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.
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.
// 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 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
# 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
// 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);
return UniquePathHelper(0, 0, r, c, A);
}
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);
}
}
// This code is contributed by Tapesh(tapeshdua420)
// 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 ] ]
console.log(uniquePathsWithObstacles(A))
// This code is contributed by shinjanpatra
Output
2
Time Complexity: O(2m*n)
Auxiliary Space: O(m+n) , because worst path will be when we iterate along corner cells, Recursion will not have all the paths so Space can’t be O(n*m) it should be O(n+m)
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++ 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 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 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# 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 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++ 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 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 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# 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 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 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.
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.
// 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][c - 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 program for the above approach
import java.io.*;
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][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 rajsanghavi9.
# 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][c - 1]
# Driver Code
A = [ [ 0, 0, 0 ], [ 0, 1, 0 ], [ 0, 0, 0 ] ]
print(uniquePathsWithObstacles(A))
# This code is contributed by shinjanpatra
// C# program for the above approach
using System;
using System.Collections.Generic;
class Program {
static int uniquePathsWithObstacles(int[, ] A)
{
int r = A.GetLength(0);
int c = A.GetLength(1);
// If obstacle is at starting position
if (A[0, 0] != 0)
return 0;
// Initializing starting position
A[0, 0] = 1;
for (int j = 1; j < c; j++) {
if (A[0, j] == 0) {
A[0, j] = A[0, j - 1];
}
else {
A[0, j] = 0;
}
}
// first row 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, c - 1];
}
// Driver code
public static void Main(String[] args)
{
int[, ] A = new int[3, 3] { { 0, 0, 0 },
{ 0, 1, 0 },
{ 0, 0, 0 } };
Console.WriteLine(uniquePathsWithObstacles(A));
}
}
// This code is contributed by Tapesh (tapeshdua420)
<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][c - 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)
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);
#include <bits/stdc++.h>
#define int long long
using namespace std;
int n, m;
int path(vector<vector<int> >& dp,
vector<vector<int> >& grid, int i, int j)
{
if (i < n && j < m && grid[i][j] == 1)
return 0;
if (i == n - 1 && j == m - 1)
return 1;
if (i >= n || j >= m)
return 0;
if (dp[i][j] != -1)
return dp[i][j];
int left = path(dp, grid, i + 1, j);
int right = path(dp, grid, i, j + 1);
return dp[i][j] = left + right;
}
int uniquePathsWithObstacles(vector<vector<int> >& grid)
{
n = grid.size();
m = grid[0].size();
if (n == 1 && m == 1 && grid[0][0] == 0)
return 1;
if (n == 1 && m == 1 && grid[0][0] == 1)
return 0;
vector<vector<int> > dp(n, vector<int>(m, -1));
// for(auto it:dp){
// for(auto vt:it)cout<<vt<<" ";
// cout<<endl;
// }
path(dp, grid, 0, 0);
// for(auto it:dp){
// for(auto vt:it)cout<<vt<<" ";
// cout<<endl;
// }
if (dp[0][0] == -1)
return 0;
return dp[0][0];
}
// Driver Code
signed main()
{
