Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Unique paths covering every non-obstacle block exactly once in a grid

  • Difficulty Level : Hard
  • Last Updated : 28 Sep, 2021

Given a grid grid[][] with 4 types of blocks: 

  • 1 represents the starting block. There is exactly one starting block.
  • 2 represents the ending block. There is exactly one ending block.
  • 0 represents empty block we can walk over.
  • -1 represents obstacles that we cannot walk over.

The task is to count the number of paths from the starting block to the ending block such that every non-obstacle block is covered exactly once.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Examples:  



Input: grid[][] = { 
{1, 0, 0, 0}, 
{0, 0, 0, 0}, 
{0, 0, 2, -1} } 
Output:
Following are the only paths covering all the non-obstacle blocks: 

Input: grid[][] = { 
{1, 0, 0, 0}, 
{0, 0, 0, 0}, 
{0, 0, 0, 2} } 
Output:

Approach: We can use simple DFS here with backtracking. We can check that a particular path has covered all the non-obstacle blocks by counting all the blocks encountered in the way and finally comparing it with the total number of blocks available and if they match, then we add it as a valid solution.

Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function for dfs.
// i, j ==> Current cell indexes
// vis ==> To mark visited cells
// ans ==> Result
// z ==> Current count 0s visited
// z_count ==> Total 0s present
void dfs(int i, int j, vector<vector<int> >& grid,
         vector<vector<bool> >& vis, int& ans,
         int z, int z_count)
{
    int n = grid.size(), m = grid[0].size();
 
    // Mark the block as visited
    vis[i][j] = 1;
    if (grid[i][j] == 0)
 
        // update the count
        z++;
 
    // If end block reached
    if (grid[i][j] == 2) {
 
        // If path covered all the non-
        // obstacle blocks
        if (z == z_count)
            ans++;
        vis[i][j] = 0;
        return;
    }
 
    // Up
    if (i >= 1 && !vis[i - 1][j] && grid[i - 1][j] != -1)
        dfs(i - 1, j, grid, vis, ans, z, z_count);
 
    // Down
    if (i < n - 1 && !vis[i + 1][j] && grid[i + 1][j] != -1)
        dfs(i + 1, j, grid, vis, ans, z, z_count);
 
    // Left
    if (j >= 1 && !vis[i][j - 1] && grid[i][j - 1] != -1)
        dfs(i, j - 1, grid, vis, ans, z, z_count);
 
    // Right
    if (j < m - 1 && !vis[i][j + 1] && grid[i][j + 1] != -1)
        dfs(i, j + 1, grid, vis, ans, z, z_count);
 
    // Unmark the block (unvisited)
    vis[i][j] = 0;
}
 
// Function to return the count of the unique paths
int uniquePaths(vector<vector<int> >& grid)
{
    int z_count = 0; // Total 0s present
    int n = grid.size(), m = grid[0].size();
    int ans = 0;
    vector<vector<bool> > vis(n, vector<bool>(m, 0));
    int x, y;
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
 
            // Count non-obstacle blocks
            if (grid[i][j] == 0)
                z_count++;
            else if (grid[i][j] == 1) {
 
                // Starting position
                x = i, y = j;
            }
        }
    }
    dfs(x, y, grid, vis, ans, 0, z_count);
    return ans;
}
 
// Driver code
int main()
{
    vector<vector<int> > grid{ { 1, 0, 0, 0 },
                               { 0, 0, 0, 0 },
                               { 0, 0, 2, -1 } };
 
    cout << uniquePaths(grid);
    return 0;
}

Java




// Java implementation of the approach
import java.util.Arrays;
class GFG
{
  static int ans = 0;
 
  // Function for dfs.
  // i, j ==> Current cell indexes
  // vis ==> To mark visited cells
  // ans ==> Result
  // z ==> Current count 0s visited
  // z_count ==> Total 0s present
  static void dfs(int i, int j, int[][] grid,
                  boolean[][] vis, int z, int z_count)
  {
    int n = grid.length, m = grid[0].length;
 
    // Mark the block as visited
    vis[i][j] = true;
    if (grid[i][j] == 0)
 
      // update the count
      z++;
 
    // If end block reached
    if (grid[i][j] == 2)
    {
 
      // If path covered all the non-
      // obstacle blocks
      if (z == z_count)
        ans++;
      vis[i][j] = false;
      return;
    }
 
    // Up
    if (i >= 1 && !vis[i - 1][j] && grid[i - 1][j] != -1)
      dfs(i - 1, j, grid, vis, z, z_count);
 
    // Down
    if (i < n - 1 && !vis[i + 1][j] && grid[i + 1][j] != -1)
      dfs(i + 1, j, grid, vis, z, z_count);
 
    // Left
    if (j >= 1 && !vis[i][j - 1] && grid[i][j - 1] != -1)
      dfs(i, j - 1, grid, vis, z, z_count);
 
