We define an island as a group of adjacent land cells connected either horizontally or vertically. The grid is considered connected if there is exactly one such group of 1s forming a single island.
Your task is to determine the minimum number of days required to turn this connected island grid into a disconnected one. On each day, you can change one land cell (1) into a water cell (0). The grid is deemed disconnected when there are either no islands left or multiple disconnected islands.
Examples
Example 1
Input:
grid = [[0,1,1,0],[0,1,1,0],[0,0,0,0]]
Output:
2
Explanation:
We need at least 2 days to get a disconnected grid. Change land grid[1][1] and grid[0][2] to water and get 2 disconnected island.
Example 2
Input:
grid = [[1,1]]
Output:
2
Explanation:
Grid of full water is also disconnected ([[1,1]] -> [[0,0]]), 0 islands.
Constraints
m == grid.lengthn == grid[i].length1 <= m, n <= 30grid[i][j]is either0or1.
Approach and Intuition
To tackle this problem, consider a few key observations:
Immediate Disconnections: If the grid itself starts with no
1s or distinct naturally disconnected islands, it's already in a disconnected state requiring zero days.Single Changes: For grids that have connected islands, investigate how making single changes in strategic locations might disconnect the island. Often, changing a cell on a thin bridge between two larger parts of an island might be enough.
Grid Edges and Corners: Generally, islands touching the grid borders need careful examination since changes here might have a deeper impact on connectivity due to fewer neighboring cells.
Minimum Cuts: Think in terms of graph theory where islands are interconnected nodes. Your task might reduce to finding the minimum "cut" in this connectivity, essentially how to remove the least nodes (or cells, in this context) to break apart the island.
Computational Bounds: Given the grid’s maximum dimensions (30x30), solutions involving examining each cell multiple times (like BFS/DFS for each cell) are computationally feasible.
Based on these observations, a plausible approach could involve:
- Using depth-first search (DFS) or breadth-first search (BFS) to identify the exact structure and connections of the current island.
- Identifying critical points or bridges within the island—cells whose removal leads to a disconnected state.
- Iteratively simulate the removal of each potentially critical cell to see if it results in a split, thereby counting the minimum number of days required.
In essence, the challenge involves transforming a connected physical structure into a disconnected one by strategically removing minimal structural components (land cells). Understanding and manipulating the connectivity efficiently is key to solving the problem optimally.
Solutions
- C++
- Java
- Python
class GridAnalysis {
private:
// Directions: right, down, left, up
const vector<vector<int>> MOVE_SET = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
struct CriticalPointDetails {
bool foundCriticalPoint;
int visitTime;
CriticalPointDetails(bool found, int time)
: foundCriticalPoint(found), visitTime(time) {}
};
public:
int disconnectDays(vector<vector<int>>& matrix) {
int rowCount = matrix.size(), colCount = matrix[0].size();
CriticalPointDetails critPoint(false, 0);
int totalLand = 0, numIslands = 0;
// Visitation and ancestry tracking for DFS
vector<vector<int>> visitTimestamp(
rowCount,
vector<int>(colCount, -1));
vector<vector<int>> lowEndpoint(
rowCount,
vector<int>(colCount, -1));
vector<vector<int>> dfsParent(
rowCount, vector<int>(colCount, -1));
// Explore the matrix for connectivity and critical points
for (int r = 0; r < rowCount; r++) {
for (int c = 0; c < colCount; c++) {
if (matrix[r][c] == 1) {
totalLand++;
if (visitTimestamp[r][c] == -1) {
exploreCriticalPoints(matrix, r, c, visitTimestamp,
lowEndpoint, dfsParent, critPoint);
numIslands++;
}
}
}
}
// Evaluate disconnection scenario
if (numIslands == 0 || numIslands > 1)
return 0;
