Problem Statement
In the given challenge, we are tasked with finding the shortest route through an n x n binary matrix named grid, with the destination being the matrix's bottom-right corner starting from the top-left corner. The matrix cells have two possible values: 0 (open path) and 1 (blocked path).
The main criteria for a "clear path" are:
- The cells on the path must contain
0. - Movement can be in any of the eight possible directions (horizontally, vertically, or diagonally adjacent).
The goal is to count the number of cells in the shortest clear path. If no such path exists, the function should return -1.
This problem is essentially a pathfinding puzzle that resembles searching for a pathway in a maze, with the complexity of movement extended to eight possible directions from each accessible cell.
Examples
Example 1
Input:
grid = [[0,1],[1,0]]
Output:
2
Explanation:
The shortest path is [0,0] → [1,1], which includes 2 cells.
Example 2
Input:
grid = [[0,0,0],[1,1,0],[1,1,0]]
Output:
4
Explanation:
The shortest path is [0,0] → [0,1] → [0,2] → [1,2] → [2,2], which includes 4 moves (5 cells), so output is 4.
Example 3
Input:
grid = [[1,0,0],[1,1,0],[1,1,0]]
Output:
-1
Explanation:
The start cell [0,0] is blocked, so no path exists.
Constraints
n == grid.lengthn == grid[i].length1 <= n <= 100grid[i][j] is 0 or 1
Approach and Intuition
Strategic Approach Based on Constraints
Use Breadth-First Search (BFS):
BFS is ideal for finding the shortest path in an unweighted graph/grid.
It proceeds level by level, ensuring the shortest path is found first.
For each cell, consider all 8 neighboring positions:
- Horizontal: left/right
- Vertical: up/down
- Diagonal: top-left, top-right, bottom-left, bottom-right
Keep a
visitedmatrix to avoid revisiting the same cell.Use a queue to process the path level by level and count the distance at each step.
Edge Case Handling:
- Immediately return
-1if either the starting cellgrid[0][0]or the ending cellgrid[n-1][n-1]is1.
- Immediately return
Efficiency:
- Since the grid size can go up to
100 x 100 = 10,000cells, the BFS approach operates in O(n²) time and is performant within the given constraints.
- Since the grid size can go up to
This BFS-driven strategy ensures both correctness and performance, leveraging level-order exploration to always return the shortest path, if any exists.
Solutions
- Java
class PathFinder {
// Option that encapsulates a potential move in the grid
class Option {
public int currentRow;
public int currentCol;
public int pathLength;
public int heuristic;
public Option(int row, int col, int pathLength, int heuristic) {
this.currentRow = row;
this.currentCol = col;
this.pathLength = pathLength;
this.heuristic = heuristic;
}
}
private static final int[][] moves =
new int[][]{{-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 1}, {1, -1}, {1, 0}, {1, 1}};
public int findShortestPath(int[][] grid) {
if (grid[0][0] != 0 || grid[grid.length - 1][grid[0].length - 1] != 0) {
return -1;
}
PriorityQueue<Option> priorityQueue = new PriorityQueue<>((o1, o2) -> o1.heuristic - o2.heuristic);
priorityQueue.add(new Option(0, 0, 1, calculateHeuristic(0, 0, grid)));
boolean[][] seen = new boolean[grid.length][grid[0].length];
while (!priorityQueue.isEmpty()) {
Option current = priorityQueue.remove();
if (current.currentRow == grid.length - 1 && current.currentCol == grid[0].length - 1) {
return current.pathLength;
}
if (seen[current.currentRow][current.currentCol]) {
continue;
}
seen[current.currentRow][current.currentCol] = true;
for (int[] step : neighbors(current.currentRow, current.currentCol, grid)) {
int newRow = step[0];
int newCol = step[1];
if (seen[newRow][newCol]) {
continue;
}
int newPathLength = current.pathLength + 1;
int newHeuristic = newPathLength + calculateHeuristic(newRow, newCol, grid);
Option newOption =
new Option(newRow, newCol, newPathLength, newHeuristic);
priorityQueue.add(newOption);
}
}
return -1;
}
private List<int[]> neighbors(int row, int col, int[][] grid) {
List<int[]> validNeighbors = new ArrayList<>();
for (int[] direction : moves) {
int nextRow = row + direction[0];
int nextCol = col + direction[1];
if (isValid(nextRow, nextCol, grid)) {
validNeighbors.add(new int[]{nextRow, nextCol});
}
}
return validNeighbors;
}
private boolean isValid(int row, int col, int[][] grid) {
return row >= 0 && col >= 0 && row < grid.length && col < grid[0].length && grid[row][col] == 0;
}
private int calculateHeuristic(int row, int col, int[][] grid) {
int remainingRows = grid.length - 1 - row;
int remainingCols = grid[0].length - 1 - col;
return Math.max(remainingRows, remainingCols);
}
}
In the Java program for finding the shortest path in a binary matrix, you're working with a grid where 0 represents a passable cell and 1 represents an impassable cell. The aim is to compute the shortest path from the top-left corner (0, 0) to the bottom-right corner using a Breadth-first Search (BFS) style approach, but with the optimization of a priority queue to facilitate faster traversal using a best-first search based on an estimated cost from the current cell to the destination.
