Unique paths in a Grid with Obstacles

Given a grid of size m * n, let us assume you are starting at (1, 1) and your goal is to reach (m, n). At any instance, if you are on (x, y), you can either go to (x, y + 1) or (x + 1, y).
Now consider if some obstacles are added to the grids. How many unique paths would there be?
An obstacle and empty space are marked as 1 and 0 respectively in the grid.

Examples:  

Input: [[0, 0, 0],
        [0, 1, 0],
        [0, 0, 0]]
Output : 2
There is only one obstacle in the middle.

Method 1: Recursion

We have discussed a problem to count the number of unique paths in a Grid when no obstacle was present in the grid. But here the situation is quite different. While moving through the grid, we can get some obstacles that we can not jump and that way to reach the bottom right corner is blocked. 

C++

// C++ code to find number of unique paths
// in a Matrix
#include<bits/stdc++.h>
using namespace std;

int  UniquePathHelper(int i, int j, int r, int c, vector<vector<int>>& A){
    // boundary condition or constraints
    if(i == r || j == c){
      return 0 ;
    }

    if(A[i][j] == 1){
      return 0 ;
    }

    // base case
    if(i == r-1 && j == c-1){
      return 1 ;
    }

    return  UniquePathHelper(i+1, j, r, c, A) +
            UniquePathHelper(i, j+1, r, c, A) ;
}


int uniquePathsWithObstacles(vector<vector<int>>& A)
{

    int r = A.size(), c = A[0].size();


    return UniquePathHelper(0, 0, r, c, A) ;
}

// Driver code
int main()
{
   vector<vector<int>> A = { { 0, 0, 0 },
                             { 0, 1, 0 },
                             { 0, 0, 0 } };

   cout << uniquePathsWithObstacles(A) << " \n";
}

Java

// Java code to find number of unique paths
// in a Matrix
import java.io.*;

class GFG {

  static int UniquePathHelper(int i, int j, int r, int c,
                              int[][] A)
  {
    // boundary condition or constraints
    if (i == r || j == c) {
      return 0;
    }

    if (A[i][j] == 1) {
      return 0;
    }

    // base case
    if (i == r - 1 && j == c - 1) {
      return 1;
    }

    return UniquePathHelper(i + 1, j, r, c, A)
      + UniquePathHelper(i, j + 1, r, c, A);
  }

  static int uniquePathsWithObstacles(int[][] A)
  {

    int r = A.length, c = A[0].length;

    return UniquePathHelper(0, 0, r, c, A);
  }

  // Driver Code
  public static void main(String[] args)
  {
    int[][] A
      = { { 0, 0, 0 }, { 0, 1, 0 }, { 0, 0, 0 } };

    System.out.print(uniquePathsWithObstacles(A));
  }
}

// This code is contributed by nipun_aggarwal

Python3

# Python code to find number of unique paths
# in a Matrix
def  UniquePathHelper(i, j, r, c, A):

    # boundary condition or constraints
    if(i == r or j == c):
      return 0

    if(A[i][j] == 1):
      return 0

    # base case
    if(i == r-1 and j == c-1):
      return 1

    return  UniquePathHelper(i+1, j, r, c, A) + UniquePathHelper(i, j+1, r, c, A)

def uniquePathsWithObstacles(A):
    r,c = len(A),len(A[0])

    return UniquePathHelper(0, 0, r, c, A)

# Driver code
A = [ [ 0, 0, 0 ],
      [ 0, 1, 0 ],
      [ 0, 0, 0 ] ]

print(uniquePathsWithObstacles(A))

# This code is contributed by shinjanpatra

Javascript

<script>

// JavaScript code to find number of unique paths
// in a Matrix
function  UniquePathHelper(i, j, r, c, A){

    // boundary condition or constraints
    if(i == r || j == c)
      return 0

    if(A[i][j] == 1)
      return 0

    // base case
    if(i == r-1 && j == c-1)
      return 1

    return  UniquePathHelper(i+1, j, r, c, A) + UniquePathHelper(i, j+1, r, c, A)
}

function uniquePathsWithObstacles(A){
    let r = A.length, c = A[0].length

    return UniquePathHelper(0, 0, r, c, A)
}

// Driver code
let A = [ [ 0, 0, 0 ],
      [ 0, 1, 0 ],
      [ 0, 0, 0 ] ]

document.write(uniquePathsWithObstacles(A))

// This code is contributed by shinjanpatra

</script>

Output

2 

Time Complexity: O(2^m*n)

Auxiliary Space: O(m*n)

Method 2: Using DP

1) Top-Down

The most efficient solution to this problem can be achieved using dynamic programming. Like every dynamic problem concept, we will not recompute the subproblems. A temporary 2D matrix will be constructed and value will be stored using the top-down approach. 

