Problem #63: Unique Paths II

Intuition & Strategy Breakdown

Use dynamic programming. dp[i][j] stores paths to cell (i, j). If cell has an obstacle, dp[i][j] = 0; otherwise dp[i][j] = dp[i-1][j] + dp[i][j-1].

Python Solution

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid):
        if not obstacleGrid or obstacleGrid[0][0] == 1:
            return 0
        m, n = len(obstacleGrid), len(obstacleGrid[0])
        dp = [[0] * n for _ in range(m)]
        dp[0][0] = 1
        for i in range(m):
            for j in range(n):
                if obstacleGrid[i][j] == 1:
                    dp[i][j] = 0
                    continue
                if i > 0:
                    dp[i][j] += dp[i - 1][j]
                if j > 0:
                    dp[i][j] += dp[i][j - 1]
        return dp[m - 1][n - 1]

Java Solution

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length, n = obstacleGrid[0].length;
        if (obstacleGrid[0][0] == 1) return 0;
        int[][] dp = new int[m][n];
        dp[0][0] = 1;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (obstacleGrid[i][j] == 1) {
                    dp[i][j] = 0;
                    continue;
                }
                if (i > 0) dp[i][j] += dp[i - 1][j];
                if (j > 0) dp[i][j] += dp[i][j - 1];
            }
        }
        return dp[m - 1][n - 1];
    }
}