- The
PathFinderclass contains an innerOptionclass representing a state in the search, which holds the current row and column, the path length from the start, and a heuristic estimate of the remaining path length. - The program initiates by confirming the start and end points are passable. If either is impassable (value of 1), the function returns -1, indicating the path is not available.
- A priority queue handles instances of
Option, sorted by their heuristic value. This heuristic is a simple yet effective estimation using the Manhattan or Chebyshev distance, favoring moves that directly approach the target. - The state-space exploration starts from the top-left corner, and for each cell, it evaluates possible moves in all 8 directions (horizontally, vertically, and diagonally). Each move generates a new
Optionthat is then pushed into the priority queue if it leads to an unvisited cell that is within the grid bounds and passable. - A boolean grid named
seentracks visited cells to prevent re-processing and looping. - The algorithm proceeds until the destination is reached, where it returns the accrued path length. If the queue exhausts without reaching the destination, -1 is returned.
Overall, the program leverages a combination of heuristic-driven BFS and precise state management to efficiently determine the shortest path, or assert that such a path does not exist due to blockages.
- Python
class Solution:
def minPathInMatrix(self, matrix: List[List[int]]) -> int:
final_row = len(matrix) - 1
final_col = len(matrix[0]) - 1
movements = [
(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 1), (1, -1), (1, 0), (1, 1)]
def find_neighbors(row, col):
for delta_row, delta_col in movements:
new_r = row + delta_row
new_c = col + delta_col
if not(0 <= new_r <= final_row and 0 <= new_c <= final_col):
continue
if matrix[new_r][new_c] != 0:
continue
yield (new_r, new_c)
def heuristic(row, col):
return max(final_row - row, final_col - col)
if matrix[0][0] or matrix[final_row][final_col]:
return -1
seen = set()
heap = [(1 + heuristic(0, 0), 1, (0, 0))]
while heap:
total_dist, dist_so_far, coord = heapq.heappop(heap)
if coord in seen:
continue
if coord == (final_row, final_col):
return dist_so_far
seen.add(coord)
for neighbor in find_neighbors(*coord):
if neighbor in seen:
continue
total_dist = heuristic(*neighbor) + dist_so_far + 1
entry = (total_dist, dist_so_far + 1, neighbor)
heapq.heappush(heap, entry)
return -1
The solution entails finding the shortest path from the top-left to the bottom-right corner in a binary matrix, where 1 represents blocked cells and 0 represents free cells. The algorithm proceeds by implementing an A* search strategy using a priority queue.
- Initializes the final row and column indices.
- Defines possible movements (eight directions), and valid neighbors are evaluated using the
find_neighborsfunction. - Uses a heuristic function to estimate the shortest path distance, which uses the maximum of the differences in row and column indices.
- Checks if the start or end is blocked; if either is, returns
-1indicating no path exists. - Utilizes a min-heap priority queue to keep track of paths being explored based on estimated total distance, prioritizing paths with lower estimated distances.
- Processes each cell coordinate from the heap; if it's the destination, returns the path length.
- For each valid neighbor not previously visited, calculates the total distance, updates the path length, and pushes it onto the heap.
- If the heap is exhausted without finding a path to the destination, returns
-1.
This implementation efficiently manages the exploration of paths, dynamically adjusting to find the shortest path in potentially large matrices.
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