if (totalLand == 1) return 1;
if (critPoint.foundCriticalPoint)
return 1;
return 2;
}
private:
void exploreCriticalPoints(vector<vector<int>>& matrix, int r, int c,
vector<vector<int>>& visitTimestamp,
vector<vector<int>>& lowEndpoint,
vector<vector<int>>& dfsParent,
CriticalPointDetails& critDetails) {
int rowCount = matrix.size(), colCount = matrix[0].size();
visitTimestamp[r][c] = critDetails.visitTime;
critDetails.visitTime++;
lowEndpoint[r][c] = visitTimestamp[r][c];
int childCnt = 0;
// Check adjacent cells
for (const auto& move : MOVE_SET) {
int nextR = r + move[0];
int nextC = c + move[1];
if (isLandCellValid(matrix, nextR, nextC)) {
if (visitTimestamp[nextR][nextC] == -1) {
childCnt++;
dfsParent[nextR][nextC] = r * colCount + c;
exploreCriticalPoints(matrix, nextR, nextC, visitTimestamp,
lowEndpoint, dfsParent, critDetails);
// Check and update low times
lowEndpoint[r][c] = min(lowEndpoint[r][c],
lowEndpoint[nextR][nextC]);
// Articulation condition
if (lowEndpoint[nextR][nextC] >= visitTimestamp[r][c] &&
dfsParent[r][c] != -1) {
critDetails.foundCriticalPoint = true;
}
} else if (nextR * colCount + nextC != dfsParent[r][c]) {
lowEndpoint[r][c] = min(lowEndpoint[r][c],
visitTimestamp[nextR][nextC]);
}
}
}
// Root condition
if (dfsParent[r][c] == -1 && childCnt > 1) {
critDetails.foundCriticalPoint = true;
}
}
// Validate land cell
bool isLandCellValid(const vector<vector<int>>& matrix, int r, int c) {
int rowCount = matrix.size(), colCount = matrix[0].size();
return (r >= 0 && c >= 0 && r < rowCount && c < colCount &&
matrix[r][c] == 1);
}
};
This solution explores how to determine the minimum number of days required to disconnect all land in a grid, where 1 represents land and 0 represents water. The implementation uses Depth-First Search (DFS) in C++ to analyze the number of disconnected sections or islands and to identify critical points or articulation points in the grid.
The main functionality is encapsulated in the disconnectDays method, which first calculates the total number of land cells and the number of initially disconnected islands. It employs a DFS approach to:
- Track visitation times and low link values.
- Identify articulation points that, if removed, increase the number of disconnected islands.
Key points of the implementation include:
Initialization: Arrays for visit timestamps, low endpoints, and DFS parent tracking are set up. These are critical for the DFS traversal and articulation point identification.
DFS Exploration: The
exploreCriticalPointshelper method is invoked recursively to traverse the grid. Through this traversal, the method checks adjacent cells (right, down, left, up) for connectivity and updates low endpoints.Articulation Point Check: During the DFS traversal, if a node meets certain conditions, it is marked as a critical point. This involves checking if any child node can connect to an ancestor node directly or through other children nodes.
Edge Cases Handling:
- If no islands exist or if there are multiple unconnected islands from the start, the function returns
0, as they are already disconnected or in isolation. - If there is only one land cell (total land count is
1), disconnecting it necessitates removing it, which can be done in1day. - If an articulation point exists (indicating a critical connection point among the land cells), the minimum days to disconnect the island are
1. - Otherwise, if no articulation points exist in a connected mass of land, then at least
2days are required, implying the need for more complex maneuvers to create disconnections.
- If no islands exist or if there are multiple unconnected islands from the start, the function returns
This approach relies heavily on understanding the structure of the grid and uses classical graph theory techniques (like articulation points) efficiently in the context of a 2D matrix. The algorithm's correctness across various topographical complexities in the grid makes it robust and reliable for solving the given problem.
