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Shortest path in grid with obstacles

Given a grid of n rows and m columns. Currently, you are at point (0, 0) and must reach point (n-1, m-1). If any cell contains 0, it is a free cell, if it contains 1, there is an obstacle and you can’t pass through the cell, and if it contains 2, that means it is free and if you step into this cell, you can pass through any adjacent cell of it that contains obstacles. In one step you can go one unit in any four directions (Inside the grid), the task is to find the minimum number of steps to reach the destination if it is impossible return -1.

Examples:

Input: n = 2, m = 3, grid[][] = {{0, 2, 1}, {0, 1, 0}}
Output: 3
Explanation: You can reach (0, 1) cell in one step, then you can pass through the blocked cell (0,2) and reach (1, 2) in 3 steps.

Input: n = 2, m = 3, grid = {{0, 0, 1}, {0, 1, 0}}
Output: -1
Explanation: You can’t reach your destination.

Finding minimum steps using BFS (Breadth-First Search)

The idea is to use Breadth-First Search (BFS) to find the minimum number of steps required to reach the destination. The BFS algorithm helps explore all possible paths from the starting point (0,0) to the destination point (n-1, m-1).

Follow the steps to solve the problem:

  • First, check whether the starting point (0, 0) or the destination point (n-1, m-1) is blocked (value 1). If either of them is blocked, it means that not able to reach the destination.
  • Use two arrays, dx and dy, to represent the four possible directions (up, right, down, and left).
  • Use a queue to perform BFS traversal on the grid.
  • Also, use a 2D vector dis to keep track of the minimum number of steps required to reach. initialize all values to -1 to indicate that they are not yet visited.
  • During BFS traversal, process each cell in the queue one by one. For each cell, explore its four neighboring cells using dx and dy.
    • If the neighboring cell is within the grid bounds, and it is either a free cell (value 0) or a special cell (value 2), proceed to process it.
    • If the neighboring cell is a special cell (value 2), update the grid to mark the adjacent blocked cells as free (value 0). This allows us to bypass obstacles by stepping into the special cell.
  • Then continue the BFS traversal from the neighboring cell and keep updating the minimum steps needed to reach each cell in the dis array.
  • Finally, look at the value in the “dis” vector for the bottom-right cell (n-1, m-1). This is the shortest way to find it. If it is still -1, it means there is no way at all.

Below is the implementation for the above approach:

C++

// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;
 
int possiblePath(int n, int m, vector<vector<int> >& grid)
{
 
    // Check if the source or destination
    // cell is blocked
    if (grid[0][0] == 1 || grid[n - 1][m - 1] == 1) {
 
        // Return -1 to indicate no path
        return -1;
    }
 
    // Create a queue to store the
    // cells to explore
    queue<pair<int, int> > q;
 
    // Add the source cell to the queue
    // and mark its distance as 0
    q.push({ 0, 0 });
 
    // Define two arrays to represent the
    // four directions of movement
    int dx[4] = { -1, 0, 1, 0 };
    int dy[4] = { 0, 1, 0, -1 };
 
    // Create a 2D vector to store the
    // distance of each cell
    // from the source
    vector<vector<int> > dis(n, vector<int>(m, -1));
 
    // Set the distance of the source
    // cell as 0
    dis[0][0] = 0;
 
    // Loop until the queue is empty or
    // the destination is reached
    while (!q.empty()) {
 
        // Get the front cell from the
        // queue and remove it
        pair<int, int> p = q.front();
        q.pop();
 
        // Loop through the four directions
        // of movement
        for (int i = 0; i < 4; i++) {
 
            // Calculate the coordinates
            // of the neighboring cell
            int x = p.first + dx[i];
            int y = p.second + dy[i];
            // Check if the neighboring
            // cell is inside the grid
            // and not visited before
            if (x >= 0 && x < n && y >= 0 && y < m
                && dis[x][y] == -1) {
 
                // Check if the neighboring
                // cell is free or special
                if (grid[x][y] == 0 || grid[x][y] == 2) {
 
                    // Set the distance of the
                    // neighboring cell as one
                    // more than the current cell
                    dis[x][y] = dis[p.first][p.second] + 1;
 
                    // Add the neighboring cell
                    // to the queue for
                    // further exploration
                    q.push({ x, y });
                }
 
