Unique Paths II

Medium Dynamic Programming 2D DP Original on LeetCode

Unique Paths II is a medium dynamic programming problem solved with the 2d dp pattern. The best approach, optimal (one row), runs in O(m · n) time and O(n) space. Below are 2 approaches in Java, from memoised recursion up.

Problem

Same as Unique Paths, but some cells contain obstacles (1) that the robot cannot enter. Count the paths from the top-left to the bottom-right.

Examples

Example 1

Input: obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
Output: 2

Example 2

Input: obstacleGrid = [[0,1],[0,0]]
Output: 1

Constraints

1 <= m, n <= 100; 1 marks an obstacle.

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

Approach	Time	Space
Memoised recursion	`O(m · n)`	`O(m · n)`
Optimal (one row)	`O(m · n)`	`O(n)`

1Memoised recursion

TimeO(m · n)

SpaceO(m · n)

paths(i, j) = 0 on an obstacle; otherwise the sum of paths from above and from the left, cached.

Out of bounds or obstacle → 0; (0, 0) → 1.

Java

class Solution {
    private Integer[][] memo;

    public int uniquePathsWithObstacles(int[][] g) {
        memo = new Integer[g.length][g[0].length];
        return paths(g, g.length - 1, g[0].length - 1);
    }

    private int paths(int[][] g, int i, int j) {
        if (i < 0 || j < 0 || g[i][j] == 1) return 0;
        if (i == 0 && j == 0) return 1;
        if (memo[i][j] != null) return memo[i][j];
        return memo[i][j] = paths(g, i - 1, j) + paths(g, i, j - 1);
    }
}

2Optimal (one row)

TimeO(m · n)

SpaceO(n)

dp[j] = ways to reach column j in the current row. An obstacle sets it to 0; otherwise add the value to the left.

dp[0] = 1 if the start is open.
For each cell: obstacle → dp[j] = 0; else if j > 0, dp[j] += dp[j - 1].

Java

class Solution {
    public int uniquePathsWithObstacles(int[][] g) {
        int n = g[0].length;
        int[] dp = new int[n];
        dp[0] = g[0][0] == 1 ? 0 : 1;
        for (int[] row : g)
            for (int j = 0; j < n; j++) {
                if (row[j] == 1) dp[j] = 0;
                else if (j > 0) dp[j] += dp[j - 1];
            }
        return dp[n - 1];
    }
}

Edge cases to test

Start or end is an obstacle (0 paths)
An obstacle in the first row blocks every cell to its right

Hints

Hint 1

Same DP as Unique Paths, but a blocked cell has 0 ways.

FAQ

What is the best time complexity for Unique Paths II?

Optimal (one row) runs in O(m · n) time and O(n) extra space.

Which pattern does Unique Paths II use?

It is a dynamic programming problem that uses the 2d dp pattern. Other problems with the same pattern: Minimum Path Sum, Unique Paths.

Is there a brute force solution for Unique Paths II?

Yes. Memoised recursion takes O(m · n) time and O(m · n) space. paths(i, j) = 0 on an obstacle; otherwise the sum of paths from above and from the left, cached.

Which edge cases should I test for Unique Paths II?

Start or end is an obstacle (0 paths); An obstacle in the first row blocks every cell to its right.