63. Unique Paths II

Question

A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below).

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

Note: m and n will be at most 100.

Example 1:

Input:
[
  [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

Summary

This is a follow up problem of 62. Unique Paths

Current position(m,n) paths equal to position(m,n-1) + position(m-1,n) paths;

position(m,n) = position(m,n-1) + position(m-1,n);

If position(m,n-1) or position(m-1, n) has obstacle, then don’t count it.

There is a special case here:

If the final position(m,n) has obstacle, then the initial path = 0.

Solution

class Solution {
public:
    map<pair<int,int>, int> memo;
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid[0].size()-1;
        int n = obstacleGrid.size()-1;
        memo[make_pair(0,0)] = (obstacleGrid[n][m] == 0 ? 1 : 0);
        return dp(m, n, obstacleGrid);
    }

    int dp(int col, int row, vector<vector<int>>& obstacleGrid) {
        if (memo.find(make_pair(col,row)) != memo.end())
            return memo[make_pair(col,row)];

        int path = 0;
        if (col > 0 && obstacleGrid[row][col-1] != 1)
            path += dp(col-1, row, obstacleGrid);
        if (row > 0 && obstacleGrid[row-1][col] != 1)
            path += dp(col, row-1, obstacleGrid);

        memo[make_pair(col,row)] = path;
        return path;
    }
};