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
Unique paths covering every non-obstacle block exactly once in a grid using DFS and Backtraking:
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 <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 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 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# 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. |
<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> |
2
Time Complexity: O(row * cols)!
Auxiliary Space: O(row * cols)