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 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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++ implementation of the approach #include 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 >& grid,          vector >& vis, int& ans,          int z, int z_count) {     int n = grid.size(), m = grid.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 >& grid) {     int z_count = 0; // Total 0s present     int n = grid.size(), m = grid.size();     int ans = 0;     vector > vis(n, vector(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 > grid{ { 1, 0, 0, 0 },                                { 0, 0, 0, 0 },                                { 0, 0, 2, -1 } };        cout << uniquePaths(grid);     return 0; }

Output:

2

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