63. Unique Paths II

Problem Statement

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.

Intuition

If there is an obstacle, 

Dont write any condition

else similar to PAths1

Links

https://leetcode.com/problems/unique-paths-ii/description/

Video Links

https://www.youtube.com/watch?v=TmhpgXScLyY&ab_channel=takeUforward

Approach 1:

Memoization

C++

class Solution {
public:
    int path(vector<vector<int>>& arr, int i, int j, vector<vector<int>> &dp){
        if(i==0 and j==0)
            return arr[0][0] == 0 ? 1 : 0;

        if(i<0 or j<0)
            return 0;

        if(dp[i][j] != -1)
            return dp[i][j];

        int l = 0, r = 0;

        if(arr[i][j] == 0){
            if( i>0 and arr[i-1][j] == 0)
                l = path(arr,i-1,j,dp);

            if(j>0 and arr[i][j-1] == 0)
                r = path(arr,i,j-1,dp);
        }
        return dp[i][j] = l+r;

    }

    int uniquePathsWithObstacles(vector<vector<int>>& arr) {
        int m = arr.size();
        int n = arr[0].size();

        vector<vector<int>> dp(m,vector<int>(n,-1));

        return path(arr,m-1,n-1,dp);
    }
};

Approach 2:

Tabulation

C++

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& arr) {
        int m = arr.size();
        int n = arr[0].size();

        vector<vector<int>> dp(m,vector<int>(n,-1));
        dp[0][0] = arr[0][0] == 0 ? 1 : 0;

        for(int i=0; i<m; i++) {
            for(int j=0; j<n; j++) {
                if(i==0 and j==0)   continue;

                int l = 0, r = 0;

                if(arr[i][j] == 0){
                    if(i>0 and arr[i-1][j] == 0)
                        l = dp[i-1][j];

                    if(j>0 and arr[i][j-1] == 0)
                        r = dp[i][j-1];
                }
                dp[i][j] = l+r;
            }
        }
        

        return dp[m-1][n-1];
    }
};

Approach 3:

C++

Approach 4:

C++

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