Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Unique Paths II in C++
Suppose there is a robot is located at the top-left corner of a n x m grid (n rows and m columns). The robot can only move either down side or right side at any point in time. The robot wants to reach the bottom-right corner of the grid (marked 'END in the diagram below).
Some cell in the grid is marked, that will be considered as obstacles. So we have to find how many possible unique paths are there? For example if the grid is like [[0,0,0],[0,1,0],[0,0,0]], then the grid will be like below −
| Robo | ||
| Obs | ||
| END |
The output will be 2, So there are total 2 different ways to reach from start position to the end position. These paths are −
- Right → Right → Down → Down
- Down → Down → Right → Right
Let us see the steps −
- a := number of rows, b := number of columns
- if grid[a – 1,b - 1] is non zero, then return 0
- create one table whose order is a x b, and name is DP
- for i := b – 1 down to 0
- if grid[a – 1, i], then break, otherwise DP[a – 1, i] := 1
- for i := a – 1 down to 0
- if grid[i, b - 1], then break, otherwise DP[i, b – 1] := 1
- for i := a – 2 down to 0
- for j := b – 2 down to 0
- DP[i, j] := DP[i + 1, j] + DP[i, j + 1] when grid[i, j] is 0, otherwise 0
- for j := b – 2 down to 0
- return DP[0, 0]
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int a = obstacleGrid.size();
int b = obstacleGrid[0].size();
if(!a || !b) return 0;
if(obstacleGrid[a - 1][b - 1])return 0;
vector < vector <lli> > dp(a, vector <lli>(b));
for(int i = b - 1; i >= 0; i--)if(obstacleGrid[a-1][i]) break; else dp[a-1][i] = 1;
for(int i = a - 1; i >= 0; i--)if(obstacleGrid[i][b - 1]) break; else dp[i][b-1] = 1 ;
for(int i = a-2; i >= 0; i--){
for(int j = b-2 ; j >= 0; j--)dp[i][j] = !obstacleGrid[i][j]? dp[i+1][j] + dp[i][j+1] : 0;
}
return dp[0][0];
}
};
main(){
Solution ob;
vector<vector<int>> v = {{0,0,0},{0,1,0},{0,0,0}};
cout << ob.uniquePathsWithObstacles(v);
}Input
[[0,0,0],[0,1,0],[0,0,0]]
Output
2
Updated on: 04-May-2020
157 Views
Advertisements