Unique Paths I, II

A robot is located at the top-left corner of am 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).

How many possible unique paths are there?

Have you met this question in a real interview? 

Yes

Example

1,1	1,2	1,3	1,4	1,5	1,6	1,7
2,1
3,1						3,7

Above is a 3 x 7 grid. How many possible unique paths are there?

Note

m and n will be at most 100.

public class Solution {

    /**

     * @param n, m: positive integer (1 <= n ,m <= 100)

     * @return an integer

     */

    public int uniquePaths(int m, int n) {

        // write your code here 

        int[] num = new int[n];

       Arrays.fill(num, 1);

       

       for(int i = 1; i < m; i++){

           for(int j = 1; j < n; j++){

                num[j] = num[j-1] + num[j]; 

           }

       }

       return num[n-1];

    }

}

Unique Paths II

Easy Unique Paths II

27%

Accepted

Follow up for "Unique Paths":

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

An obstacle and empty space is marked as1 and 0 respectively in the grid.

Have you met this question in a real interview? 

Yes

Example

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]

The total number of unique paths is 2.

Note

m and n will be at most 100.

public class Solution {

    /**

     * @param obstacleGrid: A list of lists of integers

     * @return: An integer

     */

    public int uniquePathsWithObstacles(int[][] obstacleGrid) {

        // write your code here

        int m = obstacleGrid.length, n = obstacleGrid[0].length;

        int[] num = new int[n];

        num[0] = obstacleGrid[0][0] == 0 ? 1 : 0;

        for(int i = 1 ; i < n; i++){

            num[i] = num[i-1] == 0 ? 0 : obstacleGrid[0][i] == 0 ? 1 : 0;  

        }

       

       for(int i = 1; i < m; i++){

           num[0] = num[0] == 0 ? 0 : obstacleGrid[i][0] == 0 ? 1 : 0;

           for(int j = 1; j < n; j++){

                 num[j] = obstacleGrid[i][j] == 1 ? 0 : num[j] + num[j-1];

           }

       }

       return num[n-1];

    }

}