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

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

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:
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: 

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// 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;
}
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# 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
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Output: 
2

 

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