1. Home
  2. Unique paths in a Grid with Obstacles using Dynamic Programming

Unique paths in a Grid with Obstacles using Dynamic Programming

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 paths[i][j]
           = 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";
}
Java 
// Java code to find number of unique paths
// in a Matrix
import java.util.*;

public class Main {
    public static void main(String[] args)
    {
        int[][] A
            = { { 0, 0, 0 }, { 0, 1, 0 }, { 0, 0, 0 } };
        System.out.println(uniquePathsWithObstacles(A));
    }

    public static int uniquePathsWithObstacles(int[][] A)
    {
        int r = A.length;
        int c = A[0].length;
        // create a 2D-matrix and initializing
        // with value 0
        int[][] paths = new int[r][c];

        for (int i = 0; i < r; i++) {
            Arrays.fill(paths[i], -1);
        }

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

    public static int UniquePathHelper(int i, int j, int r,
                                       int c, int[][] A,
                                       int[][] paths)
    {
        // boundary condition or constraints
        if (i == r || j == c) {
            return 0;
        }
        else if (A[i][j] == 1) {
            return 0;
        }
        // base case
        else if (i == r - 1 && j == c - 1) {
            return 1;
        }
        else if (paths[i][j] != -1) {

            return paths[i][j];
        }
        else {
            return paths[i][j]
                = UniquePathHelper(i + 1, j, r, c, A, paths)
                  + UniquePathHelper(i, j + 1, r, c, A,
                                     paths);
        }
    }
}

// This code is contributed by Tapesh(tapeshdua420)
Python 
# 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]

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


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

class Program {

    // Driver code
    static void Main(string[] args)
    {
        int[, ] A = new int[3, 3] { { 0, 0, 0 },
                                    { 0, 1, 0 },
                                    { 0, 0, 0 } };
        Console.WriteLine(uniquePathsWithObstacles(A));
    }

    static int uniquePathsWithObstacles(int[, ] A)
    {
        int r = A.GetLength(0);
        int c = A.GetLength(1);

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

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

    static int UniquePathHelper(int i, int j, int r, int c,
                                int[, ] A, 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 paths[i, j]
            = UniquePathHelper(i + 1, j, r, c, A, paths)
              + UniquePathHelper(i, j + 1, r, c, A, paths);
    }
}

// This code is contributed by Tapesh(tapeshdua420)
Javascript 
// 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 (paths[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 paths[i][j]
           = 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 ] ]
console.log(uniquePathsWithObstacles(A))

// This code is contributed by shinjanpatra

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

Below is the implementation of the above approach:

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][c - 1];//it is not visible in the article, instread only paths[r - 1] is visible so please change it
}

// 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][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
  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 
    // 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][c -1];
    }
      
    let A = [ [ 0, 0, 0 ],
             [ 0, 1, 0 ],
             [ 0, 0, 0 ] ];
 
    console.log(uniquePathsWithObstacles(A));
    
    // This code is contributed by suresh07.

Output
2

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

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 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.

Similar Reads

Unique Paths in a Grid with Obstacles using Recursion:

We have discussed the problem of counting 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 the way to reach the bottom right corner is blocked....

Unique paths in a Grid with Obstacles using Dynamic Programming:

1) Top-Down...

Unique paths in a Grid with Obstacles using Space Optimization of Dynamic Programming solution:

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

Unique paths in a Grid with Obstacles using 2D Dynamic Programming:

As Per Problem tell us that we can move in two ways  can either go to (x, y + 1) or (x + 1, y). So we just calculate all possible outcome in both ways and store in 2d dp vector and return the dp[0][0] i.e all possible ways that takes you from (0,0) to (n-1,m-1);...

Categories

HTML CSS3 Bootstrap Foundation Sass JavaScript TypeScript JQuery AngularJS React NodeJS Highcharts PHP Python Python3 Django Linux Docker Ruby Java C C ++ Perl Servlet JSP Lua Scala Go ASP C # SQL MySQL SQLite MongoDB Redis Android Swift ionic Kotlin XML tutorial

Contact US