MEDIUM Dynamic Programming 2D
63. Unique Paths II
📖 Problem
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 is marked as 1 and 0 respectively in the 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.
🧠 Visual Learning Aid
1 Model the input into the right structure
2 Choose the core technique and invariant
3 Execute step-by-step with a sample
4 Validate complexity and edge cases
JS/TS Refreshers
- •Array methods (`push`, `pop`, `shift`, `slice`)
- •Object/Map/Set usage patterns
- •Function parameter and return typing
- •Array DP table updates
- •State transition thinking
- •Base case initialization
Logical Thinking Concepts
- •Define invariants before coding
- •Check edge cases first (`[]`, single element, duplicates)
- •Estimate time/space before implementation
- •Apply Dynamic Programming reasoning pattern
💡 Approach
- → Use DP where dp[i][j] represents number of ways to reach cell (i,j)
- → If cell is obstacle, dp[i][j] = 0
- → Otherwise, dp[i][j] = dp[i-1][j] + dp[i][j-1]
- → Time: O(m * n), Space: O(m * n) or O(n) optimized
🛠️ Hints & Pitfalls
Hints
- •Use DP where dp[i][j] represents number of ways to reach cell (i,j)
- •If cell is obstacle, dp[i][j] = 0
- •Otherwise, dp[i][j] = dp[i-1][j] + dp[i][j-1]
Common Pitfalls
- •Time: O(m * n), Space: O(m * n) or O(n) optimized
🧪 Test Cases
Hidden tests on submit: 1
Test Case 1
Not run Input:
uniquePathsWithObstacles([[0,0,0],[0,1,0],[0,0,0]]); Expected:
2 Test Case 2
Not run Input:
uniquePathsWithObstacles([[0,1],[0,0]]); Expected:
1 Test Case 3
Not run Input:
uniquePathsWithObstacles([[0,0]]); Expected:
1 📝 Code Editor
📤 Output