                // Check if the neighboring
                // cell is special
                if (grid[x][y] == 2) {
 
                    // Loop through the four
                    // directions of movement again
                    for (int j = 0; j < 4; j++) {
 
                        // Calculate the coordinates
                        // of the adjacent cell
                        int xx = x + dx[j];
                        int yy = y + dy[j];
 
                        // Check if the adjacent
                        // cell is inside the grid
                        if (xx >= 0 && xx < n && yy >= 0
                            && yy < m) {
 
                            // Check if the adjacent
                            // cell is blocked
                            if (grid[xx][yy] == 1) {
 
                                // Change the adjacent
                                // cell to free
                                grid[xx][yy] = 0;
                            }
                        }
                    }
                }
            }
        }
    }
 
    // Return the distance of the
    // destination cell from the source
    return dis[n - 1][m - 1];
}
 
// Drivers code
int main()
{
 
    int n = 3;
    int m = 4;
    vector<vector<int> > grid = { { 0, 1, 2, 1 },
                                  { 2, 1, 0, 0 },
                                  { 0, 2, 1, 0 } };
 
    int result = possiblePath(n, m, grid);
 
    // Function Call
    cout << "Output: " << result << endl;
 
    return 0;
}

Java

// Java code for the above approach:
import java.util.*;
 
public class Main {
 
    public static int possiblePath(int n, int m, int[][] grid) {
 
        // Check if the source or destination cell is blocked
        if (grid[0][0] == 1 || grid[n - 1][m - 1] == 1) {
 
            // Return -1 to indicate no path
            return -1;
        }
 
        // Create a queue to store the cells to explore
        Queue<int[]> q = new LinkedList<>();
 
        // Add the source cell to the queue and mark its distance as 0
        q.add(new int[]{0, 0});
 
        // Define two arrays to represent the four directions of movement
        int[] dx = {-1, 0, 1, 0};
        int[] dy = {0, 1, 0, -1};
 
        // Create a 2D array to store the distance of each cell
        // from the source
        int[][] dis = new int[n][m];
        for (int i = 0; i < n; i++) {
            Arrays.fill(dis[i], -1);
        }
 
        // Set the distance of the source cell as 0
        dis[0][0] = 0;
 
        // Loop until the queue is empty or the destination is reached
        while (!q.isEmpty()) {
 
            // Get the front cell from the queue and remove it
            int[] p = q.poll();
            int x = p[0];
            int y = p[1];
 
            // Loop through the four directions of movement
            for (int i = 0; i < 4; i++) {
 
                // Calculate the coordinates of the neighboring cell
                int xx = x + dx[i];
                int yy = y + dy[i];
 
                // Check if the neighboring cell is inside the grid
                // and not visited before
                if (xx >= 0 && xx < n && yy >= 0 && yy < m && dis[xx][yy] == -1) {
 
                    // Check if the neighboring cell is free or special
                    if (grid[xx][yy] == 0 || grid[xx][yy] == 2) {
 
                        // Set the distance of the neighboring cell as one
                        // more than the current cell
                        dis[xx][yy] = dis[x][y] + 1;
 
                        // Add the neighboring cell to the queue for
                        // further exploration
                        q.add(new int[]{xx, yy});
                    }
 
                    // Check if the neighboring cell is special
                    if (grid[xx][yy] == 2) {
 
                        // Loop through the four directions of movement again
                        for (int j = 0; j < 4; j++) {
 
                            // Calculate the coordinates of the adjacent cell
                            int xxx = xx + dx[j];
                            int yyy = yy + dy[j];
 
                            // Check if the adjacent cell is inside the grid
                            if (xxx >= 0 && xxx < n && yyy >= 0 && yyy < m) {
 
                                // Check if the adjacent cell is blocked
                                if (grid[xxx][yyy] == 1) {
 
                                    // Change the adjacent cell to free
                                    grid[xxx][yyy] = 0;
                                }
                            }
                        }
                    }
                }
            }
        }
 
        // Return the distance of the destination cell from the source
        return dis[n - 1][m - 1];
    }
 
    public static void main(String[] args) {
 
        int n = 3;
        int m = 4;
        int[][] grid = {
            {0, 1, 2, 1},
            {2, 1, 0, 0},
            {0, 2, 1, 0}
        };
 
        int result = possiblePath(n, m, grid);
 