vector<vector<int> > v{ { 0, 0, 0 },
{ 0, 1, 0 },
{ 0, 0, 0 } };
cout << uniquePathsWithObstacles(v) << " \n";
return 0;
}
// Java code for the above approach
import java.util.*;
class Main {
static int n, m;
static int path(int[][] dp, int[][] grid, int i, int j)
{
if (i < n && j < m && grid[i][j] == 1)
return 0;
if (i == n - 1 && j == m - 1)
return 1;
if (i >= n || j >= m)
return 0;
if (dp[i][j] != -1)
return dp[i][j];
int left = path(dp, grid, i + 1, j);
int right = path(dp, grid, i, j + 1);
return dp[i][j] = left + right;
}
static int uniquePathsWithObstacles(int[][] grid)
{
n = grid.length;
m = grid[0].length;
if (n == 1 && m == 1 && grid[0][0] == 0)
return 1;
if (n == 1 && m == 1 && grid[0][0] == 1)
return 0;
int[][] dp = new int[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
dp[i][j] = -1;
}
}
path(dp, grid, 0, 0);
if (dp[0][0] == -1)
return 0;
return dp[0][0];
}
public static void main(String[] args)
{
int[][] v
= { { 0, 0, 0 }, { 0, 1, 0 }, { 0, 0, 0 } };
System.out.println(uniquePathsWithObstacles(v));
}
}
// This code is contributed by pradeepkumarppk2003
# Python code for the above approach
def uniquePathsWithObstacles(grid):
n = len(grid)
m = len(grid[0])
if n == 1 and m == 1 and grid[0][0] == 0:
return 1
if n == 1 and m == 1 and grid[0][0] == 1:
return 0
dp = [[-1 for j in range(m)] for i in range(n)]
def path(dp, grid, i, j):
if i < n and j < m and grid[i][j] == 1:
return 0
if i == n - 1 and j == m - 1:
return 1
if i >= n or j >= m:
return 0
if dp[i][j] != -1:
return dp[i][j]
left = path(dp, grid, i + 1, j)
right = path(dp, grid, i, j + 1)
dp[i][j] = left + right
return dp[i][j]
path(dp, grid, 0, 0)
if dp[0][0] == -1:
return 0
return dp[0][0]
# Driver Code
grid = [[0, 0, 0], [0, 1, 0], [0, 0, 0]]
print(uniquePathsWithObstacles(grid))
# This code is contributed by lokeshpotta20.
using System;
namespace UniquePathsWithObstacles {
class Program {
static void Main(string[] args)
{
int[][] grid
= new int[][] { new int[] { 0, 0, 0 },
new int[] { 0, 1, 0 },
new int[] { 0, 0, 0 } };
Console.WriteLine(UniquePathsWithObstacles(grid));
}
static int UniquePathsWithObstacles(int[][] grid)
{
int n = grid.Length;
int m = grid[0].Length;
if (n == 1 && m == 1 && grid[0][0] == 0)
return 1;
if (n == 1 && m == 1 && grid[0][0] == 1)
return 0;
int[][] dp = new int[n][];
for (int i = 0; i < n; i++) {
dp[i] = new int[m];
for (int j = 0; j < m; j++) {
dp[i][j] = -1;
}
}
Path(dp, grid, 0, 0);
if (dp[0][0] == -1)
return 0;
return dp[0][0];
}
static int Path(int[][] dp, int[][] grid, int i, int j)
{
int n = grid.Length;
int m = grid[0].Length;
if (i < n && j < m && grid[i][j] == 1)
return 0;
if (i == n - 1 && j == m - 1)
return 1;
if (i >= n || j >= m)
return 0;
if (dp[i][j] != -1)
return dp[i][j];
int left = Path(dp, grid, i + 1, j);
int right = Path(dp, grid, i, j + 1);
return dp[i][j] = left + right;
}
}
}
let n,m;
function path(dp, grid, i, j)
{
if(i<n && j<m && grid[i][j]==1)
return 0;
if(i==n-1 && j==m-1)
return 1;
if(i>=n || j>=m)
return 0;
if(dp[i][j]!=-1)
return dp[i][j];
let left=path(dp,grid,i+1,j);
let right=path(dp,grid,i,j+1);
return dp[i][j]=left+right;
}
function uniquePathsWithObstacles(grid)
{
n=grid.length;
m=grid[0].length;
if(n==1 && m==1 && grid[0][0]==0)
return 1;
if(n==1 && m==1 && grid[0][0]==1)
return 0;
let dp=new Array(n);
for(let i=0; i<n; i++)
dp[i]=new Array(m).fill(-1);
//for(let i=0; i<dp.length; i++){
// for(let j=0; j<dp[0].length; j++)
// console.log(dp[i][j]);
// cout<<endl;
// }
path(dp,grid,0,0);
//for(let i=0; i<dp.length; i++){
// for(let j=0; j<dp[0].length; j++)
// console.log(dp[i][j]);
// cout<<endl;
//}
if(dp[0][0]==-1)
return 0;
return dp[0][0];
}
// Driver Code
let v=[[ 0, 0, 0 ],[ 0, 1, 0 ],[ 0, 0, 0 ]];
console.log(uniquePathsWithObstacles(v));
Output
2
Time Complexity: O(M*N),For traversing all possible ways.
Auxiliary Space: O(M*N),For storing in 2D Dp Vector.