C++

// C++ code to find number of unique paths
// in a Matrix
#include <bits/stdc++.h>
using namespace std;

int UniquePathHelper(int i, int j, int r, int c,
                     vector<vector<int> >& A,
                     vector<vector<int> >& paths)
{
    // boundary condition or constraints
    if (i == r || j == c) {
        return 0;
    }

    if (A[i][j] == 1) {
        return 0;
    }

    // base case
    if (i == r - 1 && j == c - 1) {
        return 1;
    }

    if (paths[i][j] != -1) {
        return paths[i][j];
    }

    return UniquePathHelper(i + 1, j, r, c, A, paths)
           + UniquePathHelper(i, j + 1, r, c, A, paths);
}

int uniquePathsWithObstacles(vector<vector<int> >& A)
{

    int r = A.size(), c = A[0].size();

    // create a 2D-matrix and initializing
    // with value 0

    vector<vector<int> > paths(r, vector<int>(c, -1));

    return UniquePathHelper(0, 0, r, c, A, paths);
}

// Driver code
int main()
{
    vector<vector<int> > A
        = { { 0, 0, 0 }, { 0, 1, 0 }, { 0, 0, 0 } };

    cout << uniquePathsWithObstacles(A) << " \n";
}

Python3

# Python code to find number of unique paths
# in a Matrix


def UniquePathHelper(i, j, r, c, A, paths):
    # boundary condition or constraints
    if (i == r or j == c):
        return 0

    if (A[i][j] == 1):
        return 0

    # base case
    if (i == r - 1 and j == c - 1):
        return 1

    if (paths[i][j] != -1):
        return paths[i][j]

    return UniquePathHelper(i + 1, j, r, c, A, paths) + UniquePathHelper(i, j + 1, r, c, A, paths)

def uniquePathsWithObstacles(A):

    r,c = len(A),len(A[0])

    # create a 2D-matrix and initializing
    # with value 0

    paths = [[-1 for i in range(c)]for j in range(r)]

    return UniquePathHelper(0, 0, r, c, A, paths)

# Driver code

A  = [ [ 0, 0, 0 ], [ 0, 1, 0 ], [ 0, 0, 0 ] ]

print(uniquePathsWithObstacles(A))

# code is contributed by shinjanpatra

Javascript

<script>

// JavaScript code to find number of unique paths
// in a Matrix
function UniquePathHelper(i, j, r, c, A, paths)
{

    // boundary condition or constraints
    if (i == r || j == c) {
        return 0;
    }

    if (A[i][j] == 1) {
        return 0;
    }

    // base case
    if (i == r - 1 && j == c - 1) {
        return 1;
    }

    if (paths[i][j] != -1) {
        return paths[i][j];
    }

    return UniquePathHelper(i + 1, j, r, c, A, paths)
           + UniquePathHelper(i, j + 1, r, c, A, paths);
}

function uniquePathsWithObstacles(A)
{

    let r = A.length, c = A[0].length;

    // create a 2D-matrix and initializing
    // with value 0
    let paths = new Array(c);
    for(let i = 0; i < r; i++){
        paths[i] = new Array(c).fill(-1);
    }

    return UniquePathHelper(0, 0, r, c, A, paths);
}

// Driver code
let A  = [ [ 0, 0, 0 ], [ 0, 1, 0 ], [ 0, 0, 0 ] ]
document.write(uniquePathsWithObstacles(A))

// This code is contributed by shinjanpatra

</script>

Output

2 

Time Complexity: O(m*n) 

Auxiliary Space: O(m*n)

2) Bottom-Up

 A temporary 2D matrix will be constructed and value will be stored using the bottom-up approach. 

Approach

• Create a 2D matrix of the same size as the given matrix to store the results.
• Traverse through the created array row-wise and start filling the values in it.
• If an obstacle is found, set the value to 0.
• For the first row and column, set the value to 1 if an obstacle is not found.
• Set the sum of the right and the upper values if an obstacle is not present at that corresponding position in the given matrix
• Return the last value of the created 2d matrix

C++

// C++ code to find number of unique paths
// in a Matrix
#include<bits/stdc++.h>
using namespace std;

int uniquePathsWithObstacles(vector<vector<int>>& A)
{

    int r = A.size(), c = A[0].size();

    // create a 2D-matrix and initializing
    // with value 0
    vector<vector<int>> paths(r, vector<int>(c, 0));