        // Function Call
        System.out.println("Output: " + result);
    }
}

Python

from collections import deque
 
def possiblePath(n, m, grid):
    # Check if the source or destination cell is blocked
    if grid[0][0] == 1 or grid[n - 1][m - 1] == 1:
        # Return -1 to indicate no path
        return -1
 
    # Create a queue to store the cells to explore
    q = deque()
    # Add the source cell to the queue and mark its distance as 0
    q.append((0, 0))
 
    # Define two arrays to represent the four directions of movement
    dx = [-1, 0, 1, 0]
    dy = [0, 1, 0, -1]
 
    # Create a 2D list to store the distance of each cell from the source
    dis = [[-1 for _ in range(m)] for _ in range(n)]
 
    # Set the distance of the source cell as 0
    dis[0][0] = 0
 
    # Loop until the queue is empty or the destination is reached
    while q:
        # Get the front cell from the queue and remove it
        p = q.popleft()
 
        # Loop through the four directions of movement
        for i in range(4):
            # Calculate the coordinates of the neighboring cell
            x = p[0] + dx[i]
            y = p[1] + dy[i]
            # Check if the neighboring cell is inside the grid and not visited before
            if 0 <= x < n and 0 <= y < m and dis[x][y] == -1:
                # Check if the neighboring cell is free or special
                if grid[x][y] == 0 or grid[x][y] == 2:
                    # Set the distance of the neighboring cell as one more than the current cell
                    dis[x][y] = dis[p[0]][p[1]] + 1
                    # Add the neighboring cell to the queue for further exploration
                    q.append((x, y))
                # Check if the neighboring cell is special
                if grid[x][y] == 2:
                    # Loop through the four directions of movement again
                    for j in range(4):
                        # Calculate the coordinates of the adjacent cell
                        xx = x + dx[j]
                        yy = y + dy[j]
                        # Check if the adjacent cell is inside the grid
                        if 0 <= xx < n and 0 <= yy < m:
                            # Check if the adjacent cell is blocked
                            if grid[xx][yy] == 1:
                                # Change the adjacent cell to free
                                grid[xx][yy] = 0
 
    # Return the distance of the destination cell from the source
    return dis[n - 1][m - 1]
 
# Driver code
if __name__ == "__main__":
    n = 3
    m = 4
    grid = [
        [0, 1, 2, 1],
        [2, 1, 0, 0],
        [0, 2, 1, 0]
    ]
 
    result = possiblePath(n, m, grid)
 
    # Function Call
    print(result)

C#

// C# implementation of the above approach
using System;
using System.Collections.Generic;
 
public class MainClass
{
    public static int PossiblePath(int n, int m, int[][] grid)
    {
        // Check if the source or destination cell is blocked
        if (grid[0][0] == 1 || grid[n - 1][m - 1] == 1)
        {
            // Return -1 to indicate no path
            return -1;
        }
 
        // Create a queue to store the cells to explore
        Queue<int[]> q = new Queue<int[]>();
 
        // Add the source cell to the queue and mark its distance as 0
        q.Enqueue(new int[] { 0, 0 });
 
        // Define two arrays to represent the four directions of movement
        int[] dx = { -1, 0, 1, 0 };
        int[] dy = { 0, 1, 0, -1 };
 
        // Create a 2D array to store the distance of each cell
        // from the source
        int[][] dis = new int[n][];
        for (int i = 0; i < n; i++)
        {
            dis[i] = new int[m];
            for (int j = 0; j < m; j++)
            {
                dis[i][j] = -1;
            }
        }
 
        // Set the distance of the source cell as 0
        dis[0][0] = 0;
 
        // Loop until the queue is empty or the destination is reached
        while (q.Count > 0)
        {
            // Get the front cell from the queue and remove it
            int[] p = q.Dequeue();
            int x = p[0];
            int y = p[1];
 
            // Loop through the four directions of movement
            for (int i = 0; i < 4; i++)
            {
                // Calculate the coordinates of the neighboring cell
                int xx = x + dx[i];
                int yy = y + dy[i];
 
                // Check if the neighboring cell is inside the grid
                // and not visited before
                if (xx >= 0 && xx < n && yy >= 0 && yy < m && dis[xx][yy] == -1)
                {
                    // Check if the neighboring cell is free or special
                    if (grid[xx][yy] == 0 || grid[xx][yy] == 2)
                    {
                        // Set the distance of the neighboring cell as one
                        // more than the current cell
                        dis[xx][yy] = dis[x][y] + 1;
 