class Solution {
// Grid movement offsets: up, right, left, down
private static final int[][] OFFSETS = {
{ 0, 1 },
{ 1, 0 },
{ 0, -1 },
{ -1, 0 },
};
public int findMinDays(int[][] grid) {
int numRows = grid.length, numCols = grid[0].length;
ArticulationPointInfo apInfo = new ArticulationPointInfo(false, 0);
int totalCells = 0, numberOfIslands = 0;
// Storage for cell state during DFS
int[][] visitedTime = new int[numRows][numCols];
int[][] earliestBackReach = new int[numRows][numCols];
int[][] parents = new int[numRows][numCols];
// Default initialization
for (int i = 0; i < numRows; i++) {
Arrays.fill(visitedTime[i], -1);
Arrays.fill(earliestBackReach[i], -1);
Arrays.fill(parents[i], -1);
}
// Traverse each cell in grid
for (int i = 0; i < numRows; i++) {
for (int j = 0; j < numCols; j++) {
if (grid[i][j] == 1) {
totalCells++;
if (visitedTime[i][j] == -1) {
dfsArticulation(
grid,
i,
j,
visitedTime,
earliestBackReach,
parents,
apInfo
);
numberOfIslands++;
}
}
}
}
// Evaluate possibilities based on grid status
if (numberOfIslands == 0 || numberOfIslands > 1) return 0;
if (totalCells == 1) return 1;
if (apInfo.hasArticulationPoint) return 1;
return 2;
}
private void dfsArticulation(
int[][] grid,
int i,
int j,
int[][] visitedTime,
int[][] earliestBackReach,
int[][] parents,
ArticulationPointInfo apInfo
) {
int numRows = grid.length, numCols = grid[0].length;
visitedTime[i][j] = apInfo.time;
apInfo.time++;
earliestBackReach[i][j] = visitedTime[i][j];
int childCounter = 0;
// Process each possible movement from current cell
for (int[] offset : OFFSETS) {
int newRow = i + offset[0];
int newCol = j + offset[1];
if (isValidCell(grid, newRow, newCol)) {
if (visitedTime[newRow][newCol] == -1) {
childCounter++;
parents[newRow][newCol] = i * numCols + j;
dfsArticulation(
grid,
newRow,
newCol,
visitedTime,
earliestBackReach,
parents,
apInfo
);
earliestBackReach[i][j] = Math.min(
earliestBackReach[i][j],
earliestBackReach[newRow][newCol]
);
if (earliestBackReach[newRow][newCol] >= visitedTime[i][j] && parents[i][j] != -1)
apInfo.hasArticulationPoint = true;
} else if (newRow * numCols + newCol != parents[i][j]) {
earliestBackReach[i][j] = Math.min(
earliestBackReach[i][j],
visitedTime[newRow][newCol]
);
}
}
}
if (parents[i][j] == -1 && childCounter > 1) {
apInfo.hasArticulationPoint = true;
}
}
private boolean isValidCell(int[][] grid, int row, int col) {
return (
row >= 0 &&
row < grid.length &&
col >= 0 &&
col < grid[0].length &&
grid[row][col] == 1
);
}
private class ArticulationPointInfo {
boolean hasArticulationPoint;
int time;
ArticulationPointInfo(boolean apExists, int initTime) {
this.hasArticulationPoint = apExists;
this.time = initTime;
}
}
}
This Java solution aims to determine the minimum number of days required to disconnect an island on a grid. The grid consists of cells marked as either land (1) or water (0). The algorithm needs to check whether removing one or two specific cells can lead to the island being disconnected from adjacent land cells.
The solution uses Depth First Search (DFS) and articulation points concepts from graph theory. An articulation point in this context is a land cell that, if removed, would increase the number of disjoint land sections, potentially disconnecting parts of the island. To identify such points, the algorithm implements a DFS to explore the grid, marking cells with their visitation time and the earliest reachable time via a back edge.
Here's a breakdown of the approach:
- Initialize variables to track the state of the grid, including arrays for the visitation order (
visitedTime) and the earliest time of reachable ancestors (earliestBackReach). - Traverse each cell in the grid:
- For each unvisited land cell (
1), perform a DFS to classify the articulation points and possible disconnections. - Increment counters to track total cells and the number of unique islands.
- For each unvisited land cell (
- Depending on the number of islands and the presence of articulation points, determine the minimum days to disconnect:
- If there are no islands or multiple islands from the start, no days are required (
return 0). - If the entire grid consists of a single cell, one day is required (
return 1). - If there's an articulation point, disconnecting can be achieved in one day (
return 1). - Otherwise, if no articulation points exist and all land cells are interconnected as one island, two days will be required (
return 2).
- If there are no islands or multiple islands from the start, no days are required (
Additionally, a private helper function (isValidCell) ensures that cells investigated during DFS are within grid boundaries and are land cells. A custom class (ArticulationPointInfo) helps track articulation points and the progression of time during the DFS.