    // Initializing the left corner if
    // no obstacle there
    if (A[0][0] == 0)
        paths[0][0] = 1;

    // Initializing first column of
    // the 2D matrix
    for(int i = 1; i < r; i++)
    {
        // If not obstacle
        if (A[i][0] == 0)
            paths[i][0] = paths[i-1][0];
    }

    // Initializing first row of the 2D matrix
    for(int j = 1; j < c; j++)
    {

        // If not obstacle
        if (A[0][j] == 0)
            paths[0][j] = paths[0][j - 1];
    }

    for(int i = 1; i < r; i++)
    {
        for(int j = 1; j < c; j++)
        {

            // If current cell is not obstacle
            if (A[i][j] == 0)
                paths[i][j] = paths[i - 1][j] +
                              paths[i][j - 1];
        }
    }

    // Returning the corner value
    // of the matrix
    return paths[r - 1];
}

// Driver code
int main()
{
   vector<vector<int>> A = { { 0, 0, 0 },
                             { 0, 1, 0 },
                             { 0, 0, 0 } };

   cout << uniquePathsWithObstacles(A) << " \n";
}

// This code is contributed by ajaykr00kj

Java

// Java code to find number of unique paths
// in a Matrix
public class Main
{
  static int uniquePathsWithObstacles(int[][] A)
  {

    int r = 3, c = 3;

    // create a 2D-matrix and initializing
    // with value 0
    int[][] paths = new int[r];
    for(int i = 0; i < r; i++)
    {
      for(int j = 0; j < c; j++)
      {
        paths[i][j] = 0;
      }
    }

    // Initializing the left corner if
    // no obstacle there
    if (A[0][0] == 0)
      paths[0][0] = 1;

    // Initializing first column of
    // the 2D matrix
    for(int i = 1; i < r; i++)
    {
      // If not obstacle
      if (A[i][0] == 0)
        paths[i][0] = paths[i - 1][0];
    }

    // Initializing first row of the 2D matrix
    for(int j = 1; j < c; j++)
    {

      // If not obstacle
      if (A[0][j] == 0)
        paths[0][j] = paths[0][j - 1];
    }

    for(int i = 1; i < r; i++)
    {
      for(int j = 1; j < c; j++)
      {

        // If current cell is not obstacle
        if (A[i][j] == 0)
          paths[i][j] = paths[i - 1][j] +
          paths[i][j - 1];
      }
    }

    // Returning the corner value
    // of the matrix
    return paths[r - 1];
  }

  // Driver code
  public static void main(String[] args) {
    int[][] A = { { 0, 0, 0 },
                 { 0, 1, 0 },
                 { 0, 0, 0 } };

    System.out.print(uniquePathsWithObstacles(A));
  }
}

// This code is contributed by divyeshrabadiya07.

Python

# Python code to find number of unique paths in a
# matrix with obstacles.

def uniquePathsWithObstacles(A):

    # create a 2D-matrix and initializing with value 0
    paths = [[0]*len(A[0]) for i in A]

    # initializing the left corner if no obstacle there
    if A[0][0] == 0:
        paths[0][0] = 1

    # initializing first column of the 2D matrix
    for i in range(1, len(A)):

        # If not obstacle
        if A[i][0] == 0:
            paths[i][0] = paths[i-1][0]

    # initializing first row of the 2D matrix
    for j in range(1, len(A[0])):

        # If not obstacle
        if A[0][j] == 0:
            paths[0][j] = paths[0][j-1]

    for i in range(1, len(A)):
        for j in range(1, len(A[0])):

            # If current cell is not obstacle
            if A[i][j] == 0:
                paths[i][j] = paths[i-1][j] + paths[i][j-1]

    # returning the corner value of the matrix
    return paths[-1][-1]


# Driver Code
A = [[0, 0, 0], [0, 1, 0], [0, 0, 0]]
print(uniquePathsWithObstacles(A))

C#

// C# code to find number of unique paths
// in a Matrix
using System;
class GFG {

  static int uniquePathsWithObstacles(int[,] A)
  {

    int r = 3, c = 3;

    // create a 2D-matrix and initializing
    // with value 0
    int[,] paths = new int[r,c];
    for(int i = 0; i < r; i++)
    {
      for(int j = 0; j < c; j++)
      {
        paths[i, j] = 0;
      }
    }

    // Initializing the left corner if
    // no obstacle there
    if (A[0, 0] == 0)
      paths[0, 0] = 1;

