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

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 an 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.

Examples:

Input: grid[][] = { 
{1, 0, 0, 0}, 
{0, 0, 0, 0}, 
{0, 0, 2, -1} } 
Output: 2 
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: 4 

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)

The Recursive DFS.

C++

#include<bits/stdc++.h>
#include<iostream>
#define ll long long
//ll mod=1e9+7;
using namespace std;

int res = 0, empty = 1;
    void path(vector<vector<int>>& grid, int x, int y, int count) {
        if (x < 0 || x >= grid.size() || y < 0 || y >= grid[0].size() || grid[x][y] == -1) return;

        if (grid[x][y] == 2) {
            if(empty == count) res++;
            return;
        }

        grid[x][y] = -1;

        path(grid, x+1, y, count+1);
        path(grid, x-1, y, count+1);
        path(grid, x, y+1, count+1);
        path(grid, x, y-1, count+1);

        grid[x][y] = 0;
      }
      int uniquePathsIII(vector<vector<int>>& grid) {
            int start_x, start_y;
            for (int i = 0; i < grid.size(); i++) {
                for (int j = 0; j < grid[0].size(); j++) {
                    if (grid[i][j] == 1) start_x = i, start_y = j;
                    else if (grid[i][j] == 0) empty++;
                }
            }

            path(grid, start_x, start_y, 0);
            return res;
        }
signed main(){
   vector<vector<int>>grid{{1, 0, 0, 0},
                           {0, 0, 0, 0},
                           {0, 0, 2, -1}};
   cout<<uniquePathsIII(grid)<<endl;
    return 0;
}

Python3

# function to count all possible unique paths in the grid
# from start to end, considering empty cells and obstacles
def uniquePathsIII(grid):
    res = 0
    empty = 1

    # recursive function to traverse through the grid
    def path(x, y, count):
        nonlocal res
        # check if current cell is out of grid or is an obstacle
        if x < 0 or x >= len(grid) or y < 0 or y >= len(grid[0]) or grid[x][y] == -1:
            return

        # check if current cell is the end cell
        if grid[x][y] == 2:
            if empty == count:
                res += 1
            return

        # mark current cell as visited
        grid[x][y] = -1

        # recursively call path function for all possible directions
        path(x+1, y, count+1)
        path(x-1, y, count+1)
        path(x, y+1, count+1)
        path(x, y-1, count+1)

        # mark current cell as unvisited
        grid[x][y] = 0

    # initialize start coordinates and empty cell count
    for i in range(len(grid)):
        for j in range(len(grid[0])):
            if grid[i][j] == 1:
                start_x, start_y = i, j
            elif grid[i][j] == 0:
                empty += 1

    # call path function starting from the start cell
    path(start_x, start_y, 0)
    return res

# test the function
grid = [[1, 0, 0, 0], [0, 0, 0, 0], [0, 0, 2, -1]]
print(uniquePathsIII(grid))

# This code is contributed by lokeshpotta20.

C#

using System;

public class Gfg
{
    static int res = 0, empty = 1;
    static void path(int[ , ] grid, int x, int y, int count)
    {
        if (x < 0 || x >= grid.GetLength(0) || y < 0 || y >= grid.GetLength(1) || grid[x, y] == -1)
            return;

        if (grid[x, y] == 2) {
            if(empty == count)
                res++;
            return;
        }

        grid[x, y] = -1;

        path(grid, x+1, y, count+1);
        path(grid, x-1, y, count+1);
        path(grid, x, y+1, count+1);
        path(grid, x, y-1, count+1);

        grid[x, y] = 0;
     }
    static int uniquePathsIII(int[, ] grid)
    {
        int start_x=-1, start_y=-1;
        for (int i = 0; i < grid.GetLength(0); i++) {
            for (int j = 0; j < grid.GetLength(1); j++) {
                if (grid[i, j] == 1)
                {
                    start_x = i;
                    start_y = j;
                }
                else if (grid[i, j] == 0)
                    empty++;
            }
        }

        path(grid, start_x, start_y, 0);
        return res;
    }
    public static void Main(string[] args)
    {
        int[ , ] grid={{1, 0, 0, 0},{0, 0, 0, 0},{0, 0, 2, -1}};
        Console.WriteLine(uniquePathsIII(grid));
    }
}

// This code is contributed by ritaagarwal.

Javascript

let mod=1e9+7;

let res = 0, empty = 1;
function path(grid, x, y, count)
{
    if (x < 0 || x >= grid.length || y < 0 || y >= grid[0].length || grid[x][y] == -1)
        return;

    if (grid[x][y] == 2) {
        if(empty == count)
            res++;
        return;
    }

    grid[x][y] = -1;

    path(grid, x+1, y, count+1);
    path(grid, x-1, y, count+1);
    path(grid, x, y+1, count+1);
    path(grid, x, y-1, count+1);

    grid[x][y] = 0;
}
function uniquePathsIII(grid) {
    let start_x, start_y;
    for (let i = 0; i < grid.length; i++) {
        for (let j = 0; j < grid[0].length; j++) {
            if (grid[i][j] == 1)
                start_x = i, start_y = j;
            else if (grid[i][j] == 0)
                empty++;
        }
    }

    path(grid, start_x, start_y, 0);
    return res;
}
let grid= [[1, 0, 0, 0],[0, 0, 0, 0],[0, 0, 2, -1]];
console.log(uniquePathsIII(grid));

// This code is contributed by poojaagarwal2.

Java

// Java code for the above approach

import java.io.*;

class GFG {

    static int res = 0, empty = 1;
    static void path(int[][] grid, int x, int y, int count)
    {
        if (x < 0 || x >= grid.length || y < 0
            || y >= grid[0].length || grid[x][y] == -1)
            return;

        if (grid[x][y] == 2) {
            if (empty == count)
                res++;
            return;
        }

        grid[x][y] = -1;

        path(grid, x + 1, y, count + 1);
        path(grid, x - 1, y, count + 1);
        path(grid, x, y + 1, count + 1);
        path(grid, x, y - 1, count + 1);

        grid[x][y] = 0;
    }
    static int uniquePathsIII(int[][] grid)
    {
        int start_x = -1, start_y = -1;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) {
                    start_x = i;
                    start_y = j;
                }
                else if (grid[i][j] == 0)
                    empty++;
            }
        }

        path(grid, start_x, start_y, 0);
        return res;
    }

    public static void main(String[] args)
    {
        int[][] grid = { { 1, 0, 0, 0 },
                         { 0, 0, 0, 0 },
                         { 0, 0, 2, -1 } };
        System.out.println(uniquePathsIII(grid));
    }
}

// This code is contributed by lokeshmvs21.

Output

2

Since we are exploring all possible paths by making grid[x][y]=0 at end of the call, so complexity will be exponential. 

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