How do you approach Unique Paths II?

Use a 1-D rolling array over columns.

What are common mistakes on Unique Paths II?

Iterating the dimensions in an order that reads a cell before it has been written. Getting the base case wrong. It carries the whole recurrence, and the error stays invisible until a large input. Rolling down to one dimension without reversing the inner loop, which lets a 0/1 item be taken twice.

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Unique Paths II

MediumAlgorithms

Description

A robot starts at the top-left of an m x n grid and wants to reach the bottom-right, moving only right or down. Some cells contain obstacles (marked 1); the robot cannot enter them.

Return the number of unique paths. If the start or end is blocked, the answer is 0.

Example 1

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

Output: 2

Explanation: Two paths route around the center obstacle.

Constraints

- m == obstacleGrid.length - n == obstacleGrid[i].length - 1 <= m, n <= 100 - obstacleGrid[i][j] is 0 or 1

Unique Paths II: approach, complexity and pitfalls

A medium dynamic programming problem, graded against 7 test cases (5 of them hidden).

The pattern: Dynamic Programming

Overlapping subproblems solved once and reused - the pattern candidates fear most.

Reach for it when you see: "Number of ways", "minimum/maximum cost", or a recursion that recomputes the same state.

If you are stuck

Show the 2-step hint ladder

Use a 1-D rolling array over columns.
An obstacle cell contributes 0 paths; otherwise add the path count from the left.

The optimal approach

Show the approach: The approach

Rolling 1-D DP. `dp[j]` accumulates paths into column `j` for the current row. Obstacles zero out their cell; the rolling sum from above (previous `dp[j]`) plus the left (`dp[j-1]`) gives each cell's path count.

Time: O(m·n). Space: O(n).

The full reference solution in every supported language stays in the editor above - reveal it there once you have had a real attempt.

What the tests check

Read off this problem's own test suite, so these are the cases a submission actually has to survive.

Single element. A one-element case is included. Two-pointer and window solutions frequently return the wrong thing when the two indices start out equal.
Duplicate values. Repeated values appear in the input, so any logic that assumes distinct elements - or that skips over equal neighbours - is exercised.
Zero. Zero appears in the input, which is the value most likely to break a truthiness check used in place of an explicit comparison.

Common Dynamic Programming mistakes

These apply to the pattern as a whole, not just this problem.

Iterating the dimensions in an order that reads a cell before it has been written.
Getting the base case wrong. It carries the whole recurrence, and the error stays invisible until a large input.
Rolling down to one dimension without reversing the inner loop, which lets a 0/1 item be taken twice.