63. Unique Paths II

Problems #63 (Unique Paths II | Array, Dynamic Programming, Matrix)

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.

Solutions

Dynamic Programming

Code

• Java

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if(obstacleGrid[0][0] == 1) return 0;
        int endRow = obstacleGrid.length;
        int endCol = obstacleGrid[0].length;
        int[][] grid = new int[endRow][endCol];
        for(int i = 0; i < endRow; i++){
            if(obstacleGrid[i][0] == 1){
                grid[i][0] = 0;
                break;
            }
            else
                grid[i][0] = 1;
        }
        for(int j = 0; j < endCol; j++){
            if(obstacleGrid[0][j] == 1){
                grid[0][j] = 0;
                break;
            }
            else
                grid[0][j] = 1; 
        }
        for(int i = 1; i < endRow; i++){
            for(int j = 1; j < endCol; j++){
                if(obstacleGrid[i][j] == 1)
                    grid[i][j] = 0;
                else
                    grid[i][j] = grid[i - 1][j] + grid[i][j - 1];
            }
        }
        return grid[endRow - 1][endCol - 1];
    }
}

Complexity

• Time: O(m * n)
• Space: O(1)