                        // Add the neighboring cell to the queue for
                        // further exploration
                        q.Enqueue(new int[] { xx, yy });
                    }
 
                    // Check if the neighboring cell is special
                    if (grid[xx][yy] == 2)
                    {
                        // Loop through the four directions of movement again
                        for (int j = 0; j < 4; j++)
                        {
                            // Calculate the coordinates of the adjacent cell
                            int xxx = xx + dx[j];
                            int yyy = yy + dy[j];
 
                            // Check if the adjacent cell is inside the grid
                            if (xxx >= 0 && xxx < n && yyy >= 0 && yyy < m)
                            {
                                // Check if the adjacent cell is blocked
                                if (grid[xxx][yyy] == 1)
                                {
                                    // Change the adjacent cell to free
                                    grid[xxx][yyy] = 0;
                                }
                            }
                        }
                    }
                }
            }
        }
 
        // Return the distance of the destination cell from the source
        return dis[n - 1][m - 1];
    }
 
    public static void Main(string[] args)
    {
        int n = 3;
        int m = 4;
        int[][] grid = new int[][] {
            new int[] { 0, 1, 2, 1 },
            new int[] { 2, 1, 0, 0 },
            new int[] { 0, 2, 1, 0 }
        };
 
        int result = PossiblePath(n, m, grid);
 
        // Function Call
        Console.WriteLine("Output: " + result);
    }
}
 
// This code is contributed by Tapesh(tapeshdua420)

Javascript

function GFG(n, m, grid) {
    // Check if the source or destination cell is blocked
    if (grid[0][0] === 1 || grid[n - 1][m - 1] === 1) {
        // Return -1 to indicate no path
        return -1;
    }
    // Create a queue to store the cells to explore
    const queue = [];
    queue.push([0, 0]);
    // Define two arrays to represent
    // the four directions of movement
    const dx = [-1, 0, 1, 0];
    const dy = [0, 1, 0, -1];
    const dis = new Array(n);
    for (let i = 0; i < n; i++) {
        dis[i] = new Array(m).fill(-1);
    }
    // Set the distance of the source cell as 0
    dis[0][0] = 0;
    // Loop until the queue is empty or
    // the destination is reached
    while (queue.length > 0) {
        // Get the front cell from the queue and remove it
        const p = queue.shift();
        const x = p[0];
        const y = p[1];
        // Loop through the four directions of movement
        for (let i = 0; i < 4; i++) {
            // Calculate the coordinates of the neighboring cell
            const xx = x + dx[i];
            const yy = y + dy[i];
            if (xx >= 0 && xx < n && yy >= 0 && yy < m && dis[xx][yy] === -1) {
                // Check if the neighboring cell is free or special
                if (grid[xx][yy] === 0 || grid[xx][yy] === 2) {
                    dis[xx][yy] = dis[x][y] + 1;
                    // Add the neighboring cell to the queue for further exploration
                    queue.push([xx, yy]);
                }
                // Check if the neighboring cell is special
                if (grid[xx][yy] === 2) {
                    for (let j = 0; j < 4; j++) {
                        // Calculate the coordinates of the adjacent cell
                        const xxx = xx + dx[j];
                        const yyy = yy + dy[j];
                        // Check if the adjacent cell is inside the grid
                        if (xxx >= 0 && xxx < n && yyy >= 0 && yyy < m) {
                            // Check if the adjacent cell is blocked
                            if (grid[xxx][yyy] === 1) {
                                // Change the adjacent cell to free
                                grid[xxx][yyy] = 0;
                            }
                        }
                    }
                }
            }
        }
    }
    // Return the distance of the destination cell
    // from the source
    return dis[n - 1][m - 1];
}
// Driver code
const n = 3;
const m = 4;
const grid = [
    [0, 1, 2, 1],
    [2, 1, 0, 0],
    [0, 2, 1, 0]
];
const result = GFG(n, m, grid);
// Function Call
console.log("Output: " + result);

Output

Output: 5

Time complexity: O(N * M)
Auxiliary Space: O(N * M)


 


Reffered: https://www.geeksforgeeks.org

DSA

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Category:
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Sub Category:
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