    // Initializing first column of
    // the 2D matrix
    for(int i = 1; i < r; i++)
    {
      // If not obstacle
      if (A[i, 0] == 0)
        paths[i, 0] = paths[i - 1, 0];
    }

    // Initializing first row of the 2D matrix
    for(int j = 1; j < c; j++)
    {

      // If not obstacle
      if (A[0, j] == 0)
        paths[0, j] = paths[0, j - 1];
    }

    for(int i = 1; i < r; i++)
    {
      for(int j = 1; j < c; j++)
      {

        // If current cell is not obstacle
        if (A[i, j] == 0)
          paths[i, j] = paths[i - 1, j] +
          paths[i, j - 1];
      }
    }

    // Returning the corner value
    // of the matrix
    return paths[r - 1, c - 1];
  }

  // Driver code
  static void Main() {
    int[,] A = { { 0, 0, 0 },
                { 0, 1, 0 },
                { 0, 0, 0 } };

    Console.WriteLine(uniquePathsWithObstacles(A));
  }
}

// This code is contributed by divyesh072019.

Javascript

<script>
    // Javascript code to find number of unique paths
    // in a Matrix

    function uniquePathsWithObstacles(A)
    {

      let r = 3, c = 3;

      // create a 2D-matrix and initializing
      // with value 0
      let paths = new Array(r);
      for(let i = 0; i < r; i++)
      {
          paths[i] = new Array(c);
        for(let j = 0; j < c; j++)
        {
          paths[i][j] = 0;
        }
      }

      // Initializing the left corner if
      // no obstacle there
      if (A[0][0] == 0)
        paths[0][0] = 1;

      // Initializing first column of
      // the 2D matrix
      for(let i = 1; i < r; i++)
      {
        // If not obstacle
        if (A[i][0] == 0)
          paths[i][0] = paths[i - 1][0];
      }

      // Initializing first row of the 2D matrix
      for(let j = 1; j < c; j++)
      {

        // If not obstacle
        if (A[0][j] == 0)
          paths[0][j] = paths[0][j - 1];
      }

      for(let i = 1; i < r; i++)
      {
        for(let j = 1; j < c; j++)
        {

          // If current cell is not obstacle
          if (A[i][j] == 0)
            paths[i][j] = paths[i - 1][j] +
            paths[i][j - 1];
        }
      }

      // Returning the corner value
      // of the matrix
      return paths[r - 1];
    }

    let A = [ [ 0, 0, 0 ],
             [ 0, 1, 0 ],
             [ 0, 0, 0 ] ];

    document.write(uniquePathsWithObstacles(A));

    // This code is contributed by suresh07.
</script>

Output

2 

This article is contributed by Rishabh Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Method 3: Space Optimization of DP solution.

In this method, we will use the given ‘A’ 2D matrix to store the previous answer using the bottom-up approach.

Approach

• Start traversing through the given ‘A’ 2D matrix row-wise and fill the values in it.
• For the first row and the first column set the value to 1 if an obstacle is not found.
• For the first row and first column, if an obstacle is found then start filling 0 till the last index in that particular row or column.
• Now start traversing from the second row and column ( eg: A[ 1 ][ 1 ]).
• If an obstacle is found, set 0 at particular Grid ( eg: A[ i ][ j ] ), otherwise set sum of upper and left values at A[ i ][ j ].
• Return the last value of the 2D matrix.

Below is the implementation of the above approach. 

C++

// CPP program for the above approach

#include <bits/stdc++.h>
using namespace std;

int uniquePathsWithObstacles(vector<vector<int> >& A)
{

    int r = A.size();
    int c = A[0].size();

    // If obstacle is at starting position
    if (A[0][0])
        return 0;

    //  Initializing starting position
    A[0][0] = 1;

    // first row all are '1' until obstacle
    for (int j = 1; j < c; j++) {

        if (A[0][j] == 0) {
            A[0][j] = A[0][j - 1];
        }
        else {
            // No ways to reach at this index
            A[0][j] = 0;
        }
    }

    // first column all are '1' until obstacle
    for (int i = 1; i < r; i++) {

        if (A[i][0] == 0) {
            A[i][0] = A[i - 1][0];
        }
        else {
            // No ways to reach at this index
            A[i][0] = 0;
        }
    }

    for (int i = 1; i < r; i++) {

        for (int j = 1; j < c; j++) {
            // If current cell has no obstacle
            if (A[i][j] == 0) {

                A[i][j] = A[i - 1][j] + A[i][j - 1];
            }
            else {
                // No ways to reach at this index
                A[i][j] = 0;
            }
        }
    }

