63. Unique Paths II

Description

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

Constraints

• m == obstacleGrid.length

• n == obstacleGrid[i].length

• 1 <= m, n <= 100

• obstacleGrid[i][j] is 0 or 1.

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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

Solutions

Solution 1

/**
 * Time complexity : O(M*N). The rectangular grid given to us is of 
 *    size M*N and we process each cell just once.
 * Space complexity : O(1). We are utilizing the obstacleGrid as the DP array.
 *    Hence, no extra space.
 */

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if(obstacleGrid == null || obstacleGrid.length == 0 || 
                obstacleGrid[0].length == 0 || obstacleGrid[0][0] == 1) {
            return 0;
        }
        
        int rows = obstacleGrid.length, cols = obstacleGrid[0].length;
        int[][] ways = new int[rows][cols];
        
        int i = 0;
        while(i < cols && obstacleGrid[0][i] == 0) {
            ways[0][i++] = 1;
        }
        
        i = 1;
        while(i < rows && obstacleGrid[i][0] == 0) {
            ways[i++][0] = 1;
        }
        
        for(i = 1; i < rows; i++) {
            for(int j = 1; j < cols; j++) {
                if(obstacleGrid[i][j] == 0) {
                    ways[i][j] = ways[i-1][j] + ways[i][j-1];
                }
            }
        }
        return ways[rows-1][cols-1];
    }
}

Follow up