LeetCode #63 — MEDIUM

Unique Paths II

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

[[0,0,0],[0,1,0],[0,0,0]]

Example 2

[[0,1],[0,0]]

Core Insight

What unlocks the optimal approach

Use dynamic programming since, from each cell, you can move to the right or down.
assume dp[i][j] is the number of unique paths to reach (i, j). dp[i][j] = dp[i][j -1] + dp[i - 1][j]. Be careful when you encounter an obstacle. set its value in dp to 0.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length, n = obstacleGrid[0].length;
        int[] dp = new int[n];
        dp[0] = obstacleGrid[0][0] == 1 ? 0 : 1;

        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                if (obstacleGrid[r][c] == 1) {
                    dp[c] = 0; // Blocked cell: no path through it.
                } else if (c > 0) {
                    dp[c] += dp[c - 1];
                }
            }
        }
        return dp[n - 1];
    }
}

func uniquePathsWithObstacles(obstacleGrid [][]int) int {
    m, n := len(obstacleGrid), len(obstacleGrid[0])
    dp := make([]int, n)
    if obstacleGrid[0][0] == 0 {
        dp[0] = 1
    }

    for r := 0; r < m; r++ {
        for c := 0; c < n; c++ {
            if obstacleGrid[r][c] == 1 {
                dp[c] = 0
            } else if c > 0 {
                dp[c] += dp[c-1]
            }
        }
    }

    return dp[n-1]
}

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m, n = len(obstacleGrid), len(obstacleGrid[0])
        dp = [0] * n
        dp[0] = 0 if obstacleGrid[0][0] == 1 else 1

        for r in range(m):
            for c in range(n):
                if obstacleGrid[r][c] == 1:
                    dp[c] = 0
                elif c > 0:
                    dp[c] += dp[c - 1]

        return dp[-1]

impl Solution {
    pub fn unique_paths_with_obstacles(obstacle_grid: Vec<Vec<i32>>) -> i32 {
        let m = obstacle_grid.len();
        let n = obstacle_grid[0].len();
        let mut dp = vec![0; n];
        dp[0] = if obstacle_grid[0][0] == 1 { 0 } else { 1 };

        for r in 0..m {
            for c in 0..n {
                if obstacle_grid[r][c] == 1 {
                    dp[c] = 0;
                } else if c > 0 {
                    dp[c] += dp[c - 1];
                }
            }
        }

        dp[n - 1]
    }
}

function uniquePathsWithObstacles(obstacleGrid: number[][]): number {
  const m = obstacleGrid.length;
  const n = obstacleGrid[0].length;
  const dp: number[] = Array(n).fill(0);
  dp[0] = obstacleGrid[0][0] === 1 ? 0 : 1;

  for (let r = 0; r < m; r++) {
    for (let c = 0; c < n; c++) {
      if (obstacleGrid[r][c] === 1) {
        dp[c] = 0;
      } else if (c > 0) {
        dp[c] += dp[c - 1];
      }
    }
  }

  return dp[n - 1];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n)

Space

O(m × n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.