    // returning the bottom right
    // corner of Grid
    return A[r - 1];
}

// Driver Code

int main()
{

    vector<vector<int> > A
        = { { 0, 0, 0 }, { 0, 1, 0 }, { 0, 0, 0 } };

    cout << uniquePathsWithObstacles(A) << "\n";

    return 0;
}
// This code is contributed by hemantraj712

Java

// Java program for the above approach
class GFG {

    static int uniquePathsWithObstacles(int[][] A)
    {

        int r = A.length;
        int c = A[0].length;

        // If obstacle is at starting position
        if (A[0][0] != 0)
            return 0;

        //  Initializing starting position
        A[0][0] = 1;

        // first row all are '1' until obstacle
        for (int j = 1; j < c; j++) {

            if (A[0][j] == 0) {
                A[0][j] = A[0][j - 1];
            }
            else {
                // No ways to reach at this index
                A[0][j] = 0;
            }
        }

        // first column all are '1' until obstacle
        for (int i = 1; i < r; i++) {

            if (A[i][0] == 0) {
                A[i][0] = A[i - 1][0];
            }
            else {
                // No ways to reach at this index
                A[i][0] = 0;
            }
        }

        for (int i = 1; i < r; i++)
        {

            for (int j = 1; j < c; j++)
            {

                // If current cell has no obstacle
                if (A[i][j] == 0) {

                    A[i][j] = A[i - 1][j] + A[i][j - 1];
                }
                else {
                    // No ways to reach at this index
                    A[i][j] = 0;
                }
            }
        }

        // returning the bottom right
        // corner of Grid
        return A[r - 1];
    }

  // Driver code
    public static void main(String[] args)
    {
        int[][] A
            = { { 0, 0, 0 }, { 0, 1, 0 }, { 0, 0, 0 } };

        System.out.print(uniquePathsWithObstacles(A));
    }
}

// This code is contributed by rajsanghavi9.

Python3

# Python program for the above approach


def uniquePathsWithObstacles(A):

    r = len(A)
    c = len(A[0])

    # If obstacle is at starting position
    if (A[0][0]):
        return 0

    #  Initializing starting position
    A[0][0] = 1

    # first row all are '1' until obstacle
    for j in range(1,c):

        if (A[0][j] == 0):
            A[0][j] = A[0][j - 1]
        else:
            # No ways to reach at this index
            A[0][j] = 0

    # first column all are '1' until obstacle
    for i in range(1,r):

        if (A[i][0] == 0):
            A[i][0] = A[i - 1][0]
        else:
            # No ways to reach at this index
            A[i][0] = 0

    for i in range(1,r):

        for j in range(1,c):
            # If current cell has no obstacle
            if (A[i][j] == 0):

                A[i][j] = A[i - 1][j] + A[i][j - 1]
            else:
                # No ways to reach at this index
                A[i][j] = 0

    # returning the bottom right
    # corner of Grid
    return A[r - 1]

# Driver Code

A = [ [ 0, 0, 0 ], [ 0, 1, 0 ], [ 0, 0, 0 ] ]

print(uniquePathsWithObstacles(A))

# This code is contributed by shinjanpatra

Javascript

<script>

// JavaScript program for the above approach
function uniquePathsWithObstacles(A){

    let r = A.length
    let c = A[0].length

    // If obstacle is at starting position
    if (A[0][0])
        return 0

    //  Initializing starting position
    A[0][0] = 1

    // first row all are '1' until obstacle
    for(let j = 1; j < c; j++)
    {
        if (A[0][j] == 0)
            A[0][j] = A[0][j - 1]
        else
            // No ways to reach at this index
            A[0][j] = 0
    }

    // first column all are '1' until obstacle
    for(let i = 1; i < r; i++){

        if (A[i][0] == 0)
            A[i][0] = A[i - 1][0]
        else
            // No ways to reach at this index
            A[i][0] = 0
    }

   for(let i = 1; i < r; i++){

      for(let j = 1; j < c; j++){
            // If current cell has no obstacle
            if (A[i][j] == 0)

                A[i][j] = A[i - 1][j] + A[i][j - 1]
            else
                // No ways to reach at this index
                A[i][j] = 0
      }
   }

    // returning the bottom right
    // corner of Grid
    return A[r - 1]
}

// Driver Code
let A = [ [ 0, 0, 0 ], [ 0, 1, 0 ], [ 0, 0, 0 ] ]
document.write(uniquePathsWithObstacles(A),"</br>")

// This code is contributed by shinjanpatra

</script>

Output

2

Time Complexity: O(m*n)  
Auxiliary Space: O(1)