    // Right
    if (j < m - 1 && !vis[i][j + 1] && grid[i][j + 1] != -1)
      dfs(i, j + 1, grid, vis, z, z_count);
 
    // Unmark the block (unvisited)
    vis[i][j] = false;
  }
 
  // Function to return the count of the unique paths
  static int uniquePaths(int[][] grid)
  {
    int z_count = 0; // Total 0s present
    int n = grid.length, m = grid[0].length;
 
    boolean[][] vis = new boolean[n][m];
    for (int i = 0; i < n; i++)
    {
      Arrays.fill(vis[i], false);
    }
    int x = 0, y = 0;
    for (int i = 0; i < n; ++i)
    {
      for (int j = 0; j < m; ++j)
      {
 
        // Count non-obstacle blocks
        if (grid[i][j] == 0)
          z_count++;
        else if (grid[i][j] == 1)
        {
 
          // Starting position
          x = i;
          y = j;
        }
      }
    }
    dfs(x, y, grid, vis, 0, z_count);
    return ans;
  }
 
  // Driver code
  public static void main(String[] args)
  {
 
    int[][] grid = { { 1, 0, 0, 0 }, { 0, 0, 0, 0 }, { 0, 0, 2, -1 } };
    System.out.println(uniquePaths(grid));
  }
}
 
// This code is contributed by sanjeev2552

Python3




# Python3 implementation of the approach
 
# Function for dfs.
# i, j ==> Current cell indexes
# vis ==> To mark visited cells
# ans ==> Result
# z ==> Current count 0s visited
# z_count ==> Total 0s present
def dfs(i, j, grid, vis, ans, z, z_count):
 
    n = len(grid)
    m = len(grid[0])
 
    # Mark the block as visited
    vis[i][j] = 1
     
    if (grid[i][j] == 0):
 
        # Update the count
        z += 1
 
    # If end block reached
    if (grid[i][j] == 2):
 
        # If path covered all the non-
        # obstacle blocks
        if (z == z_count):
            ans += 1
             
        vis[i][j] = 0
         
        return grid, vis, ans
         
    # Up
    if (i >= 1 and not vis[i - 1][j] and
                      grid[i - 1][j] != -1):
        grid, vis, ans = dfs(i - 1, j, grid,
                             vis, ans, z,
                             z_count)
 
    # Down
    if (i < n - 1 and not vis[i + 1][j] and
                         grid[i + 1][j] != -1):
        grid, vis, ans = dfs(i + 1, j, grid,
                             vis, ans, z,
                             z_count)
 
    # Left
    if (j >= 1 and not vis[i][j - 1] and
                      grid[i][j - 1] != -1):
        grid, vis, ans = dfs(i, j - 1, grid,
                             vis, ans, z,
                             z_count)
 
    # Right
    if (j < m - 1 and not vis[i][j + 1] and
                         grid[i][j + 1] != -1):
        grid, vis, ans = dfs(i, j + 1, grid,
                             vis, ans, z,
                             z_count)
 
    # Unmark the block (unvisited)
    vis[i][j] = 0
     
    return grid, vis, ans
 
# Function to return the count
# of the unique paths
def uniquePaths(grid):
     
    # Total 0s present
    z_count = 0
    n = len(grid)
    m = len(grid[0])
    ans = 0
     
    vis = [[0 for j in range(m)]
              for i in range(n)]
     
    x = 0
    y = 0
     
    for i in range(n):
        for j in range(m):
             
            # Count non-obstacle blocks
            if grid[i][j] == 0:
                z_count += 1
                 
            elif (grid[i][j] == 1):
                 
                # Starting position
                x = i
                y = j
         
    grid, vis, ans = dfs(x, y, grid,
                         vis, ans, 0,
                         z_count)
                          
    return ans
 
# Driver code
if __name__=='__main__':
 
    grid = [ [ 1, 0, 0, 0 ],
             [ 0, 0, 0, 0 ],
             [ 0, 0, 2, -1 ] ]
              
    print(uniquePaths(grid))
     
# This code is contributed by rutvik_56

C#




// C# implementation of the approach
using System;
class GFG {
     
  static int ans = 0;
  
  // Function for dfs.
  // i, j ==> Current cell indexes
  // vis ==> To mark visited cells
  // ans ==> Result
  // z ==> Current count 0s visited
  // z_count ==> Total 0s present
  static void dfs(int i, int j, int[,] grid,
                  bool[,] vis, int z, int z_count)
  {
    int n = grid.GetLength(0), m = grid.GetLength(1);
  
    // Mark the block as visited
    vis[i,j] = true;
    if (grid[i,j] == 0)
  
      // update the count
      z++;
  