This algorithm is efficient in identifying critical points in the grid that can be targeted to disconnect the island in the minimum number of days, relying on classical graph traversal techniques optimized for the specific requirements of a grid-based layout.
class GridManager:
# Navigation for adjacent cells: east, south, west, north
MOVE = [[0, 1], [1, 0], [0, -1], [-1, 0]]
def countDays(self, matrix):
row_count, col_count = len(matrix), len(matrix[0])
dfs_info = {"contains_cut_point": False, "counter": 0}
land_count = 0
island_num = 0
# Store visitation time and other properties for each cell
entry_time = [
[-1] * col_count for _ in range(row_count)
]
low_time = [
[-1] * col_count for _ in range(row_count)
]
predecessors = [
[-1] * col_count for _ in range(row_count)
]
def _detect_cut_points(r, c):
entry_time[r][c] = dfs_info["counter"]
dfs_info["counter"] += 1
low_time[r][c] = entry_time[r][c]
offspring = 0
# Travel across valid adjacent cells
for move in self.MOVE:
next_r, next_c = r + move[0], c + move[1]
if (
0 <= next_r < row_count
and 0 <= next_c < col_count
and matrix[next_r][next_c] == 1
):
if entry_time[next_r][next_c] == -1:
offspring += 1
predecessors[next_r][next_c] = (
r * col_count + c
)
_detect_cut_points(next_r, next_c)
# Minimize low_time based on descendant’s reachability
low_time[r][c] = min(
low_time[r][c],
low_time[next_r][next_c],
)
# Check for critical vertex condition
if (
low_time[next_r][next_c]
>= entry_time[r][c]
and predecessors[r][c] != -1
):
dfs_info["contains_cut_point"] = True
elif next_r * col_count + next_c != predecessors[r][c]:
# Adjust low time for a back edge
low_time[r][c] = min(
low_time[r][c],
entry_time[next_r][next_c],
)
# Root special case for cut-vertex
if predecessors[r][c] == -1 and offspring > 1:
dfs_info["contains_cut_point"] = True
# Analyze the grid to identify islands and critical points
for i in range(row_count):
for j in range(col_count):
if matrix[i][j] == 1:
land_count += 1
if entry_time[i][j] == -1:
_detect_cut_points(i, j)
island_num += 1
# Assess how many days needed to sever the grid
if island_num == 0 or island_num >= 2:
return 0 # Disconnected already or there's no land
if land_count == 1:
return 1 # Single land tile scenario
if dfs_info["contains_cut_point"]:
return 1 # Presence of a cut-point demands immediate attention
return 2 # General case, need to remove two arbitrary land parcels
The given Python code defines a class GridManager that is designed to determine the minimum number of days required to disconnect all land sections of an island represented as a grid matrix. Here's how the algorithm works, focusing on its primary functions and processes:
- Initialize support data structures for Depth-First Search (DFS) such as
entry_time,low_time, andpredecessorsto help with the cycle detection and cut vertex identification. - Define a nested function
_detect_cut_points()to perform a DFS and identify cut points in the graph. Cut points are vertices that, when removed, increase the number of connected components. - Within
_detect_cut_points(r, c), explore adjacent cells and recursively call the function for unvisited land cells, updatinglow_timeandentry_time. - Check for critical vertex conditions where removing a vertex would separate the graph. This helps identify if an immediate action (removal) can separate the island.
- Main method
countDays(matrix)initializes the grid exploration:- Count the total land and launch DFS from unvisited cells.
- Detect cut points or validate if the graph (or subgraphs) already consists of separate components.
- Based on whether islands are already separate, the total land amount or presence of cut points, the method returns:
0if the island configuration is already disconnected or there's no land.1if there's only one piece of land or if there's a cut-point, which means immediate disconnection can occur by removing one specific land piece.2in the more general scenario, implying two removals are necessary for disconnection.
This solution effectively utilizes graph traversal techniques to assess the structural vulnerability of land connections in a grid, targeting minimum interventions for maximal disconnection. This approach is directly applicable to scenarios requiring insights into connectivity and segmentation of nodes in a network or grid.
No comments yet.