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# 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:
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 ``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[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 >& grid)``{``    ``int` `z_count = 0; ``// Total 0s present``    ``int` `n = grid.size(), m = grid[0].size();``    ``int` `ans = 0;``    ``vector > 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 > 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

 ``

Output:

`2`

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

The Recursive DFS.

## C++

 `#include``#include``#define ll long long``//ll mod=1e9+7;``using` `namespace` `std;` `int` `res = 0, empty = 1;``    ``void` `path(vector>& 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>& 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>grid{{1, 0, 0, 0},``                           ``{0, 0, 0, 0},``                           ``{0, 0, 2, -1}};``   ``cout<

## 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)