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# Unique paths in a Grid with Obstacles

• Difficulty Level : Easy
• Last Updated : 08 Jul, 2021

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.```

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 which we can not jump and that way to reach the bottom right corner is blocked.
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 bottom up approach.

### Approach

• Create a 2D matrix of same size of 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 obstacle is not found.
• Set the sum of the right and the upper values if 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``using` `namespace` `std;` `int` `uniquePathsWithObstacles(vector>& A)``{``    ` `    ``int` `r = A.size(), c = A.size();``    ` `    ``// create a 2D-matrix and initializing``    ``// with value 0``    ``vector> paths(r, vector<``int``>(c, 0));``    ` `    ``// Initializing the left corner if``    ``// no obstacle there``    ``if` `(A == 0)``        ``paths = 1;``        ` `    ``// Initializing first column of``    ``// the 2D matrix``    ``for``(``int` `i = 1; i < r; i++)``    ``{``        ``// If not obstacle``        ``if` `(A[i] == 0)``            ``paths[i] = paths[i-1];``    ``}``    ` `    ``// Initializing first row of the 2D matrix``    ``for``(``int` `j = 1; j < c; j++)``    ``{``        ` `        ``// If not obstacle``        ``if` `(A[j] == 0)``            ``paths[j] = paths[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> 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

 ``
Output
`2 `

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### Space Optimization of DP solution.

In this method, we will use 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 first column set the value to 1 if obstacle is not found.
• For the first row and first column ,if obstacle is found then start filling 0 till last index in that particular row or column.
• Now start traversing from 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 2D matrix.

Below is the implementation of the above approach.

## C++

 `// CPP program for the above approach` `#include ``using` `namespace` `std;` `int` `uniquePathsWithObstacles(vector >& A)``{` `    ``int` `r = A.size();``    ``int` `c = A.size();` `    ``// If obstacle is at starting position``    ``if` `(A)``        ``return` `0;` `    ``//  Initializing starting position``    ``A = 1;` `    ``// first row all are '1' until obstacle``    ``for` `(``int` `j = 1; j < c; j++) {` `        ``if` `(A[j] == 0) {``            ``A[j] = A[j - 1];``        ``}``        ``else` `{``            ``// No ways to reach at this index``            ``A[j] = 0;``        ``}``    ``}` `    ``// first column all are '1' until obstacle``    ``for` `(``int` `i = 1; i < r; i++) {` `        ``if` `(A[i] == 0) {``            ``A[i] = A[i - 1];``        ``}``        ``else` `{``            ``// No ways to reach at this index``            ``A[i] = 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;``            ``}``        ``}``    ``}` `    ``// returing the bottom right``    ``// corner of Grid``    ``return` `A[r - 1];``}` `// Driver Code` `int` `main()``{` `    ``vector > A``        ``= { { 0, 0, 0 }, { 0, 1, 0 }, { 0, 0, 0 } };` `    ``cout << uniquePathsWithObstacles(A) << ``"\n"``;` `    ``return` `0;``}``// This code is contributed by hemantraj712`
Output
`2`

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

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