Grid Unique Paths with Obstacles
A medium Dynamic Programming problem included in Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.
- Topic
- Dynamic Programming
- Sheets
- 1
- Core for
- 9 roles
- Platform
- LeetCode
The problem
A robot is on an m x n grid with some obstacles. Cells with obstacles are marked as 1 and empty cells as 0. The robot can only move down or right. Find the number of unique paths from top-left to bottom-right avoiding obstacles.
Example 1
- Input
- obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
- Output
- 2
- Why
- Two unique paths avoiding the obstacle in the center.
Example 2
- Input
- obstacleGrid = [[0,1],[0,0]]
- Output
- 1
- Why
- Only one path avoiding the obstacle at position (0,1).
Example 3
- Input
- obstacleGrid = [[1]]
- Output
- 0
- Why
- The starting cell is an obstacle, so no path exists.
Constraints
- m == obstacleGrid.length
- n == obstacleGrid[i].length
- 1 <= m, n <= 100
- obstacleGrid[i][j] is 0 or 1
How to think about it
Updated 2026-09-09Obstacles act as dead ends carrying exactly 0 paths forward. Any cell with an obstacle has its path count forced to 0; every other cell sums the paths arriving from its top and left neighbors.
Approaches, worst first
Recursive search with obstacle checks
time O(2^(m+n)) · space O(m + n)
Recurse over down and right moves, returning 0 whenever an obstacle or boundary is hit. Explores duplicate cells exponentially.
2D dynamic programming grid
time O(m * n) · space O(m * n)
dp[i][j] is paths to reach cell (i, j). If grid[i][j] == 1, dp[i][j] = 0; otherwise dp[i][j] = dp[i-1][j] + dp[i][j-1]. Carefully initialize borders until the first obstacle.
1D rolling rowWrite this one
time O(m * n) · space O(n)
Maintain array dp of size n. Initialize dp[0] = 1 if grid[0][0] == 0 else 0. For each row and column, if obstacle set dp[j] = 0; else if j > 0 add dp[j-1] to dp[j].
Where people lose marks · 3
- Obstacle at the start (grid[0][0] == 1) or target (grid[m-1][n-1] == 1): must immediately yield 0.
- Obstacle along the top row or first column: all subsequent cells along that border are blocked and must have 0 paths, not 1.
- Integer overflow: while test cases typically fit within 32-bit signed or 64-bit integers, intermediate grids can accumulate large sums.
The theory behind it
Dynamic Programming — the ground this problem stands on. All Dynamic Programming problems
What Dynamic Programming is
Dynamic programming is a method for solving a complex problem by breaking it into overlapping subproblems, solving each subproblem only once, and remembering the answers in a lookup table. Instead of recalculating identical questions over and over, future steps look up previous answers directly. By assembling these saved pieces from the bottom up or storing them during recursion, a task that would take billions of steps finishes in a fraction of a second.
When to reach for it
Reach for dynamic programming when questions ask for the maximum profit, minimum cost, total number of distinct ways to achieve a goal, or whether a target can be formed. Signals include overlapping choices where making a choice now affects what choices remain later, but greedy picking fails to find the true global optimum. If drawing a recursive decision tree reveals the same subproblem states repeating across branches, dynamic programming is needed.
How the pattern works
Identify the state variables that uniquely describe a subproblem, such as an array index and remaining capacity. Write the base cases first, representing states whose answers are known without calculation. Next, write the recurrence relation that expresses the current state using previously solved states, taking the minimum, maximum, or sum among your options. Build the solution either top-down by caching recursive returns in a memo table, or bottom-up by filling an array in topological dependency order. When each state depends only on the previous row, compress storage down to a single array.
What each operation costs
| Operation | Time |
|---|---|
| fill dynamic programming table of n states | O(n) |
| solve two-dimensional grid of m by n states | O(m * n) |
| space-optimized state transition keeping one row | O(n) |
What usually goes wrong with Dynamic Programming
- Filling a bottom-up table in an order where the current cell needs values that have not been computed yet, reading uninitialized zeros.
- Failing to initialize base cases properly, such as filling a minimization table with zeros instead of infinity, which traps the answer at zero.
- Overwriting values in a 1D space-optimized knapsack array by scanning in the wrong direction, allowing the same item to be chosen multiple times.
Which roles need this problem
Dynamic Programming is a core topic for these 9 roles — if you're targeting one of them, this problem is early in your path, not optional.
Secondary for 5 more roles, including Game Developer, Cryptography Engineer, Performance Engineer.
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