63. Unique Paths II | LeetCode Solution

You are given an m x n integer array grid. There is a robot initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m-1][n-1]). The robot can only move either down or right at any point in time.

An obstacle and space are marked as 1 or 0 respectively in grid. A path that the robot takes cannot include any square that is an obstacle.

Return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The testcases are generated so that the answer will be less than or equal to 2 * 109.

 

Example 1:

Input: obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
Output: 2
Explanation: There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right

Example 2:

Input: obstacleGrid = [[0,1],[0,0]]
Output: 1

 

Constraints:

  • m == obstacleGrid.length
  • n == obstacleGrid[i].length
  • 1 <= m, n <= 100
  • obstacleGrid[i][j] is 0 or 1.

class Solution {
public:
    
   
    int helper(int a,int b,int m,int n,vector<vector<int>>&dp,vector<vector<int>>&grid){
         if(a>m-1 || b>n-1){
            return 0;
        }
        if(grid[a][b]==1)return 0;
        
        if(a==m-1 && b==n-1){
            return 1;
        }
       
        int ans=0;
           if(dp[a+1][b]==-1){
            dp[a+1][b]=helper(a+1,b,m,n,dp,grid);
        }
         if(dp[a][b+1]==-1){
             dp[a][b+1]=helper(a,b+1,m,n,dp,grid);
         }
            ans=ans+dp[a+1][b]+ dp[a][b+1];
            
        
        return ans;
    }
    
    int uniquePathsWithObstacles(vector<vector<int>>& grid) {
        int n=grid.size();
        int m=grid[0].size();
        vector<vector<int>>dp(n+1,vector<int>(m+1,-1));
        return helper(0,0,n,m,dp,grid);
    }
};