    // If end block reached
    if (grid[i,j] == 2)
    {
  
      // If path covered all the non-
      // obstacle blocks
      if (z == z_count)
        ans++;
      vis[i,j] = false;
      return;
    }
  
    // Up
    if (i >= 1 && !vis[i - 1,j] && grid[i - 1,j] != -1)
      dfs(i - 1, j, grid, vis, z, z_count);
  
    // Down
    if (i < n - 1 && !vis[i + 1,j] && grid[i + 1,j] != -1)
      dfs(i + 1, j, grid, vis, z, z_count);
  
    // Left
    if (j >= 1 && !vis[i,j - 1] && grid[i,j - 1] != -1)
      dfs(i, j - 1, grid, vis, z, z_count);
  
    // Right
    if (j < m - 1 && !vis[i,j + 1] && grid[i,j + 1] != -1)
      dfs(i, j + 1, grid, vis, z, z_count);
  
    // Unmark the block (unvisited)
    vis[i,j] = false;
  }
  
  // Function to return the count of the unique paths
  static int uniquePaths(int[,] grid)
  {
    int z_count = 0; // Total 0s present
    int n = grid.GetLength(0), m = grid.GetLength(1);
  
    bool[,] vis = new bool[n,m];
    for (int i = 0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
        {
            vis[i,j] = false;
        }
    }
    int x = 0, y = 0;
    for (int i = 0; i < n; ++i)
    {
      for (int j = 0; j < m; ++j)
      {
  
        // Count non-obstacle blocks
        if (grid[i,j] == 0)
          z_count++;
        else if (grid[i,j] == 1)
        {
  
          // Starting position
          x = i;
          y = j;
        }
      }
    }
    dfs(x, y, grid, vis, 0, z_count);
    return ans;
  }
   
  // Driver code
  static void Main() {
    int[,] grid = { { 1, 0, 0, 0 }, { 0, 0, 0, 0 }, { 0, 0, 2, -1 } };
    Console.WriteLine(uniquePaths(grid));
  }
}
 
// This code is contributed by divyesh072019.

Javascript




<script>
    // Javascript implementation of the approach
    let ans = 0;
  
    // Function for dfs.
    // i, j ==> Current cell indexes
    // vis ==> To mark visited cells
    // ans ==> Result
    // z ==> Current count 0s visited
    // z_count ==> Total 0s present
    function dfs(i, j, grid, vis, z, z_count)
    {
      let n = grid.length, m = grid[0].length;
 
      // Mark the block as visited
      vis[i][j] = true;
      if (grid[i][j] == 0)
 
        // update the count
        z++;
 
      // If end block reached
      if (grid[i][j] == 2)
      {
 
        // If path covered all the non-
        // obstacle blocks
        if (z == z_count)
          ans++;
        vis[i][j] = false;
        return;
      }
 
      // Up
      if (i >= 1 && !vis[i - 1][j] && grid[i - 1][j] != -1)
        dfs(i - 1, j, grid, vis, z, z_count);
 
      // Down
      if (i < n - 1 && !vis[i + 1][j] && grid[i + 1][j] != -1)
        dfs(i + 1, j, grid, vis, z, z_count);
 
      // Left
      if (j >= 1 && !vis[i][j - 1] && grid[i][j - 1] != -1)
        dfs(i, j - 1, grid, vis, z, z_count);
 
      // Right
      if (j < m - 1 && !vis[i][j + 1] && grid[i][j + 1] != -1)
        dfs(i, j + 1, grid, vis, z, z_count);
 
      // Unmark the block (unvisited)
      vis[i][j] = false;
    }
 
    // Function to return the count of the unique paths
    function uniquePaths(grid)
    {
      let z_count = 0; // Total 0s present
      let n = grid.length, m = grid[0].length;
 
      let vis = new Array(n);
      for (let i = 0; i < n; i++)
      {
          vis[i] = new Array(m);
        for(let j = 0; j < m; j++)
        {
            vis[i][j] = false;
        }
      }
      let x = 0, y = 0;
      for (let i = 0; i < n; ++i)
      {
        for (let j = 0; j < m; ++j)
        {
 
          // Count non-obstacle blocks
          if (grid[i][j] == 0)
            z_count++;
          else if (grid[i][j] == 1)
          {
 
            // Starting position
            x = i;
            y = j;
          }
        }
      }
      dfs(x, y, grid, vis, 0, z_count);
      return ans;
    }
     
    let grid = [ [ 1, 0, 0, 0 ], [ 0, 0, 0, 0 ], [ 0, 0, 2, -1 ] ];
    document.write(uniquePaths(grid));
 
// This code is contributed by decode2207.
</script>
Output: 
2

 

Time Complexity: O(row * cols)
Auxiliary Space: O(row * cols)




My Personal Notes arrow_drop_up
Recommended Articles
Page :